'tt'i
v
MATHEMATICAL FORMULAS*
Quadratic Formula
Derivatives and Integrals
If ax 2 + bx + c = 0, then x —
— b ± Vfc2 — 4 ac 2 a
, . nx
(1 + x)n = 1 + - +
v 2 1!
Binomial Theorem
n(n — l)x2 2!~
(x2<l)
Products of Vectors
Let 8 be the smaller of the two angles between a and b . Then
a ■ b = b ■ a = axbx + ayby + azbz = ab cos 8
7? X b — —b X 7? =
br b„
by b.
ax a, bY b.
+ k
ax a ,
5,
d .
— — sin x = cos x dx
d
—— cos x = —sin x dx
sin x dx =
J cos x dx =
dx
f
f
f
dx
Vx2 + a2 x dx
(x2 + fl2)3/2 dx
(x2 + a2)312
ln(x + Vx2 + a2)
1
(x2 + n2)1/2 X
; fl2(x2 + fl2)1/2
Cramer’s Rule
Two simultaneous equations in unknowns x and y,
fljx + b^y = Ci and a2x + b2y = c2.
= ( aybz ~ byaz)i + ( azbx - bzax )] + ( axby - bxay) k \a X b\ = ab sin 8
Trigonometric Identities
sin a ± sin /3 = 2 sin a ± fi) cos ~ 0) cos a + cos 0 = 2 cos y(a + 0) cos - 0)
* See Appendix E for a more complete list.
have the solutions
Cl bi
c2 b2
cxb2 - c2by
a , h,
CN
1
<N
1
a2 b2
ai ci
a2 c2
axc2 — ci2c1
ai bi
aib2 — a2bi
a2 b2
SI PREFIXES*
Factor
Prefix
Symbol
Factor
Prefix
Symbol
1024
yotta
Y
io-1
deci
d
1021
zetta
Z
io-2
centi
c
1018
exa
E
io-3
milli
m
1015
peta
P
io-6
micro
1012
tera
T
io-9
nano
n
109
giga
G
io-12
pico
p
106
mega
M
io-15
femto
f
103
kilo
k
lO-rs
atto
a
102
hecto
h
io-21
zepto
z
101
deka
da
io-24
yocto
y
*In all cases, the first syllable is accented, as in na-no-me-ter.
—cos x
sinx
EXTENDED
FUNDAMENTALS OF PHYSICS
This page intentionally left blank
EXTENDED
Halliday & Resnick
FUNDAMENTALS OF PHYSICS
JEARL WALKER
CLEVELAND STATE UNIVERSITY
Wiley
EXECUTIVE EDITOR Stuart Johnson
SENIOR PRODUCT DESIGNER Geraldine Osnato
CONTENT EDITOR Alyson Rentrop
ASSOCIATE MARKETING DIRECTOR Christine Kushner
TEXT and COVER DESIGNER Madelyn Lesure
PAGE MAKE-UP Lee Goldstein
PHOTO EDITOR Jennifer Atkins
COPYEDITOR Helen Walden
PROOFREADER Lilian Brady
SENIOR PRODUCTION EDITOR Elizabeth Swain
COVER IMAGE © 2007 CERN
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Library of Congress Cataloging-in-Publication Data
Walker, Jearl
Fundamentals of physics / Jearl Walker, David Halliday, Robert Resnick — 10th edition.
volumes cm Includes index.
ISBN 978-1-118-23072-5 (Extended edition)
Binder-ready version ISBN 978-1-118-23061-9 (Extended edition)
1. Physics — Textbooks. I. Resnick, Robert. II. Halliday, David. III. Title.
QC21.3.H35 2014 530— dc23
Printed in the United States of America 10 987654321
2012035307
CONTENTS
1 Measurement
2 Motion Along a Straight Line
3 Vectors
4 Motion in Two and Three Dimensions
5 Force and Motion-1
6 Force and Motion-ll
7 Kinetic Energy and Work
8 Potential Energy and Conservation of Energy
9 Center of Mass and Linear Momentum
10 Rotation
11 Rolling, Torque, and Angular Momentum
12 Equilibrium and Elasticity
13 Gravitation
14 Fluids
15 Oscillations
1 6 Waves-I
17 Waves-I I
18 Temperature, Heat, and the First Law of Thermodynamics
19 The Kinetic Theory of Gases
20 Entropy and the Second Law of Thermodynamics
21 Coulomb's Law
22 Electric Fields
23 Gauss’ Law
24 Electric Potential
25 Capacitance
26 Current and Resistance
27 Circuits
28 Magnetic Fields
29 Magnetic Fields Due to Currents
30 Induction and Inductance
31 Electromagnetic Oscillations and Alternating Current
32 Maxwell’s Equations; Magnetism of Matter
33 Electromagnetic Waves
34 Images
35 Interference
36 Diffraction
37 Relativity
38 Photons and Matter Waves
39 More About Matter Waves
40 All About Atoms
41 Conduction of Electricity in Solids
42 Nuclear Physics
43 Energy from the Nucleus
44 Quarks, Leptons, and the Big Bang
Appendices / Answers to Checkpoints and Odd-Numbered Questions and Problems / Index
v
i Measurement
1-1 MEASURING THINGS, INCLUDING LENGTHS i
What Is Physics? i
Measuring Things i
The International System of Units 2
Changing Units 3
Length 3
Significant Figures and Decimal Places 4
1-2 TIME 5
Time 5
1- 3 MASS e
Mass 6
REVIEW & SUMMARY s PROBLEMS s
^gfjjotion Along a Straight Line 1 3
2- 1 POSITION, DISPLACEMENT, AND AVERAGE VELOCITY 1 3
What Is Physics? 1 3
Motion 1 4
Position and Displacement 1 4 Average Velocity and Average Speed 1 5
2-2 INSTANTANEOUS VELOCITY AND SPEED 1 s
Instantaneous Velocity and Speed 1 s
2-3 ACCELERATION 20
Acceleration 20
2-4 CONSTANT ACCELERATION 23
Constant Acceleration: A Special Case 23 Another Look at Constant Acceleration 26
2-5 FREE-FALL ACCELERATION 27
Free-Fall Acceleration 27
2- 6 GRAPHICAL INTEGRATION IN MOTION ANALYSIS 29
Graphical Integration in Motion Analysis 29
REVIEW & SUMMARY 30 QUESTIONS 31 PROBLEMS 32
3 Vectors 40
3- 1 VECTORS AND THEIR COMPONENTS 40
What Is Physics? 40 Vectors and Scalars 40 Adding Vectors Geometrically 41 Components of Vectors 42
3-2 UNIT VECTORS, ADDING VECTORS OY COMPONENTS 46
Unit Vectors 46
Adding Vectors by Components 46 Vectors and the Laws of Physics 47
3- 3 MULTIPLYING VECTORS so
Multiplying Vectors 50
REVIEW & SUMMARY 55 QUESTIONS se PROBLEMS 57
<H|otion in Two and Three Dimensions 62
4- 1 POSITION AND DISPLACEMENT 62
What Is Physics? 62 Position and Displacement 63
4-2 AVERAGE VELOCITY AND INSTANTANEOUS VELUCITY 64
Average Velocity and Instantaneous Velocity 65
4-3 AVERAGE ACCELERATION AND INSTANTANEOUS ACCELERATION 67
Average Acceleration and Instantaneous Acceleration 68
4-4 PROJECTILE MOTION 70
Projectile Motion 70
4-5 UNIFORM CIRCULAR MOTION 76
Uniform Circular Motion 76
4-6 RELATIVE MOTION IN ONE DIMENSION 78
Relative Motion in One Dimension 78
4- 7 RELATIVE MOTION IN TWO DIMENSIONS so
Relative Motion in Two Dimensions so
REVIEW & SUMMARY si QUESTIONS 82 PROBLEMS 84
5 Force and Motion-1 94
5- 1 NEWTON'S FIRST AND SECOND LAWS 94
What Is Physics? 94 Newtonian Mechanics 95 Newton’s First Law 95 Force 96 Mass 97
Newton’s Second Law 98
5-2 SOME PARTICULAR FORCES 102
Some Particular Forces 102
5-3 APPLYING NEWTON’S LAWS 106
Newton’s Third Law 1 06 Applying Newton’s Laws 1 os
REVIEWS SUMMARY 114 QUESTIONS 114 PROBLEMS ne
VI
6 Force and Motion— II 124
6-1 FRICTION 124
What Is Physics? 1 24 Friction 1 24 Properties of Friction 1 27
6-2 THE DRAG FORCE AND TERMINAL SPEED 1 30
The Drag Force and Terminal Speed 1 30
6- 3 DNIFORM CIRCDLAR MOTION 133
Uniform Circular Motion 1 33
REVIEWS SUMMARY 138 QUESTIONS 139 PROBLEMS 140
*7; Kinetic Energy and Work 1 49
7- 1 KINETIC ENERGY 149
What Is Physics? 1 49 What Is Energy? 1 49 Kinetic Energy 1 50
7-2 WORK AND KINETIC ENERGY 151
Work 151
Work and Kinetic Energy 1 52
7-3 WORK DONE BY THE GRAVITATIONAL FORCE 1 55
Work Done by the Gravitational Force 1 56
7-4 WORK DONE BY A SPRING FORCE 159
Work Done by a Spring Force 1 59
7-5 WORK DONE BY A GENERAL VARIABLE FORCE 1 62
Work Done by a General Variable Force 1 62
7- 6 POWER 166
Power 1 66
REVIEW & SUMMARY 168 QUESTIONS 169 PROBLEMS 170
<3f§j)tential Energy and Conservation of Energy 1 77
8- 1 POTENTIAL ENERGY 177
What Is Physics? 1 77
Work and Potential Energy 1 78
Path Independence of Conservative Forces 1 79
Determining Potential Energy Values 1 si
8-2 CONSERVATION OF MECHANICAL ENERGY 1 84
Conservation of Mechanical Energy 1 84
8-3 READING A POTENTIAL ENERGY CURVE 1 87
Reading a Potential Energy Curve 1 87
CONTENTS
8-4 WORK DONE ON A SYSTEM BY AN EXTERNAL FORCE 1 91
Work Done on a System by an External Force 1 92
8- 5 CONSERVATION OF ENERGY 195
Conservation of Energy 1 95
REVIEW & SUMMARY 199 QUESTIONS 200 PROBLEMS 202
9 Center of Mass and Linear Momentum 21 4
9- 1 CENTER OF MASS 214
What Is Physics? 214 The Center of Mass 21 5
9-2 NEWTON'S SECOND LAW FOR A SYSTEM OF PARTICLES 220
Newton’s Second Law for a System of Particles 220
9-3 LINEAR MOMENTUM 224
Linear Momentum 224
The Linear Momentum of a System of Particles 225
9-4 COLLISION AND IMPULSE 226
Collision and Impulse 226
9-5 CONSERVATION OF LINEAR MOMENTUM 230
Conservation of Linear Momentum 230
9-6 MOMENTUM AND KINETIC ENERGY IN COLLISIONS 233
Momentum and Kinetic Energy in Collisions 233 Inelastic Collisions in One Dimension 234
9-7 ELASTIC COLLISIONS IN ONE DIMENSION 237
Elastic Collisions in One Dimension 237
9-8 COLLISIONS IN TWO DIMENSIONS 240
Collisions in Two Dimensions 240
9- 9 SYSTEMS WITH VARYING MASS: A ROCKET 241
Systems with Varying Mass: A Rocket 241
REVIEW & SUMMARY 243 QUESTIONS 245 PROBLEMS 246
<^§|flotation 257
10- 1 ROTATIONAL VARIABLES 257
What Is Physics? 25s
Rotational Variables 259
Are Angular Quantities Vectors? 264
10-2 ROTATION WITH CONSTANT ANGULAR ACCELERATION 266
Rotation with Constant Angular Acceleration 266
10-3 RELATING THE LINEAR AND ANGULAR VARIABLES 268
Relating the Linear and Angular Variables 268
vii
viii CONTENTS
10-4 KINETIC ENERGY OF ROTATION 271
Kinetic Energy of Rotation 271
10-5 CALCULATING THE ROTATIONAL INERTIA 273
Calculating the Rotational Inertia 273
10-6 TORQUE 277
Torque 27s
10-7 NEWTON'S SECOND LAW FOR ROTATION 279
Newton’s Second Law for Rotation 279
10- 8 WORK AND ROTATIONAL KINETIC ENERGY 232
Work and Rotational Kinetic Energy 282
REVIEW & SUMMARY 285 QUESTIONS 286 PROBLEMS 287
•••^Rolling. Torque, and Aogular Momentum 295
11- 1 ROLLING AS TRANSLATION AND ROTATION COMDINED 295
What Is Physics? 295
Rolling as Translation and Rotation Combined 295
11-2 FORCES AND KINETIC ENERGY OF ROLLING 293
The Kinetic Energy of Rolling 298 The Forces of Rolling 299
11-3 THE YO-YO 301
The Yo-Yo 302
11-4 TORQUE REVISITED 302
Torque Revisited 303
11-5 ANGULAR MOMENTUM 305
Angular Momentum 305
11-6 NEWTON'S SECOND LAW IN ANGULAR FORM 307
Newton's Second Law in Angular Form 307
11-7 ANGULAR MOMENTUM OF A RIGID DODY 31 o
The Angular Momentum of a System of Particles 31 o
The Angular Momentum of a Rigid Body Rotating About a Fixed Axis 31 1
11-8 CONSERVATION OF ANGULAR MOMENTUM 31 2
Conservation of Angular Momentum 31 2
11- 9 PRECESSION OF A GYROSCOPE 317
Precession of a Gyroscope 31 7
REVIEW & SUMMARY sis QUESTIONS 319 PROBLEMS 320
<^||Equilibriuni and Elasticity 327
12- 1 EQUILIBRIUM 327
What Is Physics? 327
Equilibrium 327
The Requirements of Equilibrium 329 The Center of Gravity 330
12-2 SOME EXAMPLES OF STATIC EQUILIHRIUM 332
Some Examples of Static Equilibrium 332
12- 3 ELASTICITY 338
Indeterminate Structures 33s Elasticity 339
REVIEW & SUMMARY 343 QUESTIONS 343 PROBLEMS 345
<^Hfiravitation 354
13- 1 NEWTON'S LAW OF GRAVITATION 354 What Is Physics? 354
Newton’s Law of Gravitation 355
13-2 GRAVITATION AND THE PRINCIPLE OF SUPERPOSITION 357 Gravitation and the Principle of Superposition 357
13-3 GRAVITATION NEAR EARTH'S SURFACE 359
Gravitation Near Earth's Surface 360
13-4 GRAVITATION INSIDE EARTH 362 Gravitation Inside Earth 363
13-5 GRAVITATIONAL POTENTIAL ENERGY 364 Gravitational Potential Energy 364
13-6 PLANETS AND SATELLITES: KEPLER'S LAWS 368 Planets and Satellites: Kepler’s Laws 369
13-7 SATELLITES: ORDITS AND ENERGY 371 Satellites: Orbits and Energy 371
13- 8 EINSTEIN AND GRAVITATION 374
Einstein and Gravitation 374
REVIEW & SUMMARY 376 QUESTIONS 377 PROBLEMS 37s
14 Fluids 386
14- 1 FLUIDS, DENSITY, AND PRESSURE 386
What Is Physics? 386 What Is a Fluid? 386 Density and Pressure 387
14-2 FLUIDS AT REST sss
Fluids at Rest 389
14-3 MEASURING PRESSURE 392
Measuring Pressure 392
CONTENTS
IX
14-4 PASCAL’S PRINCIPLE 393
Pascal’s Principle 393
14-5 ARCHIMEDES’ PRINCIPLE 394
Archimedes’ Principle 395
14-6 THE EQUATION OF CONTINUITY 398
Ideal Fluids in Motion 39s The Equation of Continuity 399
14- 7 BERNOULLI’S EQUATION 401
Bernoulli’s Equation 401
REVIEW & SUMMARY 405 QUESTIONS 405 PROBLEMS 406
scillations 413
15- 1 SIMPLE HARMONIC MOTION 413
What Is Physics? 4i4
Simple Harmonic Motion 41 4
The Force Law for Simple Harmonic Motion 41 9
15-2 ENERGY IN SIMPLE HARMONIC MOTION 421
Energy in Simple Harmonic Motion 421
15-3 AN ANGULAR SIMPLE HARMONIC OSCILLATOR 423
An Angular Simple Harmonic Oscillator 423
15-4 PENDULUMS, CIRCULAR MOTION 424 Pendulums 425
Simple Harmonic Motion and Uniform Circular Motion 428
15-5 DAMPED SIMPLE HARMONIC MOTION 430
Damped Simple Harmonic Motion 430
15- 6 FORCED OSCILLATIONS AND RESONANCE 432
Forced Oscillations and Resonance 432
REVIEW & SUMMARY 434 QUESTIONS 434 PROBLEMS 436
^Waves-I 444
16- 1 TRANSVERSE WAVES 444
What Is Physics? 445
Types of Waves 445
Transverse and Longitudinal Waves 445
Wavelength and Frequency 446
The Speed of a Traveling Wave 449
16-2 WAVE SPEED ON A STRETCHED STRING 452
Wave Speed on a Stretched String 452
16-3 ENERGY AND POWER OF A WAVE TRAVELING ALONG A STRING 454
Energy and Power of a Wave Traveling Along a String 454
16-4 THE WAVE EQUATION 456
The Wave Equation 456
16-5 INTERFERENCE OF WAVES 458
The Principle of Superposition for Waves 458 Interference of Waves 459
16-6 PHASORS 462 Phasors 462
16- 7 STANDING WAVES AND RESONANCE 465
Standing Waves 465
Standing Waves and Resonance 467
REVIEW & SUMMARY 470 QUESTIONS 471 PROBLEMS 472
<^pVaves-ll 479
17- 1 SPEED OF SOUND 479
What Is Physics? 479 Sound Waves 479 The Speed of Sound 4so
17-2 TRAVELING SOUND WAVES 482
Traveling Sound Waves 482
17-3 INTERFERENCE 485
Interference 485
17-4 INTENSITY AND SOUND LEVEL 488 Intensity and Sound Level 489
17-5 SOURCES OF MUSICAL SOUND 492
Sources of Musical Sound 493
17-6 BEATS 496 Beats 497
17-7 THE DOPPLER EFFECT 498
The Doppler Effect 499
17- 8 SUPERSONIC SPEEDS, SHOCK WAVES 503
Supersonic Speeds, Shock Waves 503
REVIEW & SUMMARY 504 QUESTIONS sos PROBLEMS soe
<^pTemperature, Heat, and the First Law ot Thermodynamics 51 4
18- 1 TEMPERATURE 51 4 What Is Physics? 514 Temperature 515
The Zeroth Law of Thermodynamics 515 Measuring Temperature 51 6
18-2 THE CELSIUS AND FAHRENHEIT SCALES sis
The Celsius and Fahrenheit Scales 51 s
X
CONTENTS
18-3 THERMAL EXPANSION 520
Thermal Expansion 520
18-4 ABSORPTION OF HEAT 522
Temperature and Heat 523
The Absorption of Heat by Solids and Liquids 524
18-5 THE FIRST LAW OF THERMODYNAMICS 52s
A Closer Look at Heat and Work 52s The First Law of Thermodynamics 531 Some Special Cases of the First Law of Thermodynamics 532
18- 6 HEAT TRANSFER MECHANISMS 534
Heat Transfer Mechanisms 534
REVIEW & SUMMARY sss QUESTIONS 540 PROBLEMS 541
|pe Kinetic Theory of Gases 549
19- 1 AVOGADRO'S NUMBER 549
What Is Physics? 549 Avogadro’s Number 550
19-2 IBEAL GASES 550
Ideal Gases 551
19-3 PRESSURE, TEMPERATURE, AND RMS SPEED 554
Pressure, Temperature, and RMS Speed 554
19-4 TRANSLATIONAL KINETIC ENERGY 557
Translational Kinetic Energy 557
19-5 MEAN FREE PATH sss
Mean Free Path 55s
19-6 THE DISTRIBUTION OF MOLECULAR SPEEDS seo
The Distribution of Molecular Speeds 56i
19-7 THE MOLAR SPECIFIC HEATS OF AN IDEAL GAS 564 The Molar Specific Heats of an Ideal Gas 564
19-8 DEGREES OF FREEDOM AND MOLAR SPECIFIC HEATS 568
Degrees of Freedom and Molar Specific Heats 568 A Hint of Quantum Theory 570
19- 9 THE ADIABATIC EXPANSION OF AN IDEAL GAS 571
The Adiabatic Expansion of an Ideal Gas 571
REVIEW & SUMMARY 575 QUESTIONS 576 PROBLEMS 577
20 Entropy and the Second Law of Thermodynamics 533
20- 1 ENTROPY sss
What Is Physics? 584 Irreversible Processes and Entropy 584
Change in Entropy 585
The Second Law of Thermodynamics sss
20-2 ENTROPY IN THE REAL WORLD: ENGINES 590
Entropy in the Real World: Engines 590
20-3 REFRIGERATORS AND REAL ENGINES 595
Entropy in the Real World: Refrigerators 596 The Efficiencies of Real Engines 597
20- 4 A STATISTICAL VIEW OF ENTROPY 598 A Statistical View of Entropy 59s
REVIEW & SUMMARY 602 QUESTIONS 603 PROBLEMS 604
<^j|Coulomb’s Law 609
21- 1 COULOMB'S LAW 609
What Is Physics? 6i o Electric Charge 610 Conductors and Insulators 6i 2 Coulomb's Law 6i 3
21-2 CHARGE IS QUANTIZED 619
Charge Is Quantized 6i 9
21- 3 CHARGE IS CONSERVED 621
Charge Is Conserved 621
REVIEW & SUMMARY 622 QUESTIONS 623 PROBLEMS 624
•^pElectric Fields eso
22- 1 THE ELECTRIC FIELD eso
What Is Physics? 630 The Electric Field 63 1 Electric Field Lines 63i
22-2 THE ELECTRIC FIELD DUE TO A CHARGED PARTICLE 633 The Electric Field Due to a Point Charge 633
22-3 THE ELECTRIC FIELD DUE TO A DIPOLE 635
The Electric Field Due to an Electric Dipole 636
22-4 THE ELECTRIC FIELD DUE TO A LINE OF CHARGE 638 The Electric Field Due to Line of Charge 638
22-5 THE ELECTRIC FIELD DUE TO A CHARGED DISK 643 The Electric Field Due to a Charged Disk 643
22-6 A POINT CHARGE IN AN ELECTRIC FIELD 645 A Point Charge in an Electric Field 645
22-7 A DIPOLE IN AN ELECTRIC FIELD 647
A Dipole in an Electric Field 648
REVIEW & SUMMARY eso QUESTIONS esi PROBLEMS 652
CONTENTS
XI
23 Gauss' Law 659
23-1 ELECTRIC FLUX 659
What Is Physics 659 Electric Flux 660
23-2 GAUSS’ LAW 664
Gauss' Law 664
Gauss' Law and Coulomb’s Law 666
23-3 A CHARGED ISDLATED CDNDUCTDR 668
A Charged Isolated Conductor 668
23-4 APPLYING GAUSS' LAW: CYLINDRICAL SYMMETRY 67i
Applying Gauss’ Law: Cylindrical Symmetry 67i
23-5 APPLYING GAUSS' LAW: PLANAR SYMMETRY 673
Applying Gauss’ Law: Planar Symmetry 673
23- 6 APPLYING GAUSS' LAW: SPHERICAL SYMMETRY 675
Applying Gauss’ Law: Spherical Symmetry 675
REVIEW & SUMMARY 677 QUESTIONS 677 PROBLEMS 679
<^flectric Potential ess
24- 1 ELECTRIC POTENTIAL ess
What Is Physics? 685
Electric Potential and Electric Potential Energy 686
24-2 EQUIPOTENTIAL SURFACES AND THE ELECTRIC FIELD 690
Equipotential Surfaces 690 Calculating the Potential from the Field 69i
24-3 POTENTIAL DUE TO A CHARGED PARTICLE 694
Potential Due to a Charged Particle 694 Potential Due a Group of Charged Particles 695
24-4 POTENTIAL DUE TO AN ELECTRIC DIPOLE 697
Potential Due to an Electric Dipole 697
24-5 POTENTIAL DUE TO A CONTINUOUS CHARGE DISTRIBUTION egs
Potential Due to a Continuous Charge Distribution 698
24-6 CALCULATING THE FIELD FROM THE POTENTIAL 701
Calculating the Field from the Potential 701
24-7 ELECTRIC POTENTIAL ENERGY OF A SYSTEM OF CHARGED PARTICLES 703
Electric Potential Energy of a System of Charged Particles 703
24-8 POTENTIAL OF A CHARGED ISOLATED CONDUCTOR 706
Potential of Charged Isolated Conductor 706
REVIEW & SUMMARY 707 QUESTIONS 7os PROBLEMS 710
<^f| Capacitance 717
25-1 CAPACITANCE 717 What Is Physics? 717 Capacitance 717
25-2 CALCULATING THE CAPACITANCE 719
Calculating the Capacitance 720
25-3 CAPACITORS IN PARALLEL AND IN SERIES 723
Capacitors in Parallel and in Series 724
25-4 ENERGY STORED IN AN ELECTRIC FIELD 728
Energy Stored in an Electric Field 72s
25-5 CAPACITOR WITH A DIELECTRIC 731
Capacitor with a Dielectric 731 Dielectrics: An Atomic View 733
25- 6 DIELECTRICS AND GAUSS’ LAW 735
Dielectrics and Gauss’ Law 735
REVIEW & SUMMARY 73s QUESTIONS 73s PROBLEMS 739
<^S Current and Resistance 745
26- 1 ELECTRIC CURRENT 745 What Is Physics? 745 Electric Current 746
26-2 CURRENT DENSITY 748
Current Density 749
26-3 RESISTANCE AND RESISTIVITY 752
Resistance and Resistivity 753
26-4 OHM’S LAW 756
Ohm’s Law 756
A Microscopic View of Ohm’s Law 758
26- 5 POWER, SEMICONDUCTORS, SUPERCONDUCTORS 760
Power in Electric Circuits 760 Semiconductors 762 Superconductors 763
REVIEW & SUMMARY 763 QUESTIONS 764 PROBLEMS 765
<^p)ircuits 771
27- 1 SINGLE-LOOP CIRCUITS 771
What Is Physics? 772
“Pumping" Charges 772
Work, Energy, and Emf 773
Calculating the Current in a Single-Loop Circuit 774
Other Single-Loop Circuits 776
Potential Difference Between Two Points 777
xii CONTENTS
27-2 MULTILOOP CIRCUITS rsi
Multiloop Circuits 78i
27-3 THE AMMETER AND THE VOLTMETER 788 The Ammeter and the Voltmeter 788
27- 4 RC CIRCUITS 788
RC Circuits 789
REVIEW & SUMMARY 793 QUESTIONS 793 PROBLEMS 795
^plagnetic Fields 803
28- 1 MAGNETIC FIELDS AND THE DEFINITION OF I sos
What Is Physics? 803
What Produces a Magnetic Field? so4
The Definition of E 804
28-2 CROSSED FIELDS: DISCOVERY OF THE ELECTRON sos
Crossed Fields: Discovery of the Electron 809
28-3 CROSSED FIELDS: THE HALL EFFECT 81 o
Crossed Fields: The Hall Effect si 1
28-4 A CIRCULATING CHARGED PARTICLE si 4
A Circulating Charged Particle si 4
28-5 CYCLOTRONS AND SYNCHROTRONS si 7
Cyclotrons and Synchrotrons si s
28-6 MAGNETIC FORCE ON A CURRENT-CARRYING WIRE 820
Magnetic Force on a Current-Carrying Wire 820
28-7 TORQUE ON A CURRENT LOOP 822
Torque on a Current Loop 822
28- 8 THE MAGNETIC DIPOLE MOMENT 824 The Magnetic Dipole Moment 825
REVIEW & SUMMARY 827 QUESTIONS 827 PROBLEMS 829
29 Magnetic Fields Due to Currents 836
29- 1 MAGNETIC FIELD DUE TU A CURRENT sse
What Is Physics? 836
Calculating the Magnetic Field Due to a Current 837
29-2 FORCE BETWEEN TWO PARALLEL CURRENTS 842 Force Between Two Parallel Currents 842
29-3 AMPERE'S LAW 844
Ampere’s Law 844
29-4 SOLENOIDS AND TOROIDS 848
Solenoids and Toroids 848
29- 5 A CURRENT-CARRYING COIL AS A MAGNETIC DIPOLE ssi
A Current-Carrying Coil as a Magnetic Dipole ssi
REVIEW & SUMMARY 854 QUESTIONS sss PROBLEMS sse
^Induction and Inductance 864
30- 1 FARADAY’S LAW AND LENZ’S LAW 864
What Is Physics 864 Two Experiments 865 Faraday’s Law of Induction 865 Lenz’s Law 868
30-2 INDUCTION AND ENERGY TRANSFERS 87i Induction and Energy Transfers 87i
30-3 INDUCED ELECTRIC FIELDS 874 Induced Electric Fields 875
30-4 INDUCTORS AND INDUCTANCE 879 Inductors and Inductance 879
30-5 SELF-INDUCTION ssi
Self-Induction ssi
30-6 RL CIRCUITS 882
EL Circuits 883
30-7 ENERGY STORED IN A MAGNETIC FIELD 887 Energy Stored in a Magnetic Field 887
30-8 ENERGY DENSITY OF A MAGNETIC FIELD 889 Energy Density of a Magnetic Field 889
30- 9 MUTUAL INDUCTION 890
Mutual Induction 890
REVIEW & SUMMARY 893 QUESTIONS 893 PROBLEMS 895
31 Electromagnetic Oscillations and Alternating Current 903
31- 1 LC OSCILLATIONS 903
What Is Physics? 904 LC Oscillations, Qualitatively 904 The Electrical-Mechanical Analogy 906 LC Oscillations, Quantitatively 907
31-2 DAMPED OSCILLATIONS IN AN RLC CIRCUIT 91 o Damped Oscillations in an ELC Circuit 91 1
31-3 FORCED OSCILLATIONS OF THREE SIMPLE CIRCUITS 91 2
Alternating Current 913 Forced Oscillations 91 4 Three Simple Circuits 914
31-4 THE SERIES /?/£ CIRCUIT 921 The Series ELC Circuit 921
31-5 POWER IN ALTERNATING-CURRENT CIRCUITS 927
Power in Alternating-Current Circuits 927
31- 6 TRANSFORMERS 930
Transformers 930
REVIEW & SUMMARY 933 QUESTIONS 934 PROBLEMS 935
32 Maxwell's Equations; Magnetism of Matter 941
32- 1 GAUSS’ LAW FOR MAGNETIC FIELOS 94i
What Is Physics? 941
Gauss' Law for Magnetic Fields 942
32-2 INOUCEO MAGNETIC FIELOS 943 Induced Magnetic Fields 943
32-3 DISPLACEMENT CURRENT 946
Displacement Current 947 Maxwell’s Equations 949
32-4 MAGNETS 950
Magnets 950
32-5 MAGNETISM AND ELECTRONS 952
Magnetism and Electrons 953 Magnetic Materials 956
32-6 DIAMAGNETISM 957
Diamagnetism 957
32-7 PARAMAGNETISM 959
Paramagnetism 959
32- 8 FERROMAGNETISM 96i
Ferromagnetism 96i
REVIEW & SUMMARY 964 QUESTIONS 965 PROBLEMS 967
33 Electromagnetic Waves 972
33- 1 ELECTROMAGNETIC WAVES 972
What Is Physics? 972
Maxwell’s Rainbow 973
The Traveling Electromagnetic Wave, Qualitatively 974 The Traveling Electromagnetic Wave, Quantitatively 977
33-2 ENERGY TRANSPORT AND THE POYNTING VECTOR 930
Energy Transport and the Poynting Vector 98 1
33-3 RADIATION PRESSURE 983
Radiation Pressure 983
33-4 POLARIZATION 985
Polarization 985
CONTENTS
33-5 REFLECTION AND REFRACTION 990
Reflection and Refraction 991
33-6 TOTAL INTERNAL REFLECTION 996
Total Internal Reflection 996
33- 7 POLARIZATION GY REFLECTION 997
Polarization by Reflection 99s
REVIEW & SUMMARY 999 QUESTIONS 1000 PROBLEMS 1001
<^pmages 1010
34- 1 IMAGES AND PLANE MIRRORS 1010
What Is Physics? 1 01 o Two Types of Image 1 01 o Plane Mirrors 1012
34-2 SPHERICAL MIRRORS 101 4
Spherical Mirrors 1015 Images from Spherical Mirrors 1 01 6
34-3 SPHERICAL REFRACTING SURFACES 1 020
Spherical Refracting Surfaces 1 020
34-4 THIN LENSES 1023
Thin Lenses 1023
34-5 OPTICAL INSTRUMENTS 1030
Optical Instruments 1 030
34- 6 THREE PROOFS 1033
REVIEW & SUMMARY lose QUESTIONS 1037 PROBLEMS loss
^^Interference 1 047
35- 1 LIGHT AS A WAVE 1047
What Is Physics? 1 047 Light as a Wave 1 048
35-2 YOUNG'S INTERFERENCE EXPERIMENT 1 053
Diffraction 1 053
Young’s Interference Experiment 1 054
35-3 INTERFERENCE AND DOUBLE-SLIT INTENSITY 1059
Coherence 1 059
Intensity in Double-Slit Interference 1 060
35-4 INTERFERENCE FROM THIN FILMS 1 063
Interference from Thin Films 1 064
35-5 MICHELSON'S INTERFEROMETER 1070
Michelson’s Interferometer 1071
REVIEW & SUMMARY 1072 QUESTIONS 1072 PROBLEMS 1074
xiii
XIV
CONTENTS
36 Diffraction iosi
36-1 SINGLE-SLIT DIFFRACTION iosi
What Is Physics? i osi
Diffraction and the Wave Theory of Light i osi
Diffraction by a Single Slit: Locating the Minima 1 083
36-2 INTENSITY IN SINGLE-SLIT DIFFRACTION i ose
Intensity in Single-Slit Diffraction i os6
Intensity in Single-Slit Diffraction, Quantitatively i oss
36-3 DIFFRACTION DY A CIRCULAR APERTURE i oso
Diffraction by a Circular Aperture 1 091
36-4 DIFFRACTION DY A DOUDLE SLIT 1 094 Diffraction by a Double Slit 1 095
36-5 DIFFRACTION GRATINGS 1093
Diffraction Gratings 1 09s
36-6 GRATINGS: DISPERSION AND RESOLVING POWER 1101
Gratings: Dispersion and Resolving Power 1101
36- 7 X-RAY DIFFRACTION 1104
X-Ray Diffraction 1 1 04
REVIEW & SUMMARY 1 1 07 QUESTIONS 1 1 07 PROBLEMS 1 1 os
37 Relativity me
37- 1 SIMULTANEITY AND TIME DILATION me
What Is Physics? 1 1 1 6 The Postulates 1117 Measuring an Event ms The Relativity of Simultaneity 1 1 20 The Relativity of Time 1121
37-2 THE RELATIVITY OF LENGTH 1125
The Relativity of Length 1 1 26
37-3 THE LORENTZ TRANSFORMATION 1 1 29
The Lorentz Transformation 1 1 29
Some Consequences of the Lorentz Equations 1131
37-4 THE RELATIVITY OF VELOCITIES 1133
The Relativity of Velocities 1 1 33
37-5 DOPPLER EFFECT FOR LIGHT 1134
Doppler Effect for Light 1135
37-6 MOMENTUM AND ENERGY 1137
A New Look at Momentum 1 1 38 A New Look at Energy 1 1 38
REVIEW & SUMMARY 1 1 43 QUESTIONS 1 1 44 PROBLEMS 1 1 45
38 Photons and Matter Waves i ^ 53
38-1 THE PHOTON, THE QUANTUM OF LIGHT 1 1 53
What Is Physics? 1 1 53
The Photon, the Quantum of Light 1 1 54
38-2 THE PHOTOELECTRIC EFFECT 1155
The Photoelectric Effect 1 1 56
38-3 PHOTONS, MOMENTUM, COMPTON SCATTERING, LIGHT INTERFERENCE nss
Photons Have Momentum 1 1 59 Light as a Probability Wave 1 1 62
38-4 THE GIRTH OF QUANTUM PHYSICS 1 1 64 The Birth of Quantum Physics 1 1 65
38-5 ELECTRONS AND MATTER WAVES 1 1 66
Electrons and Matter Waves 1 1 67
38-6 SCHRODINGER'S EQUATION 1170
Schrodinger’s Equation 1 1 70
38-7 HEISENBERG’S UNCERTAINTY PRINCIPLE 1 1 72
Heisenberg’s Uncertainty Principle 1173
38-8 REFLECTION FROM A POTENTIAL STEP 1 1 74 Reflection from a Potential Step 1 1 74
38- 9 TUNNELING THROUGH A POTENTIAL BARRIER 1 1 76
Tunneling Through a Potential Barrier 1 1 76
REVIEWS, SUMMARY 1179 QUESTIONS 1 iso PROBLEMS
^§|More About Matter Waves 1 1 ae
39- 1 ENERGIES OF A TRAPPED ELECTRON use
What Is Physics? 1 1 86
String Waves and Matter Waves 1 1 87
Energies of a Trapped Electron 1 1 87
39-2 WAVE FUNCTIONS OF A TRAPPED ELECTRON 1 1 91
Wave Functions of a Trapped Electron 1 1 92
39-3 AN ELECTRON IN A FINITE WELL 1195
An Electron in a Finite Well 1 1 95
39-4 TWO- AND THREE-DIMENSIONAL ELECTRON TRAPS 1 1 97
More Electron Traps 1 1 97
Two- and Three-Dimensional Electron Traps 1 200
39-5 THE HYDROGEN ATOM 1201
The Hydrogen Atom Is an Electron Trap 1 202 The Bohr Model of Hydrogen, a Lucky Break 1 203 Schrodinger's Equation and the Hydrogen Atom 1 205 REVIEWS, SUMMARY 1213 QUESTIONS 1213 PROBLEMS
CONTENTS
XV
40 All About Atoms 1210
40-1 PROPERTIES OF ATOMS 1219
What Is Physics? 1 220
Some Properties of Atoms 1 220
Angular Momentum, Magnetic Dipole Moments 1 222
40-2 THE STERN-GERLACH EXPERIMENT 1 226 The Stern-Gerlach Experiment 1 226
40-3 MAGNETIC RESONANCE 1229
Magnetic Resonance 1 229
40-4 EXCLUSION PRINCIPLE AND MULTIPLE ELECTRONS IN A TRAP 1 230
The Pauli Exclusion Principle 1 230 Multiple Electrons in Rectangular Traps 1 231
40-5 BUILDING THE PERIODIC TABLE 1234
Building the Periodic Table 1 234
40-6 X RAYS AND THE ORDERING OF THE ELEMENTS 1 236
X Rays and the Ordering of the Elements 1 237
40- 7 LASERS 1240
Lasers and Laser Light 1241 How Lasers Work 1 242
REVIEW & SUMMARY 1245 QUESTIONS 1246 PROBLEMS 1247
^Conduction of Electricity in Solids 1 252
41- 1 THE ELECTRICAL PROPERTIES OF METALS 1 252
What Is Physics? 1 252 The Electrical Properties of Solids 1 253 Energy Levels in a Crystalline Solid 1 254 Insulators 1 254 Metals 1 255
41-2 SEMICONDUCTORS AND DOPING 1261
Semiconductors 1 262 Doped Semiconductors 1 263
41- 3 THE//-/? JUNCTION AND THE TRANSISTOR 1265
The p-n Junction 1 266 The Junction Rectifier 1 267 The Light-Emitting Diode (LED) 1 268 The Transistor 1 270
REVIEW & SUMMARY 1271 QUESTIONS 1272 PROBLEMS 1272
<^H| Nuclear Physics 1 276
42- 1 DISCOVERING THE NUCLEUS 1276
What Is Physics? 1 276 Discovering the Nucleus 1 276
42-2 SOME NUCLEAR PROPERTIES 1279
Some Nuclear Properties 1 280
42-3 RADIOACTIVE DECAY 1286
Radioactive Decay 1 286
42-4 ALPHA DECAY 1239
Alpha Decay 1 289
42-5 BETA DECAY 1292
Beta Decay 1 292
42-6 RADIOACTIVE DATING 1295
Radioactive Dating 1 295
42-7 MEASURING RADIATION DOSAGE 1 296
Measuring Radiation Dosage 1 296
42- 8 NUCLEAR MODELS 1297
Nuclear Models 1 297
REVIEW & SUMMARY 1300 QUESTIONS 1301 PROBLEMS 1302
<^pnergy from the Nucleus 1 309
43- 1 NUCLEAR FISSION 1309
What Is Physics? 1 309
Nuclear Fission: The Basic Process 1310
A Model for Nuclear Fission 1312
43-2 THE NUCLEAR REACTOR isie
The Nuclear Reactor 131 6
43-3 A NATURAL NUCLEAR REACTOR 1 320
A Natural Nuclear Reactor 1 320
43-4 THERMONUCLEAR FUSION: THE BASIC PROCESS 1 322
Thermonuclear Fusion: The Basic Process 1 322
43-5 THERMONUCLEAR FUSION IN THE SUN AND OTHER STARS 1 324 Thermonuclear Fusion in the Sun and Other Stars 1 324
43- 6 CONTROLLED THERMONUCLEAR FUSION 1326
Controlled Thermonuclear Fusion 1 326
REVIEW & SUMMARY 1329 QUESTIONS 1329 PROBLEMS 1330
<^jfluarks, Leptons, and the Big Bang 1 334
44- 1 GENERAL PROPERTIES OF ELEMENTARY PARTICLES 1 334
What Is Physics? 1 334
Particles, Particles, Particles 1 335 An Interlude 1 339
44-2 LEPTONS, HADRONS, AND STRANGENESS 1 343
The Leptons 1 343
XVI
CONTENTS
The Hadrons 1 345
Still Another Conservation Law 1 346
The Eightfold Way 1 347
44-3 QUARKS AND MESSENGER PARTICLES 1 349
The Quark Model 1349
Basic Forces and Messenger Particles 1 352
44-4 COSMOLOGY 1355
A Pause for Reflection 1 355
The Universe Is Expanding 1 356
The Cosmic Background Radiation 1 357
Dark Matter 1 35s
The Big Bang 135s
A Summing Up i36i
REVIEW & SUMMARY 1362 QUESTIONS 1352 PROBLEMS 1353
APPEN DICES
A The International System of Units (SI) a-i B Some Fundamental Constants of Physics a-3 C Some Astronomical Data a-4 D Conversion Factors a-s E Mathematical Formulas a-9 F Properties of The Elements a-i 2 G Periodic Table of The Elements a-i 5
ANSWERS
to Checkpoints and Odd-Numbered Questions and Problems an-i INDEX l-l
WHY I WROTE THIS BOOK
Fun with a big challenge. That is how I have regarded physics since the day when Sharon, one of the students in a class I taught as a graduate student, suddenly demanded of me, “What has any of this got to do with my life?” Of course I immediately responded, “Sharon, this has everything to do with your life — this is physics.”
She asked me for an example. I thought and thought but could not come up with a single one.That night I began writing the book The Flying Circus of Physics (John Wiley & Sons Inc., 1975) for Sharon but also for me because I realized her complaint was mine. I had spent six years slugging my way through many dozens of physics textbooks that were carefully written with the best of pedagogical plans, but there was something missing. Physics is the most interesting subject in the world because it is about how the world works, and yet the textbooks had been thor¬ oughly wrung of any connection with the real world. The fun was missing.
I have packed a lot of real-world physics into Fundamentals of Physics , con¬ necting it with the new edition of The Flying Circus of Physics. Much of the mate¬ rial comes from the introductory physics classes I teach, where I can judge from the faces and blunt comments what material and presentations work and what do not.
The notes I make on my successes and failures there help form the basis of this book. My message here is the same as I had with every student I’ve met since Sharon so long ago: “Yes, you can reason from basic physics concepts all the way to valid conclusions about the real world, and that understanding of the real world is where the fun is.”
I have many goals in writing this book but the overriding one is to provide in¬ structors with tools by which they can teach students how to effectively read scientific material, iden¬ tify fundamental concepts, reason through scientific questions, and solve quantitative problems. This process is not easy for either students or instructors. Indeed, the course associated with this book may be one of the most challenging of all the courses taken by a student. However, it can also be one of the most rewarding because it reveals the world’s fundamental clockwork from which all scientific and engineering applications spring.
Many users of the ninth edition (both instructors and students) sent in comments and suggestions to improve the book. These improvements are now incorporated into the narrative and problems throughout the book. The publisher John Wiley & Sons and I regard the book as an ongoing project and encourage more input from users. You can send suggestions, corrections, and positive or negative comments to John Wiley & Sons or Jearl Walker (mail address: Physics Department, Cleveland State University, Cleveland, OH 44115 USA; or the blog site at www.flyingcircusofphysics.com). We may not be able to respond to all suggestions, but we keep and study each of them.
WHAT’S NEW?
Modules and Learning Objectives “What was I supposed to learn from this section?” Students have asked me this question for decades, from the weakest student to the strongest. The problem is that even a thoughtful student may not feel confident that the important points were captured while read¬ ing a section. I felt the same way back when I was using the first edition of Halliday and Resnick while taking first-year physics.
To ease the problem in this edition, I restructured the chapters into concept modules based on a primary theme and begin each module with a list of the module’s learning objectives. The list is an explicit statement of the skills and learning points that should be gathered in reading the module. Each list is following by a brief summary of the key ideas that should also be gathered. For example, check out the first module in Chapter 16, where a student faces a truck load of concepts and terms. Rather than depending on the student’s ability to gather and sort those ideas, I now provide an explicit checklist that functions somewhat like the checklist a pilot works through before taxiing out to the runway for takeoff.
XVII
PREFACE
xviii
^wileyNs
PLUS
Links Between Homework Problems and Learning Objectives in Wiley plus, every question and prob¬ lem at the end of the chapter is linked to a learning objective, to answer the (usually unspoken) ques¬ tions, “Why am I working this problem? What am I supposed to learn from it?” By being explicit about a problem’s purpose, I believe that a student might better transfer the learning objective to other problems with a different wording but the same key idea. Such transference would help defeat the common trouble that a student learns to work a particular problem but cannot then apply its key idea to a problem in a different setting.
Rewritten Chapters My students have continued to be challenged by several key chapters and by spots in several other chapters and so, in this edition, I rewrote a lot of the material. For example, I redesigned the chapters on Gauss’ law and electric potential, which have proved to be tough-going for my students. The presentations are now smoother and more direct to the key points. In the quan¬ tum chapters, I expanded the coverage of the Schrodinger equation, including reflection of matter waves from a step potential. At the request of several instructors, I decoupled the discussion of the Bohr atom from the Schrodinger solution for the hydrogen atom so that the historical account of Bohr’s work can be bypassed. Also, there is now a module on Planck’s blackbody radiation.
rWILEY _
PLUS
PLUS
New Sample Problems and Homework Questions and Problems Sixteen new sample problems have been added to the chapters, written so as to spotlight some of the difficult areas for my students. Also, about 250 problems and 50 questions have been added to the homework sections of the chapters. Some of these problems come from earlier editions of the book, as requested by several instructors.
Video Illustrations In the eVersion of the text available in Wiley PLUS, David Maiullo of Rutgers University has created video versions of approximately 30 of the photo¬ graphs and figures from the text. Much of physics is the study of things that move and video can often provide a better representation than a static photo or figure.
Online Aid Wiley plus is not just an online grading pro¬ gram. Rather, it is a dynamic learning center stocked with many different learning aids, including just-in-time problem-solving tutorials, embedded reading quizzes to encourage reading, animated figures, hundreds of sample problems, loads of simulations and demonstrations, and over 1500 videos ranging from math reviews to mini-lectures to examples. More of these learning aids are added every semester. For this 10th edition of HRW, some of the photos involving motion have been converted into videos so that the motion can be slowed and analyzed.
These thousands of learning aids are available 24/7 and can be repeated as many times as de¬ sired. Thus, if a student gets stuck on a homework problem at, say, 2:00 AM (which appears to be a popular time for doing physics homework), friendly and helpful resources are available at the click of a mouse.
LEARNINGS TOOLS
When I learned first-year physics in the first edition of Halliday and Resnick, I caught on by repeatedly reread¬ ing a chapter. These days we better understand that students have a wide range of learning styles. So, I have produced a wide range of learning tools, both in this new edition and online in Wiley PLUS'.
Animations of one of the key figures in each chapter. Here in the book, those figures are flagged with the swirling icon. In the online chapter in WileyPLUS, a mouse click begins the animation. I have chosen the fig¬ ures that are rich in information so that a student can see the physics in action and played out over a minute or two
Angular velocity derived from angular position
PREFACE
XIX
^Weyo
PLUS
XWeyo
PLUS
instead of just being flat on a printed page. Not only does this give life to the physics, but the anima¬ tion can be repeated as many times as a student wants.
Videos I have made well over 1500 instructional videos, with more coming each semester. Students can watch me draw or type on the screen as they hear me talk about a solution, tutorial, sample prob¬ lem, or review, very much as they would experience were they sitting next to me in my office while I worked out something on a notepad. An instructor’s lectures and tutoring will always be the most valuable learning tools, but my videos are available 24 hours a day, 7 days a week, and can be repeated indefinitely.
• Video tutorials on subjects in the chapters. I chose the subjects that chal¬ lenge the students the most, the ones that my students scratch their heads about.
• Video reviews of high school math, such as basic algebraic manipulations, trig functions, and simultaneous equations.
• Video introductions to math, such as vector multiplication, that will be new to the students.
• Video presentations of every Sample Problem in the textbook chapters . My intent is to work out the physics, starting with the Key Ideas instead of just grabbing a formula. However, I also want to demonstrate how to read a sam¬ ple problem, that is, how to read technical material to learn problem-solving procedures that can be transferred to other types of problems.
• Video solutions to 20% of the end-of chapter problems. The availability and timing of these solutions are controlled by the instructor. For example, they might be available after a homework deadline or a quiz. Each solution is not simply a plug-and-chug recipe. Rather I build a solution from the Key Ideas to the first step of reasoning and to a final solution. The student learns not just how to solve a particular problem but how to tackle any problem, even those that require physics courage.
• Video examples of how to read data from graphs (more than simply reading off a number with no comprehension of the physics).
Starts from rest.
In a certain time interval, it rotates tt/4 rad at constant angular acceleration 4.0 rad/s2, reaching angular speed 4.5 rad/s.
How much time (from rest) to reach that time interval?
Interval 2: This time inteval with given data
Interval 1 : From rest to the start of that time interval
►)■■)■) a
titnn. .;i:l
Problem-Solving Help I have written a large number of resources Wiley PLUS designed to help build the students’ problem-solving skills.
for
GETTING STARTEOi Whet ■* the radius o# rotation (in maters) of a pomt on the nm of the
•umber, no tolerance
Step l : Solution Step J of GO Tutorial 10-30
What « the fool angler speed m radians par second?
• Every sample problem in the textbook is available online in both reading and video formats.
• Hundreds of additional sample problems. These are available as stand¬ alone resources but (at the discretion of the instructor) they are also linked out of the homework problems. So, if a homework problem deals with, say, forces on a block on a ramp, a link to a related sample problem is provided.
However, the sample problem is not just a replica of the homework problem and thus does not provide a solution that can be merely duplicated without comprehension.
• GO Tutorials for 15% of the end-of-chapter homework problems. In multi¬ ple steps, I lead a student through a homework problem, starting with the Key Ideas and giving hints when wrong answers are submitted. However, I pur¬ posely leave the last step (for the final answer) to the student so that they are responsible at the end. Some online tutorial systems trap a student when wrong answers are given, which can generate a lot of frustration. My GO Tutorials are not traps, because at any step along the way, a student can return to the main problem.
• Hints on every end-of-chapter homework problem are available (at the discretion of the instructor). I wrote these as true hints about the main ideas and the general procedure for a solution, not as recipes that provide an answer without any comprehension.
t on how to approach this problem, on your own. To v>ew the original question while you work, you can just drag this screen to the side. (This GO Tutorial
Step : Solution Step 1 of GO Tutorial 10-30
mt angular
acceleration eouations o I Table 10-1 modified fi (Duo = (uQ + at
(2) 0- (fo = <M0f + {of2
(3) to2 ■ u>o + 2a(0- (fo)
(4) 0- 0o - {(cog ♦ u»)f
(5) 0 - 0o ■ u)f - jaf2
Counterclockwise «s the positive direction o I rotation, and (2) If a partide moves around a rotation axis at radius (centnpetai) acceleration ar at any moment is related t along the circular path) and its angular speed at dial m
clockwise is the negative direction. ' e magnitude of its radial tangential speed v (the speed
(3) If a partide moves around a rotation axis at radius r, acceleration at (the acceleration along the circular path) at a acceleration a at that moment by
(4) If a partide moves around a rotation axis at radius r, th which * rotates is related to the defence s 4 moves along *
s = rA0
Step : Solution Step 3 ol GO Tutorial 10-30
exact number, n
Step : Solution Step 4 of GO Tutorial 10-30
Through what angidar distance does the flywheel rotate to reach the final angular speed?
Now that you know how to solve the problem, go beck and try again on your a
Ullll
XX
PREFACE
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PLUS
Evaluation Materials
• Reading questions are available within each online section. I wrote these so that they do not require analysis or any deep understanding; rather they simply test whether a student has read the section. When a student opens up a section, a randomly chosen reading question (from a bank of questions) appears at the end. The instructor can decide whether the question is part of the grading for that section or whether it is just for the benefit of the student.
• Checkpoints are available within most sections. I wrote these so that they require analysis and deci¬ sions about the physics in the section. Answers to all checkpoints are in the back of the book.
Checkpoint 1
Here are three pairs of initial and final positions, respectively, along an x axis. Which pairs give a negative displacement: (a) —3 m, +5 m; (b) —3 m, —7 m; (c) 7 m, —3 m?
• All end-of-chapter homework Problems in the book (and many more problems) are available in Wiley PL US. The instructor can construct a homework assignment and control how it is graded when the answers are submitted online. For example, the instructor controls the deadline for submission and how many attempts a student is allowed on an answer. The instructor also controls which, if any, learning aids are available with each homework problem. Such links can include hints, sample prob¬ lems, in-chapter reading materials, video tutorials, video math reviews, and even video solutions (which can be made available to the students after, say, a homework deadline).
• Symbolic notation problems that require algebraic answers are available in every chapter.
• All end-of-chapter homework Questions in the book are available for assignment in Wiley PL US. These Questions (in a multiple choice format) are designed to evaluate the students’ conceptual un¬ derstanding.
Icons for Additional Help When worked-out solutions are provided either in print or electronically for certain of the odd-numbered problems, the statements for those problems include an icon to alert both student and instructor as to where the solutions are located. There are also icons indicating which problems have GO Tutorial, an Interactive LearningWare, or a link to the The Flying Circus of Physics. An icon guide is provided here and at the beginning of each set of problems.
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign SSM Worked-out solution available in Student Solutions Manual WWW Worked-out solution is at
• - ••• Number of dots indicates level of problem difficulty ILW Interactive solution is at
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
http://www.wiley.com/college/halliday
VERSIONS OF THE TEXT
To accommodate the individual needs of instructors and students, the ninth edition of Fundamentals of Physics is available in a number of different versions.
The Regular Edition consists of Chapters 1 through 37 (ISBN 9781118230718).
The Extended Edition contains seven additional chapters on quantum physics and cosmology, Chapters 1^14 (ISBN 9781118230725).
Volume 1 — Chapters 1-20 (Mechanics and Thermodynamics), hardcover,
ISBN 9781118233764
Volume 2 — Chapters 21-44 (E&M, Optics, and Quantum Physics), hardcover,
ISBN 9781118230732
PREFACE
XXI
INSTRUCTOR SUPPLEMENTS
Instructor’s Solutions Manual by Sen-Ben Liao, Lawrence Livermore National Laboratory. This man¬ ual provides worked-out solutions for all problems found at the end of each chapter. It is available in both MSWord and PDF.
Instructor Companion Site http://www.wiley.com/college/halliday
• Instructor’s Manual This resource contains lecture notes outlining the most important topics of each chapter; demonstration experiments; laboratory and computer projects; film and video sources; answers to all Questions, Exercises, Problems, and Checkpoints; and a correlation guide to the Questions, Exercises, and Problems in the previous edition. It also contains a complete list of all problems for which solutions are available to students (SSM.WWW, and ILW).
• Lecture PowerPoint Slides These PowerPoint slides serve as a helpful starter pack for instructors, outlining key concepts and incorporating figures and equations from the text.
• Classroom Response Systems (“Clicker”) Questions by David Marx, Illinois State University. There are two sets of questions available: Reading Quiz questions and Interactive Lecture ques¬ tions. The Reading Quiz questions are intended to be relatively straightforward for any student who reads the assigned material. The Interactive Lecture questions are intended for use in an interactive lecture setting.
• Wiley Physics Simulations by Andrew Duffy, Boston University and John Gastineau, Vernier Software. This is a collection of 50 interactive simulations (Java applets) that can be used for class¬ room demonstrations.
• Wiley Physics Demonstrations by David Maiullo, Rutgers University. This is a collection of digital videos of 80 standard physics demonstrations. They can be shown in class or accessed from WileyPLUS. There is an accompanying Instructor’s Guide that includes “clicker” questions.
• Test Bank For the 10th edition, the Test Bank has been completely over-hauled by Suzanne Willis, Northern Illinois University. The Test Bank includes more than 2200 multiple-choice questions. These items are also available in the Computerized Test Bank which provides full editing features to help you customize tests (available in both IBM and Macintosh versions).
• All text illustrations suitable for both classroom projection and printing.
Online Homework and Quizzing. In addition to WileyPLUS, Fundamentals of Physics, tenth edition, also supports WebAssignPLUS and LON-CAPA, which are other programs that give instructors the ability to deliver and grade homework and quizzes online. WebAssign PLUS also offers students an online version of the text.
STUDENT SUPPLEMENTS
Student Companion Site. The web site http://www.wiley.com/college/halliday was developed specifi¬ cally for Fundamentals of Physics, tenth edition, and is designed to further assist students in the study of physics. It includes solutions to selected end-of-chapter problems (which are identified with a www icon in the text); simulation exercises; tips on how to make best use of a programmable calcu¬ lator; and the Interactive LearningWare tutorials that are described below.
Student Study Guide (ISBN 9781118230787) by Thomas Barrett of Ohio State University. The Student Study Guide consists of an overview of the chapter’s important concepts, problem solving techniques and detailed examples.
Student Solutions Manual (ISBN 9781118230664) by Sen-Ben Liao, Lawrence Livermore National Laboratory. This manual provides students with complete worked-out solutions to 15 percent of the problems found at the end of each chapter within the text. The Student Solutions Manual for the 10th edition is written using an innovative approach called TEAL which stands for Think, Express, Analyze, and Learn. This learning strategy was originally developed at the Massachusetts Institute of Technology and has proven to be an effective learning tool for students. These problems with TEAL solutions are indicated with an SSM icon in the text.
xxii PREFACE
Interactive Learningware. This software guides students through solutions to 200 of the end-of-chapter problems. These problems are indicated with an ILW icon in the text. The solutions process is devel¬ oped interactively, with appropriate feedback and access to error-specific help for the most common mistakes.
Introductory Physics with Calculus as a Second Language: (ISBN 9780471739104) Mastering Problem Solving by Thomas Barrett of Ohio State University. This brief paperback teaches the student how to approach problems more efficiently and effectively. The student will learn how to recognize common patterns in physics problems, break problems down into manageable steps, and apply appropriate techniques. The book takes the student step by step through the solutions to numerous examples.
ACKNOWLEDGMENTS
A great many people have contributed to this book. Sen-Ben Liao of Lawrence Livermore National Laboratory, James Whitenton of Southern Polytechnic State University, and Jerry Shi, of Pasadena City College, performed the Herculean task of working out solutions for every one of the homework problems in the book. At John Wiley publishers, the book received support from Stuart Johnson, Geraldine Osnato and Aly Rentrop, the editors who oversaw the entire project from start to finish. We thank Elizabeth Swain, the production editor, for pulling all the pieces together during the complex production process. We also thank Maddy Lesure for her design of the text and the cover; Lee Goldstein for her page make-up; Helen Walden for her copyediting; and Lilian Brady for her proofreading. Jennifer Atkins was inspired in the search for unusual and interesting photographs. Both the publisher John Wiley & Sons, Inc. and Jearl Walker would like to thank the following for comments and ideas about the recent editions:
Jonathan Abramson, Portland State University ; Omar Adawi, Parkland College ; Edward Adelson, The Ohio State University ; Steven R. Baker, Naval Postgraduate School ; George Caplan, Wellesley College ; Richard Kass, The Ohio State University ; M. R. Khoshbin-e-Khoshnazar, Research Institution for Curriculum Development & Educational Innovations (Tehran)', Craig Kletzing, University of Iowa, Stuart Loucks, American River College', Laurence Lurio, Northern Illinois University ; Ponn Maheswaranathan, Winthrop University; Joe McCullough, Cabrillo College', Carl E. Mungan, U. S. Naval Academy, Don N. Page, University of Alberta', Elie Riachi, Fort Scott Community College', Andrew G. Rinzler, University of Florida', Dubravka Rupnik, Louisiana State University, Robert Schabinger, Rutgers University, Ruth Schwartz, Milwaukee School of Engineering ; Carol Strong, University of Alabama at Huntsville, Nora Thornber, Raritan Valley Community College', Frank Wang, LaGuardia Community College', Graham W. Wilson, University of Kansas', Roland Winkler, Northern Illinois University, William Zacharias, Cleveland State University, Ulrich Zurcher, Cleveland State University.
Finally, our external reviewers have been outstanding and we acknowledge here our debt to each member of that team.
Maris A. Abolins, Michigan State University Edward Adelson, Ohio State University Nural Akchurin, Texas Tech
Yildirim Aktas, University of North Carolina-Charlotte
Barbara Andereck, Ohio Wesleyan University
Tetyana Antimirova, Ryerson University
Mark Arnett , Kirkwood Community College
Arun Bansil , Northeastern University
Richard Barber, Santa Clara University
Neil Basecu, Westchester Community College
An and Batra, Howard University
Kenneth Bolland, The Ohio State University
Richard Bone, Florida International University
Michael E. Browne, University of Idaho
Timothy J. Burns, Leeward Community College
Joseph Buschi, Manhattan College
Philip A. Casabella, Rensselaer Polytechnic Institute
Randall Caton, Christopher Newport College
Roger Clapp, University of South Florida
W. R. Conkie, Queen’s University
Renate Crawford, University of Massachusetts-Dartmoiith Mike Crivello, San Diego State University Robert N. Davie, Jr., St. Petersburg Junior College Cheryl K. Dellai, Glendale Community College Eric R. Dietz, California State University at Chico
N. John DiNardo, Drexel University
Eugene Dunnam, University of Florida
Robert Endorf, University of Cincinnati
F. Paul Esposito, University of Cincinnati
Jerry Finkelstein, San Jose State University
Robert H. Good, California State University-Hayward
Michael Gorman, University of Houston
Benjamin Grinstein, University of California, San Diego
John B. Gruber, San Jose State University
Ann Hanks, American River College
Randy Harris, University of Calif ornia-D avis
Samuel Harris, Purdue University
Harold B. Hart, Western Illinois University
Rebecca Hartzler, Seattle Central Community College
John Hubisz, North Carolina StateUniversity
Joey Huston , Michigan State University
David Ingram, Ohio University
Shawn Jackson, University of Tulsa
Hector Jimenez, University of Puerto Rico
Sudhakar B. Joshi, York University
Leonard M. Kahn, University of Rhode Island
Sudipa Kirtley, Rose-Hulman Institute
Leonard Kleinman, University of Texas at Austin
Craig Kletzing, University of Iowa
Peter F. Koehler, University of Pittsburgh
xxiii
XXIV
ACKNOWLEDGMENTS
Arthur Z. Kovacs, Rochester Institute of Technology
Kenneth Krane, Oregon State University
Hadley Lawler, Vanderbilt University
Priscilla Laws, Dickinson College
Edbertho Leal, Polytechnic University of Puerto Rico
Vern Lindberg, Rochester Institute of Technology
Peter Loly, University of Manitoba
James MacLaren, Tulane University
Andreas Mandelis, University of Toronto
Robert R. Marchini, Memphis State University
Andrea Markelz, University at Buffalo, SUNY
Paul Marquard, Caspar College
David Marx, Illinois State University
Dan Mazilu, Washington and LeeUniversity
James H. McGuire, Tulane University
David M. McKinstry, Eastern Washington University
Jordon Morelli, Queen’s University
Eugene Mosca, United States Naval Academy
Eric R. Murray, Georgia Institute of Technology, School of
Physics
James Napolitano, Rensselaer Polytechnic Institute Blaine Norum, University of Virginia Michael O’Shea, Kansas State University Patrick Papin, San Diego State University Kiumars Parvin, San Jose State University Robert Pelcovits, Brown University Oren P. Quist, South Dakota State University Joe Redish, University of Maryland
Timothy M. Ritter, University of North Carolina at Pembroke
Dan Styer, Oberlin College
Frank Wang, LaGuardia Community College
Robert Webb, Texas A&M University
Suzanne Willis, Northern Illinois University
Shannon Willoughby, Montana State University
c
Measurement
1 -1 MEASURING THINGS, INCLUDING LENGTHS
Learning Objectives _
After reading this module, you should be able to . . .
1 .01 Identify the base quantities in the SI system. 1.02 Name the most frequently used prefixes for SI units.
1.03 Change units (here for length, area, and volume) by using chain-link conversions.
1.04 Explain that the meter is defined in terms of the speed of light in vacuum.
Key Ideas _
• Physics is based on measurement of physical quantities. Certain physical quantities have been chosen as base quanti¬ ties (such as length, time, and mass); each has been defined in terms of a standard and given a unit of measure (such as meter, second, and kilogram). Other physical quantities are defined in terms of the base quantities and their standards and units.
• The unit system emphasized in this book is the International System of Units (SI). The three physical quantities displayed in Table 1 -1 are used in the early chapters. Standards, which must be both accessible and invariable, have been estab¬ lished for these base quantities by international agreement.
These standards are used in all physical measurement, for both the base quantities and the quantities derived from them. Scientific notation and the prefixes of Table 1 -2 are used to simplify measurement notation.
• Conversion of units may be performed by using chain-link conversions in which the original data are multiplied succes¬ sively by conversion factors written as unity and the units are manipulated like algebraic quantities until only the desired units remain.
• The meter is defined as the distance traveled by light during a precisely specified time interval.
What Is Physics?
Science and engineering are based on measurements and comparisons. Thus, we need rules about how things are measured and compared, and we need experiments to establish the units for those measurements and comparisons. One purpose of physics (and engineering) is to design and conduct those experiments.
For example, physicists strive to develop clocks of extreme accuracy so that any time or time interval can be precisely determined and compared. You may wonder whether such accuracy is actually needed or worth the effort. Here is one example of the worth: Without clocks of extreme accuracy, the Global Positioning System (GPS) that is now vital to worldwide navigation would be useless.
Measuring Things
We discover physics by learning how to measure the quantities involved in physics. Among these quantities are length, time, mass, temperature, pressure, and electric current.
We measure each physical quantity in its own units, by comparison with a standard. The unit is a unique name we assign to measures of that quantity — for example, meter (m) for the quantity length. The standard corresponds to exactly 1.0 unit of the quantity. As you will see, the standard for length, which corresponds
1
2
CHAPTER 1
MEASUREMENT
Table 1-1 Units for Three SI Base Quantities
Quantity
Unit Name
Unit Symbol
Length
meter
m
Time
second
s
Mass
kilogram
kg
Table 1 -2
Prefixes for SI Units
Factor
Prefix"
Symbol
1024
yotta-
Y
1021
zetta-
Z
1018
exa-
E
1015
peta-
P
1012
tera-
T
109
g'ga-
G
106
mega-
M
103 * *
kilo-
k
102
hecto-
h
101
deka-
da
ltr1
deci-
d
10-* 1 2
centi-
c
10-3
milli-
m
10"6
micro-
f-
10"9
nano-
n
10-12
pico-
p
ltr15
femto-
f
io-18
atto-
a
10-21
zepto-
z
10-24
yocto-
y
to exactly 1.0 m, is the distance traveled by light in a vacuum during a certain fraction of a second. We can define a unit and its standard in any way we care to. However, the important thing is to do so in such a way that scientists around the world will agree that our definitions are both sensible and practical.
Once we have set up a standard — say, for length — we must work out proce¬ dures by which any length whatever, be it the radius of a hydrogen atom, the wheelbase of a skateboard, or the distance to a star, can be expressed in terms of the standard. Rulers, which approximate our length standard, give us one such procedure for measuring length. However, many of our comparisons must be indirect. You cannot use a ruler, for example, to measure the radius of an atom or the distance to a star.
Base Quantities. There are so many physical quantities that it is a problem to organize them. Fortunately, they are not all independent; for example, speed is the ratio of a length to a time. Thus, what we do is pick out — by international agree¬ ment — a small number of physical quantities, such as length and time, and assign standards to them alone. We then define all other physical quantities in terms of these base quantities and their standards (called base standards). Speed, for example, is defined in terms of the base quantities length and time and their base standards.
Base standards must be both accessible and invariable. If we define the length standard as the distance between one’s nose and the index finger on an outstretched arm, we certainly have an accessible standard — but it will, of course, vary from person to person. The demand for precision in science and engineering pushes us to aim first for invariability. We then exert great effort to make dupli¬ cates of the base standards that are accessible to those who need them.
The International System of Units
In 1971, the 14th General Conference on Weights and Measures picked seven quantities as base quantities, thereby forming the basis of the International
System of Units, abbreviated SI from its French name and popularly known as the metric system. Table 1-1 shows the units for the three base quantities — length, mass, and time — that we use in the early chapters of this book. These units were defined to be on a “human scale.”
Many SI derived units are defined in terms of these base units. For example, the SI unit for power, called the watt (W), is defined in terms of the base units for mass, length, and time. Thus, as you will see in Chapter 7,
1 watt = 1 W = 1 kg • m2/s3, (1-1)
where the last collection of unit symbols is read as kilogram-meter squared per second cubed.
To express the very large and very small quantities we often run into in physics, we use scientific notation , which employs powers of 10. In this notation,
3 560 000 000 m = 3.56 X 109 m (1-2)
and 0.000 000 492 s = 4.92 X 10-7 * s. (1-3)
Scientific notation on computers sometimes takes on an even briefer look, as in
3.56 E9 and 4.92 E-7, where E stands for “exponent of ten.” It is briefer still on
some calculators, where E is replaced with an empty space.
As a further convenience when dealing with very large or very small mea¬
surements, we use the prefixes listed in Table 1-2. As you can see, each prefix
represents a certain power of 10, to be used as a multiplication factor. Attaching
a prefix to an SI unit has the effect of multiplying by the associated factor. Thus,
we can express a particular electric power as
"The most frequently used prefixes are shown in bold type.
1.27 X 109 watts = 1.27 gigawatts = 1.27 GW
(1-4)
1-1 MEASURING THINGS, INCLUDING LENGTHS
3
or a particular time interval as
2.35 X 10~9 s = 2.35 nanoseconds = 2.35 ns. (1-5)
Some prefixes, as used in milliliter, centimeter, kilogram, and megabyte, are probably familiar to you.
Changing Units
We often need to change the units in which a physical quantity is expressed. We do so by a method called chain-link conversion. In this method, we multiply the original measurement by a conversion factor (a ratio of units that is equal to unity). For example, because 1 min and 60 s are identical time intervals, we have
1 min , 60 s
Thus, the ratios (1 min)/(60 s) and (60 s)/(l min) can be used as conversion factors. This is not the same as writing ^ = 1 or 60 = 1; each number and its unit must be treated together.
Because multiplying any quantity by unity leaves the quantity unchanged, we can introduce conversion factors wherever we find them useful. In chain-link conversion, we use the factors to cancel unwanted units. For example, to convert 2 min to seconds, we have
2 min = (2 min)(l) = (2 mirf)^- = 120 s. (1-6)
If you introduce a conversion factor in such a way that unwanted units do not cancel, invert the factor and try again. In conversions, the units obey the same algebraic rules as variables and numbers.
Appendix D gives conversion factors between SI and other systems of units, including non-SI units still used in the United States. However, the conversion factors are written in the style of “1 min = 60 s” rather than as a ratio. So, you need to decide on the numerator and denominator in any needed ratio.
Length
In 1792, the newborn Republic of France established a new system of weights and measures. Its cornerstone was the meter, defined to be one ten-millionth of the distance from the north pole to the equator. Later, for practical reasons, this Earth standard was abandoned and the meter came to be defined as the distance between two fine lines engraved near the ends of a platinum-iridium bar, the standard meter bar, which was kept at the International Bureau of Weights and Measures near Paris. Accurate copies of the bar were sent to standardizing labo¬ ratories throughout the world. These secondary standards were used to produce other, still more accessible standards, so that ultimately every measuring device derived its authority from the standard meter bar through a complicated chain of comparisons.
Eventually, a standard more precise than the distance between two fine scratches on a metal bar was required. In 1960, a new standard for the meter, based on the wavelength of light, was adopted. Specifically, the standard for the meter was redefined to be 1 650 763.73 wavelengths of a particular orange -red light emitted by atoms of krypton-86 (a particular isotope, or type, of krypton) in a gas discharge tube that can be set up anywhere in the world. This awkward number of wavelengths was chosen so that the new standard would be close to the old meter-bar standard.
4
CHAPTER 1
MEASUREMENT
By 1983, however, the demand for higher precision had reached such a point that even the krypton-86 standard could not meet it, and in that year a bold step was taken. The meter was redefined as the distance traveled by light in a specified time interval. In the words of the 17th General Conference on Weights and Measures:
© The meter is the length of the path traveled by light in a vacuum during a time interval of 1/299 792 458 of a second.
This time interval was chosen so that the speed of light c is exactly
c = 299 792 458 m/s.
Measurements of the speed of light had become extremely precise, so it made sense to adopt the speed of light as a defined quantity and to use it to redefine the meter.
Table 1-3 Some Approximate Lengths Table 1-3 shows a wide range of lengths, from that of the universe (top line)
—— to those of some very small objects.
Measurement Length in Meters
Distance to the first
galaxies formed
2
X
1026
Distance to the
Andromeda galaxy
2
X
1022
Distance to the nearby
star Proxima Centauri
4
X
1016
Distance to Pluto
6
X
1012
Radius of Earth
6
X
106
Height of Mt. Everest
9
X
103
Thickness of this page
1
X
10“4
Length of a typical virus
1
X
10-8
Radius of a hydrogen atom
5
X
HT11
Radius of a proton
1
X
1(L15
Significant Figures and Decimal Places
Suppose that you work out a problem in which each value consists of two digits. Those digits are called significant figures and they set the number of digits that you can use in reporting your final answer. With data given in two significant figures, your final answer should have only two significant figures. However, depending on the mode setting of your calculator, many more digits might be displayed. Those extra digits are meaningless.
In this book, final results of calculations are often rounded to match the least number of significant figures in the given data. (However, sometimes an extra significant figure is kept.) When the leftmost of the digits to be discarded is 5 or more, the last remaining digit is rounded up; otherwise it is retained as is. For example, 11.3516 is rounded to three significant figures as 11.4 and 11.3279 is rounded to three significant figures as 11.3. (The answers to sample problems in this book are usually presented with the symbol = instead of ~ even if rounding is involved.)
When a number such as 3.15 or 3.15 X 103 is provided in a problem, the number of significant figures is apparent, but how about the number 3000? Is it known to only one significant figure (3 X 103)? Or is it known to as many as four significant figures (3.000 X 103)? In this book, we assume that all the zeros in such given num¬ bers as 3000 are significant, but you had better not make that assumption elsewhere.
Don’t confuse significant figures with decimal places. Consider the lengths 35.6 mm, 3.56 m, and 0.00356 m. They all have three significant figures but they have one, two, and five decimal places, respectively.
Sample Problem 1 .Ol Estimating order of magnitude, ball of string
The world’s largest ball of string is about 2 m in radius. To the nearest order of magnitude, what is the total length L of the string in the ball?
KEY IDEA
We could, of course, take the ball apart and measure the to¬ tal length L, but that would take great effort and make the
ball’s builder most unhappy. Instead, because we want only the nearest order of magnitude, we can estimate any quanti¬ ties required in the calculation.
Calculations: Let us assume the ball is spherical with radius R = 2 m. The string in the ball is not closely packed (there are uncountable gaps between adjacent sections of string). To allow for these gaps, let us somewhat overestimate
1-2 TIME
5
the cross-sectional area of the string by assuming the cross section is square, with an edge length d = 4 mm. Then, with a cross-sectional area of d2 and a length L, the string occupies a total volume of
V = (cross-sectional area)(length) = cPL.
d2L = 4 R\
4 R3 _ 4(2 m)3
d2 (4 X 10~3 m)2
= 2 X 106 m « 106 m = 103 km.
(Answer)
This is approximately equal to the volume of the ball, given by ^ttR3, which is about 4 R3 because tt is about 3. Thus, we have the following
(Note that you do not need a calculator for such a simplified calculation.) To the nearest order of magnitude, the ball contains about 1000 km of string!
^WILEYO
PLUS Additional examples, video, and practice available at WileyPLUS
1-2 TIME
Learning Objectives _
After reading this module, you should be able to .. . 1 .06 Use various measures of time, such as for motion or as
1 .05 Change units for time by using chain-link conversions. determined on different clocks.
Key Idea -
• The second is defined in terms of the oscillations of light signals are sent worldwide by radio signals keyed to atomic
emitted by an atomic (cesium-1 33) source. Accurate time clocks in standardizing laboratories.
Time
Time has two aspects. For civil and some scientific purposes, we want to know the time of day so that we can order events in sequence. In much scientific work, we want to know how long an event lasts. Thus, any time standard must be able to answer two questions: “ When did it happen?” and “What is its duration!” Table 1-4 shows some time intervals.
Any phenomenon that repeats itself is a possible time standard. Earth’s rotation, which determines the length of the day, has been used in this way for centuries; Fig. 1-1 shows one novel example of a watch based on that rotation. A quartz clock, in which a quartz ring is made to vibrate continuously, can be calibrated against Earth’s rotation via astronomical observations and used to measure time intervals in the laboratory. Flowever, the calibration cannot be carried out with the accuracy called for by modern scientific and engineering technology.
Table 1-4 Some Approximate Time Intervals
Measurement
Time Interval in Seconds
Measurement
Time Interval in Seconds
Lifetime of the proton (predicted)
3 X 104°
Time between human heartbeats 8 X 10 1 Lifetime of the muon 2 X 10"6
Age of the universe
5 X 1017
Shortest lab light pulse
1 X 10"16
Age of the pyramid of Cheops 1 X 1011 Human life expectancy 2 X 109
Lifetime of the most unstable particle
1 X 10“23
Length of a day
9 X 104
The Planck time"
1 X 10“43
“This is the earliest time after the big bang at which the laws of physics as we know them can be applied.
Steven Pitkin
Figure 1-1 When the metric system was proposed in 1792, the hour was redefined to provide a 10-hour day. The idea did not catch on. The maker of this 10-hour watch wisely provided a small dial that kept con¬ ventional 12-hour time. Do the two dials indicate the same time?
6
CHAPTER 1 MEASUREMENT
1980
1981
1982
Figure 1-2 Variations in the length of the day over a 4-year period. Note that the entire vertical scale amounts to only 3 ms (= 0.003 s).
To meet the need for a better time standard, atomic clocks have been developed. An atomic clock at the National Institute of Standards and Technology (NIST) in Boulder, Colorado, is the stan¬ dard for Coordinated Universal Time (UTC) in the United States. Its time signals are available by shortwave radio (stations WWV and WWVH) and by telephone (303-499-7111). Time signals (and related information) are also available from the United States Naval Observatory at website http://tycho.usno.navy.mil/time.html. (To set a clock extremely accurately at your particular location, you would have to account for the travel time required for these signals to reach you.)
Figure 1-2 shows variations in the length of one day on Earth over a 4-year period, as determined by comparison with a cesium (atomic) clock. Because the variation displayed by Fig. 1-2 is sea¬ sonal and repetitious, we suspect the rotating Earth when there is a difference between Earth and atom as timekeepers. The variation is due to tidal effects caused by the Moon and to large-scale winds.
The 13th General Conference on Weights and Measures in 1967 adopted a standard second based on the cesium clock:
1983
©
One second is the time taken by 9 192 631 770 oscillations of the light (of a specified wavelength) emitted by a cesium-133 atom.
Atomic clocks are so consistent that, in principle, two cesium clocks would have to run for 6000 years before their readings would differ by more than 1 s. Even such accuracy pales in comparison with that of clocks currently being developed; their precision may be 1 part in 1018 — that is, 1 s in 1 X 1018 s (which is about 3 X 1010 y).
1-3 MASS
Learning Objectives _
After readin9 this module’ Y°u should be able t0 ■ ■ ■ 1 .08 Relate density to mass and volume when the mass is
1 .07 Change units for mass by using chain-link uniformly distributed,
conversions.
Key Ideas _
• The kilogram is defined in terms of a platinum-iridium • The density p of a material is the mass per unit volume:
standard mass kept near Paris. For measurements on an m
atomic scale, the atomic mass unit, defined in terms of p =
the atom carbon-1 2, is usually used.
Mass
The Standard Kilogram
The SI standard of mass is a cylinder of platinum and iridium (Fig. 1-3) that is kept at the International Bureau of Weights and Measures near Paris and assigned, by
Figure 1-3 The international 1 kg standard of mass, a platinum-iridium cylinder 3.9 cm in height and in diameter.
1 -3 MASS
7
international agreement, a mass of 1 kilogram. Accurate copies have been sent to standardizing laboratories in other countries, and the masses of other bodies can be determined by balancing them against a copy. Table 1-5 shows some masses expressed in kilograms, ranging over about 83 orders of magnitude.
The U.S. copy of the standard kilogram is housed in a vault at NIST. It is removed, no more than once a year, for the purpose of checking duplicate copies that are used elsewhere. Since 1889, it has been taken to France twice for recomparison with the primary standard.
A Second Mass Standard
The masses of atoms can be compared with one another more precisely than they can be compared with the standard kilogram. For this reason, we have a second mass standard. It is the carbon-12 atom, which, by international agree¬ ment, has been assigned a mass of 12 atomic mass units (u). The relation between the two units is
1 u = 1.660 538 86 X 10~27 kg,
(1-7)
Table 1-5 Some Approximate Masses
Mass in
Object Kilograms
Known universe
1
X
1053
Our galaxy
2
X
1041
Sun
2
X
1030
Moon
7
X
1022
Asteroid Eros
5
X
1015
Small mountain
1
X
1012
Ocean liner
7
X
107
Elephant
5
X
103
Grape
3
X
10~3
Speck of dust
7
X
10-1°
Penicillin molecule
5
X
l0-n
Uranium atom
4
X
10“25
Proton
2
X
10“27
Electron
9
X
10“31
with an uncertainty of ±10 in the last two decimal places. Scientists can, with reasonable precision, experimentally determine the masses of other atoms rela¬ tive to the mass of carbon-12. What we presently lack is a reliable means of extending that precision to more common units of mass, such as a kilogram.
Density
As we shall discuss further in Chapter 14, density p (lowercase Greek letter rho) is the mass per unit volume:
Densities are typically listed in kilograms per cubic meter or grams per cubic centimeter. The density of water (1.00 gram per cubic centimeter) is often used as a comparison. Fresh snow has about 10% of that density; platinum has a density that is about 21 times that of water.
Sample Problem 1 .02 Density and liquefaction
A heavy object can sink into the ground during an earthquake if the shaking causes the ground to undergo liquefaction, in which the soil grains experience little friction as they slide over one another. The ground is then effectively quicksand. The possibility of liquefaction in sandy ground can be pre¬ dicted in terms of the void ratio e for a sample of the ground:
n,
e =
(1-9)
Here, F„rains is the total volume of the sand grains in the sam¬ ple and Fvojds is the total volume between the grains (in the voids). If e exceeds a critical value of 0.80, liquefaction can occur during an earthquake. What is the corresponding sand density psand? Solid silicon dioxide (the primary component of sand) has a density of pSiQi = 2.600 X 103 kg/m3.
KEY IDEA
The density of the sand psand in a sample is the mass per unit volume — that is, the ratio of the total mass msand of the sand grains to the total volume Vtotal of the sample:
_ ^sand
Psand tt
* total
(1-10)
Calculations: The total volume Ftotal of a sample is
F total Fp,
+ Fv,
Substituting for Fvoids from Eq. 1-9 and solving for Fgrains lead to
Vm
Ftotal
1 + e "
(1-11)
8
CHAPTER 1 MEASUREMENT
From Eq. 1-8, the total mass msand of the sand grains is the product of the density of silicon dioxide and the total vol¬ ume of the sand grains:
^sand Psi02 ^grains* (1-12)
Substituting this expression into Eq. 1-10 and then substitut¬ ing for Egrains from Eq. 1-11 lead to
Psand
PSiQ2 Kotal
Kota I 1 E C
Psio2 1 + e '
(1-13)
Substituting pSiQi = 2.600 X 103 kg/m3 and the critical value of e = 0.80, we find that liquefaction occurs when the sand density is less than
Psand
2.600 X 10 3 kg/m3 1.80
1.4 X 103 kg/m3.
(Answer)
A building can sink several meters in such liquefaction.
/wileyAs
'PLUS Additional examples, video, and practice available at WileyPLUS
eview & Summary
Measurement in Physics Physics is based on measurement of physical quantities. Certain physical quantities have been cho¬ sen as base quantities (such as length, time, and mass); each has been defined in terms of a standard and given a unit of measure (such as meter, second, and kilogram). Other physical quantities are defined in terms of the base quantities and their standards and units.
SI Units The unit system emphasized in this book is the International System of Units (SI). The three physical quantities displayed in Table 1-1 are used in the early chapters. Standards, which must be both accessible and invariable, have been estab¬ lished for these base quantities by international agreement. These standards are used in all physical measurement, for both the base quantities and the quantities derived from them. Scientific notation and the prefixes of Table 1-2 are used to sim¬ plify measurement notation.
Changing Units Conversion of units may be performed by us¬ ing chain-link conversions in which the original data are multiplied
successively by conversion factors written as unity and the units are manipulated like algebraic quantities until only the desired units remain.
Length The meter is defined as the distance traveled by light during a precisely specified time interval.
Time The second is defined in terms of the oscillations of light emitted by an atomic (cesium-133) source. Accurate time signals are sent worldwide by radio signals keyed to atomic clocks in stan¬ dardizing laboratories.
Mass The kilogram is defined in terms of a platinum- iridium standard mass kept near Paris. For measurements on an atomic scale, the atomic mass unit, defined in terms of the atom carbon-12, is usually used.
Density The density p of a material is the mass per unit volume:
roblems
Tutoring problem available (at instructor's discretion) in WileyPLUS and WebAssign SSM Worked-out solution available in Student Solutions Manual WWW Worked-out solution is at
• - ••• Number of dots indicates level of problem difficulty ILW Interactive solution is at A
-C3&- Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
Module 1-1 Measuring Things, Including Lengths
•1 SSM Earth is approximately a sphere of radius 6.37 X 106 m. What are (a) its circumference in kilometers, (b) its surface area in square kilometers, and (c) its volume in cubic kilometers?
•2 A gry is an old English measure for length, defined as 1/10 of a line, where line is another old English measure for length, defined as 1/12 inch. A common measure for length in the publishing busi¬ ness is a point, defined as 1/72 inch. What is an area of 0.50 gry2 in points squared (points2)?
•3 The micrometer (1 pm) is often called the micron, (a) How
many microns make up 1.0 km? (b) What fraction of a centimeter equals 1.0 pml (c) How many microns are in 1.0 yd?
•4 Spacing in this book was generally done in units of points and picas: 12 points = 1 pica, and 6 picas = 1 inch. If a figure was mis¬ placed in the page proofs by 0.80 cm, what was the misplacement in (a) picas and (b) points?
•5 SSM WWW Horses are to race over a certain English meadow for a distance of 4.0 furlongs. What is the race distance in (a) rods and (b) chains? (1 furlong = 201.168 m, 1 rod = 5.0292 m, and 1 chain = 20.117 m.)
PROBLEMS
9
••6 You can easily convert common units and measures electroni¬ cally, but you still should be able to use a conversion table, such as those in Appendix D. Table 1-6 is part of a conversion table for a system of volume measures once common in Spain; a volume of 1 fanega is equivalent to 55.501 dm3 (cubic decimeters). To complete the table, what numbers (to three significant figures) should be en¬ tered in (a) the cahiz column, (b) the fanega column, (c) the cuar- tilla column, and (d) the almude column, starting with the top blank? Express 7.00 almudes in (e) medios, (f) cahizes, and (g) cu¬ bic centimeters (cm3).
Table 1-6 Problem 6
cahiz
fanega
cuartilla
almude
medio
1 cahiz =
1
12
48
144
288
1 fanega =
1
4
12
24
1 cuartilla =
1
3
6
1 almude =
1
2
1 medio =
1
••7 ILW Hydraulic engineers in the United States often use, as a unit of volume of water, the acre-foot , defined as the volume of wa¬ ter that will cover 1 acre of land to a depth of 1 ft. A severe thun¬ derstorm dumped 2.0 in. of rain in 30 min on a town of area 26 km2. What volume of water, in acre-feet, fell on the town?
••8 ® Harvard Bridge, which connects MIT with its fraternities across the Charles River, has a length of 364.4 Smoots plus one ear. The unit of one Smoot is based on the length of Oliver Reed Smoot, Jr., class of 1962, who was carried or dragged length by length across the bridge so that other pledge members of the Lambda Chi Alpha fraternity could mark off (with paint) 1-Smoot lengths along the bridge. The marks have been repainted biannually by fraternity pledges since the initial measurement, usually during times of traffic congestion so that the police can¬ not easily interfere. (Presumably, the police were originally up¬ set because the Smoot is not an SI base unit, but these days they seem to have accepted the unit.) Figure 1-4 shows three parallel paths, measured in Smoots (S), Willies (W), and Zeldas (Z). What is the length of 50.0 Smoots in (a) Willies and (b) Zeldas?
Module 1-2 Time
•10 Until 1883, every city and town in the United States kept its own local time. Today, travelers reset their watches only when the time change equals 1.0 h. How far, on the average, must you travel in degrees of longitude between the time-zone boundaries at which your watch must be reset by 1.0 h? (Hint: Earth rotates 360° in about 24 h.)
•1 1 For about 10 years after the French Revolution, the French government attempted to base measures of time on multiples of ten: One week consisted of 10 days, one day consisted of 10 hours, one hour consisted of 100 minutes, and one minute consisted of 100 seconds. What are the ratios of (a) the French decimal week to the standard week and (b) the French decimal second to the standard second?
•1 2 The fastest growing plant on record is a Hesperoyucca whip- plei that grew 3.7 m in 14 days. What was its growth rate in micro¬ meters per second?
•1 3 © Three digital clocks A, B, and C run at different rates and do not have simultaneous readings of zero. Figure 1-6 shows si¬ multaneous readings on pairs of the clocks for four occasions. (At the earliest occasion, for example, B reads 25.0 s and C reads 92.0 s.) If two events are 600 s apart on clock A, how far apart are they on (a) clock B and (b) clock C? (c) When clock A reads 400 s, what does clock B read? (d) When clock C reads 15.0 s, what does clock B read? (Assume negative readings for prezero times.)
312 512
1 1
25.0
i
125
i
200
i
1
290
i
i
92.0
i
i
142
i
Figure 1-6
Problem 13.
•14 A lecture period (50 min) is close to 1 microcentury, (a) How long is a microcentury in minutes? (b) Using
/ actual — approximation \
percentage difference = I - - — j - I 100,
0
i
32
i
212
i
i
0
1
i
i
i
258
i
i
60
i
i
216
i
Figure 1-4 Problem 8.
••9 Antarctica is roughly semicircular, with a radius of 2000 km (Fig. 1-5). The average thickness of its ice cover is 3000 m. How many cubic centimeters of ice does Antarctica contain? (Ignore the curvature of Earth.)
I X^OOO km
3000 rtT yS
T
Figure 1-5 Problem 9.
find the percentage difference from the approximation.
•15 A fortnight is a charming English measure of time equal to 2.0 weeks (the word is a contraction of “fourteen nights”). That is a nice amount of time in pleasant company but perhaps a painful string of microseconds in unpleasant company. How many mi¬ croseconds are in a fortnight?
•16 Time standards are now based on atomic clocks. A promis¬ ing second standard is based on pulsars , which are rotating neu¬ tron stars (highly compact stars consisting only of neutrons). Some rotate at a rate that is highly stable, sending out a radio beacon that sweeps briefly across Earth once with each rotation, like a lighthouse beacon. Pulsar PSR 1937 + 21 is an example; it rotates once every 1.557 806 448 872 75 ± 3 ms, where the trailing ±3 indicates the uncertainty in the last decimal place (it does not mean ±3 ms), (a) How many rotations does PSR 1937 + 21 make in 7.00 days? (b) How much time does the pulsar take to rotate ex¬ actly one million times and (c) what is the associated uncertainty?
io
CHAPTER 1 MEASUREMENT
•17 SSM Five clocks are being tested in a laboratory. Exactly at noon, as determined by the WWV time signal, on successive days of a week the clocks read as in the following table. Rank the five clocks according to their relative value as good timekeepers, best to worst. Justify your choice.
Clock
Sun.
Mon.
Tues.
Wed.
Thurs.
Fri.
Sat.
A
12:36:40
12:36:56
12:37:12
12:37:27
12:37:44
12:37:59
12:38:14
B
11:59:59
12:00:02
11:59:57
12:00:07
12:00:02
11:59:56
12:00:03
c
15:50:45
15:51:43
15:52:41
15:53:39
15:54:37
15:55:35
15:56:33
D
12:03:59
12:02:52
12:01:45
12:00:38
11:59:31
11:58:24
11:57:17
E
12:03:59
12:02:49
12:01:54
12:01:52
12:01:32
12:01:22
12:01:12
••18 Because Earth’s rotation is gradually slowing, the length of each day increases: The day at the end of 1.0 century is 1.0 ms longer than the day at the start of the century. In 20 centuries, what is the total of the daily increases in time?
•••19 Suppose that, while lying on a beach near the equator watching the Sun set over a calm ocean, you start a stopwatch just as the top of the Sun disappears. You then stand, elevating your eyes by a height H = 1.70 m, and stop the watch when the top of the Sun again disappears. If the elapsed time is t = 11.1 s, what is the radius r of Earth?
Module 1-3 Mass
•20 © The record for the largest glass bottle was set in 1992 by a team in Millville, New Jersey — they blew a bottle with a volume of 193 U.S. fluid gallons, (a) How much short of 1.0 million cubic cen¬ timeters is that? (b) If the bottle were filled with water at the leisurely rate of 1.8 g/min, how long would the filling take? Water has a density of 1000 kg/m3.
•2 1 Earth has a mass of 5.98 X 1024 kg. The average mass of the atoms that make up Earth is 40 u. How many atoms are there in Earth?
•22 Gold, which has a density of 19.32 g/cm3, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber, (a) If a sample of gold, with a mass of 27.63 g, is pressed into a leaf of 1.000 fim thickness, what is the area of the leaf? (b) If, instead, the gold is drawn out into a cylindrical fiber of radius 2.500 fjLm, what is the length of the fiber?
•23 SSM (a) Assuming that water has a density of exactly 1 g/cm3, find the mass of one cubic meter of water in kilograms, (b) Suppose that it takes 10.0 h to drain a container of 5700 m3 of water. What is the “mass flow rate,” in kilograms per second, of wa¬ ter from the container?
••24 ® Grains of fine California beach sand are approximately spheres with an average radius of 50 /im and are made of silicon dioxide, which has a density of 2600 kg/m3. What mass of sand grains would have a total surface area (the total area of all the individual spheres) equal to the surface area of a cube 1.00 m on an edge?
••25 ■iJEP During heavy rain, a section of a mountainside mea¬ suring 2.5 km horizontally, 0.80 km up along the slope, and 2.0 m deep slips into a valley in a mud slide. Assume that the mud ends up uniformly distributed over a surface area of the valley measuring 0.40 km X 0.40 km and that mud has a density of 1900 kg/m3. What is the mass of the mud sitting above a 4.0 m2 area of the valley floor?
••26 One cubic centimeter of a typical cumulus cloud contains 50 to 500 water drops, which have a typical radius of 10 /xm. For
that range, give the lower value and the higher value, respectively, for the following, (a) How many cubic meters of water are in a cylindrical cumulus cloud of height 3.0 km and radius 1.0 km? (b) How many 1-liter pop bottles would that water fill? (c) Water has a density of 1000 kg/m3. How much mass does the water in the cloud have?
••27 Iron has a density of 7.87 g/cm3, and the mass of an iron atom is 9.27 X 10-26 kg. If the atoms are spherical and tightly packed, (a) what is the volume of an iron atom and (b) what is the distance be¬ tween the centers of adjacent atoms?
••28 A mole of atoms is 6.02 X 1023 atoms. To the nearest order of magnitude, how many moles of atoms are in a large domestic cat? The masses of a hydrogen atom, an oxygen atom, and a carbon atom are 1.0 u, 16 u, and 12 u, respectively. (Hint: Cats are some¬ times known to kill a mole.)
••29 On a spending spree in Malaysia, you buy an ox with a weight of 28.9 piculs in the local unit of weights: 1 picul = 100 gins, 1 gin = 16 tahils, 1 tahil = 10 chees, and 1 chee = 10 hoons. The weight of 1 hoon corresponds to a mass of 0.3779 g. When you arrange to ship the ox home to your astonished family, how much mass in kilograms must you declare on the shipping manifest? (Hint: Set up multiple chain-link conversions.)
••30 ® Water is poured into a container that has a small leak. The mass m of the water is given as a function of time t by m = 5.00t°-8 — 3.00f + 20.00, with t > 0, m in grams, and t in sec¬ onds. (a) At what time is the water mass greatest, and (b) what is that greatest mass? In kilograms per minute, what is the rate of mass change at (c) t = 2.00 s and (d) t = 5.00 s?
•••31 A vertical container with base area measuring 14.0 cm by 17.0 cm is being filled with identical pieces of candy, each with a volume of 50.0 mm3 and a mass of 0.0200 g. Assume that the volume of the empty spaces between the candies is negligible. If the height of the candies in the container increases at the rate of 0.250 cm/s, at what rate (kilograms per minute) does the mass of the candies in the container increase?
Additional Problems
32 In the United States, a doll house has the scale of 1 : 12 of a real house (that is, each length of the doll house is ]2 that of the real house) and a miniature house (a doll house to fit within a doll house) has the scale of 1 : 144 of a real house. Suppose a real house (Fig. 1-7) has a front length of 20 m, a depth of 12 m, a height of 6.0 m, and a standard sloped roof (vertical triangular faces on the ends) of height 3.0 m. In cubic meters, what are the volumes of the corre¬ sponding (a) doll house and (b) miniature house?
12 m - ►
Figure 1-7 Problem 32.
PROBLEMS
1 1
33 SSM A ton is a measure of volume frequently used in ship¬ ping, but that use requires some care because there are at least three types of tons: A displacement ton is equal to 7 barrels bulk, a freight ton is equal to 8 barrels bulk, and a register ton is equal to 20 barrels bulk. A barrel bulk is another measure of vol¬ ume: 1 barrel bulk = 0.1415 m3. Suppose you spot a shipping order for “73 tons” of M&M candies, and you are certain that the client who sent the order intended “ton” to refer to volume (instead of weight or mass, as discussed in Chapter 5). If the client actually meant displacement tons, how many extra U.S. bushels of the can¬ dies will you erroneously ship if you interpret the order as (a) 73 freight tons and (b) 73 register tons? (1 m3 = 28.378 U.S. bushels.)
34 Two types of barrel units were in use in the 1920s in the United States. The apple barrel had a legally set volume of 7056 cu¬ bic inches; the cranberry barrel, 5826 cubic inches. If a merchant sells 20 cranberry barrels of goods to a customer who thinks he is receiving apple barrels, what is the discrepancy in the shipment volume in liters?
35 An old English children’s rhyme states, “Little Miss Muffet sat on a tuffet, eating her curds and whey, when along came a spi¬ der who sat down beside her. . . .” The spider sat down not because of the curds and whey but because Miss Muffet had a stash of 11 tuffets of dried flies. The volume measure of a tuffet is given by 1 tuffet = 2 pecks = 0.50 Imperial bushel, where 1 Imperial bushel = 36.3687 liters (L). What was Miss Muffet’s stash in (a) pecks,
(b) Imperial bushels, and (c) liters?
36 Table 1-7 shows some old measures of liquid volume. To complete the table, what numbers (to three significant figures) should be entered in (a) the wey column, (b) the chaldron column,
(c) the bag column, (d) the pottle column, and (e) the gill column, starting from the top down? (f) The volume of 1 bag is equal to 0.1091 m3. If an old story has a witch cooking up some vile liquid in a cauldron of volume 1.5 chaldrons, what is the volume in cubic meters?
Table 1-7 Problem 36
wey
chaldron
bag
pottle
gill
1 wey =
1 chaldron =
1 bag =
1 pottle =
1 gill =
i
10/9
40/3
640
120 240
37 A typical sugar cube has an edge length of 1 cm. If you had a cubical box that contained a mole of sugar cubes, what would its edge length be? (One mole = 6.02 X 1023 units.)
38 An old manuscript reveals that a landowner in the time of King Arthur held 3.00 acres of plowed land plus a live¬ stock area of 25.0 perches by 4.00 perches. What was the total area in (a) the old unit of roods and (b) the more modern unit of square meters? Here, 1 acre is an area of 40 perches by 4 perches, 1 rood is an area of 40 perches by 1 perch, and 1 perch is the length 16.5 ft.
39 SSM A tourist purchases a car in England and ships it home to the United States. The car sticker advertised that the car’s fuel con¬ sumption was at the rate of 40 miles per gallon on the open road.
The tourist does not realize that the U.K. gallon differs from the U.S. gallon:
1 U.K. gallon = 4.546 090 0 liters 1 U.S. gallon = 3.785 411 8 liters.
For a trip of 750 miles (in the United States), how many gallons of fuel does (a) the mistaken tourist believe she needs and (b) the car actually require?
40 Using conversions and data in the chapter, determine the number of hydrogen atoms required to obtain 1.0 kg of hydrogen. A hydrogen atom has a mass of 1.0 u.
41 SSM A cord is a volume of cut wood equal to a stack 8 ft long, 4 ft wide, and 4 ft high. How many cords are in 1.0 m3?
42 One molecule of water (H20) contains two atoms of hydrogen and one atom of oxygen. A hydrogen atom has a mass of 1.0 u and an atom of oxygen has a mass of 16 u, approximately, (a) What is the mass in kilograms of one molecule of water? (b) How many mole¬ cules of water are in the world’s oceans, which have an estimated total mass of 1.4 X 1021 kg?
43 A person on a diet might lose 2.3 kg per week. Express the mass loss rate in milligrams per second, as if the dieter could sense the second-by-second loss.
44 What mass of water fell on the town in Problem 7? Water has a density of 1.0 X 103 kg/m3.
45 (a) A unit of time sometimes used in microscopic physics is the shake. One shake equals 10-8 s. Are there more shakes in a second than there are seconds in a year? (b) Humans have ex¬ isted for about 106 years, whereas the universe is about 1010 years old. If the age of the universe is defined as 1 “universe day,” where a universe day consists of “universe seconds” as a normal day consists of normal seconds, how many universe seconds have humans existed?
46 A unit of area often used in measuring land areas is the hectare, defined as 104 m2. An open-pit coal mine consumes 75 hectares of land, down to a depth of 26 m, each year. What volume of earth, in cubic kilometers, is removed in this time?
47 SSM An astronomical unit (AU) is the average distance between Earth and the Sun, approximately 1.50 X 108 km. The speed of light is about 3.0 X 108 m/s. Express the speed of light in astronomical units per minute.
48 The common Eastern mole, a mammal, typically has a mass of 75 g, which corresponds to about 7.5 moles of atoms. (A mole of atoms is 6.02 X 1023 atoms.) In atomic mass units (u), what is the average mass of the atoms in the common Eastern mole?
49 A traditional unit of length in Japan is the ken (1 ken = 1.97 m). What are the ratios of (a) square kens to square meters and (b) cubic kens to cubic meters? What is the volume of a cylin¬ drical water tank of height 5.50 kens and radius 3.00 kens in (c) cu¬ bic kens and (d) cubic meters?
50 You receive orders to sail due east for 24.5 mi to put your sal¬ vage ship directly over a sunken pirate ship. However, when your divers probe the ocean floor at that location and find no evidence of a ship, you radio back to your source of information, only to discover that the sailing distance was supposed to be 24.5 nautical miles, not regular miles. Use the Length table in Appendix D to calculate how far horizontally you are from the pirate ship in kilometers.
1 2
CHAPTER 1 MEASUREMENT
51 The cubit is an ancient unit of length based on the distance between the elbow and the tip of the middle finger of the mea¬ surer. Assume that the distance ranged from 43 to 53 cm, and suppose that ancient drawings indicate that a cylindrical pillar was to have a length of 9 cubits and a diameter of 2 cubits. For the stated range, what are the lower value and the upper value, respectively, for (a) the cylinder’s length in meters, (b) the cylin¬ der’s length in millimeters, and (c) the cylinder’s volume in cubic meters?
52 As a contrast between the old and the modern and between the large and the small, consider the following: In old rural England 1 hide (between 100 and 120 acres) was the area of land needed to sustain one family with a single plough for one year. (An area of 1 acre is equal to 4047 m2.) Also, 1 wapentake was the area of land needed by 100 such families. In quantum physics, the cross-sectional area of a nucleus (defined in terms of the chance of a particle hitting and being absorbed by it) is measured in units of barns, where 1 barn is 1 X 10-28 m2. (In nuclear physics jargon, if a nucleus is “large,” then shooting a particle at it is like shooting a bullet at a barn door, which can hardly be missed.) What is the ratio of 25 wapentakes to 11 barns?
53 SSM An astronomical unit (AU) is equal to the average distance from Earth to the Sun, about 92.9 X 106 mi. A parsec (pc) is the distance at which a length of 1 AU would subtend an angle of exactly 1 second of arc (Fig. 1-8). A light-year (ly) is the distance that light, trav¬ eling through a vacuum with a speed of 186 000 mi/s, would cover in 1.0 year. Express the Earth-Sun distance in (a) parsecs and (b) light-years.
54 The description for a certain brand of house paint claims a cov¬ erage of 460 ft2/gal. (a) Express this quantity in square meters per liter, (b) Express this quantity in an SI unit (see Appendices A and D). (c) What is the inverse of the original quantity, and (d) what is its physical significance?
55 Strangely, the wine for a large wedding reception is to be served in a stunning cut-glass receptacle with the interior dimen¬ sions of 40 cm X 40 cm X 30 cm (height). The receptacle is to be initially filled to the top. The wine can be purchased in bottles of the sizes given in the following table. Purchasing a larger bottle in¬ stead of multiple smaller bottles decreases the overall cost of the wine. To minimize the cost, (a) which bottle sizes should be pur¬ chased and how many of each should be purchased and, once the receptacle is filled, how much wine is left over in terms of (b) stan¬ dard bottles and (c) liters?
1 standard bottle 1 magnum = 2 standard bottles 1 jeroboam = 4 standard bottles 1 rehoboam = 6 standard bottles 1 methuselah = 8 standard bottles 1 Salmanazar = 12 standard bottles 1 balthazar = 16 standard bottles = 11.356 L 1 nebuchadnezzar = 20 standard bottles
56 The corn-hog ratio is a financial term used in the pig market and presumably is related to the cost of feeding a pig until it is large enough for market, ft is defined as the ratio of the market price of a pig with a mass of 3.108 slugs to the market price of a U.S. bushel of corn. (The word “slug” is derived from an old German word that means “to hit”; we have the same meaning for “slug” as a verb in modern English.) A U.S. bushel is equal to 35.238 L. If the corn-hog ratio is listed as 5.7 on the market ex¬ change, what is it in the metric units of
price of 1 kilogram of pig price of 1 liter of corn
(Hint: See the Mass table in Appendix D.)
57 You are to fix dinners for 400 people at a convention of Mexican food fans. Your recipe calls for 2 jalapeno peppers per serving (one serving per person). However, you have only ha- banero peppers on hand. The spiciness of peppers is measured in terms of the scoville heat unit (SHU). On average, one jalapeno pepper has a spiciness of 4000 SHU and one habanero pepper has a spiciness of 300 000 SHU. To get the desired spiciness, how many habanero peppers should you substitute for the jalapeno peppers in the recipe for the 400 dinners?
58 A standard interior staircase has steps each with a rise (height) of 19 cm and a run (horizontal depth) of 23 cm. Research suggests that the stairs would be safer for descent if the run were, instead, 28 cm. For a particular staircase of total height 4.57 m, how much farther into the room would the staircase extend if this change in run were made?
59 In purchasing food for a political rally, you erroneously order shucked medium-size Pacific oysters (which come 8 to 12 per U.S. pint) instead of shucked medium-size Atlantic oysters (which come 26 to 38 per U.S. pint). The filled oyster container shipped to you has the interior measure of 1.0 m X 12 cm X 20 cm, and a U.S. pint is equivalent to 0.4732 liter. By how many oysters is the order short of your anticipated count?
60 An old English cookbook carries this recipe for cream of net¬ tle soup: "Boil stock of the following amount: 1 breakfastcup plus 1 teacup plus 6 tablespoons plus 1 dessertspoon. Using gloves, separate nettle tops until you have 0.5 quart; add the tops to the boiling stock. Add 1 tablespoon of cooked rice and 1 saltspoon of salt. Simmer for 15 min.” The following table gives some of the conversions among old (premetric) British measures and among common (still premetric) U.S. measures. (These measures just scream for metrication.) For liquid measures, 1 British teaspoon = 1 U.S. teaspoon. For dry measures, 1 British teaspoon = 2 U.S. tea¬ spoons and 1 British quart = 1 U.S. quart. In U.S. measures, how much (a) stock, (b) nettle tops, (c) rice, and (d) salt are required in the recipe?
Old British Measures
U.S. Measures
teaspoon = 2 saltspoons dessertspoon = 2 teaspoons tablespoon = 2 dessertspoons teacup = 8 tablespoons breakfastcup = 2 teacups
tablespoon = 3 teaspoons half cup = 8 tablespoons cup = 2 half cups
An angle of exactly 1 second
1 pc -
1 pc
Figure 1-8 Problem 53.
I ,
AU
c
2
Motion Along a Straight Line
2-1 POSITION, DISPLACEMENT, AND AVERAGE VELOCITY
Learning Objectives _
After reading this module, you should be able to ...
2.01 Identify that if all parts of an object move in the same di¬ rection and at the same rate, we can treat the object as if it were a (point-like) particle. (This chapter is about the mo¬ tion of such objects.)
2.02 Identify that the position of a particle is its location as read on a scaled axis, such as an x axis.
2.03 Apply the relationship between a particle's displacement and its initial and final positions.
Key Ideas _
• The position x of a particle on an x axis locates the particle with respect to the origin, or zero point, of the axis.
• The position is either positive or negative, according to which side of the origin the particle is on, or zero if the particle is at the origin. The positive direction on an axis is the direction of increasing positive numbers; the opposite direction is the negative direction on
the axis.
• The displacement Ax of a particle is the change in its position:
Ax = x2 — Xi.
• Displacement is a vector quantity. It is positive if the particle has moved in the positive direction of the x axis and negative if the particle has moved in the negative direction.
2.04 Apply the relationship between a particle's average velocity, its displacement, and the time interval for that displacement.
2.05 Apply the relationship between a particle's average speed, the total distance it moves, and the time interval for the motion.
2.06 Given a graph of a particle's position versus time,
determine the average velocity between any two particular times.
• When a particle has moved from position x1 to position x2 during a time interval At = t2 — th its average velocity during that interval is
Ax x2 — x1
Vavg = "IT" = t2-tl'
• The algebraic sign of vavg indicates the direction of motion (vavg is a vector quantity). Average velocity does not depend on the actual distance a particle moves, but instead depends on its original and final positions.
• On a graph of x versus f, the average velocity for a time in¬ terval At is the slope of the straight line connecting the points on the curve that represent the two ends of the interval.
• The average speed savg of a particle during a time interval At depends on the total distance the particle moves in that time interval:
total distance
What Is Physics?
One purpose of physics is to study the motion of objects — how fast they move, for example, and how far they move in a given amount of time. NASCAR engineers are fanatical about this aspect of physics as they determine the performance of their cars before and during a race. Geologists use this physics to measure tectonic-plate motion as they attempt to predict earthquakes. Medical researchers need this physics to map the blood flow through a patient when diagnosing a partially closed artery, and motorists use it to determine how they might slow sufficiently when their radar detector sounds a warning. There are countless other examples. In this chapter, we study the basic physics of motion where the object (race car, tectonic plate, blood cell, or any other object) moves along a single axis. Such motion is called one-dimensional motion.
13
14
CHAPTER 2 MOTION ALONG A STRAIGHT LINE
Motion
The world, and everything in it, moves. Even seemingly stationary things, such as a roadway, move with Earth’s rotation. Earth’s orbit around the Sun, the Sun’s orbit around the center of the Milky Way galaxy, and that galaxy’s migration relative to other galaxies. The classification and comparison of motions (called kinematics) is often challenging. What exactly do you measure, and how do you compare?
Before we attempt an answer, we shall examine some general properties of motion that is restricted in three ways.
1. The motion is along a straight line only. The line may be vertical, horizontal, or slanted, but it must be straight.
2. Forces (pushes and pulls) cause motion but will not be discussed until Chapter 5. In this chapter we discuss only the motion itself and changes in the motion. Does the moving object speed up, slow down, stop, or reverse direction? If the motion does change, how is time involved in the change?
3. The moving object is either a particle (by which we mean a point-like object such as an electron) or an object that moves like a particle (such that every portion moves in the same direction and at the same rate). A stiff pig slipping down a straight playground slide might be considered to be moving like a par¬ ticle; however, a tumbling tumbleweed would not.
Positive direction
Negative direction
-2 -1 0 Origin^
1
■ x (m)
Figure 2-1 Position is determined on an axis that is marked in units of length (here meters) and that extends indefinitely in opposite directions. The axis name, here x, is always on the positive side of the origin.
Position and Displacement
To locate an object means to find its position relative to some reference point, of¬ ten the origin (or zero point) of an axis such as the x axis in Fig. 2-1. The positive direction of the axis is in the direction of increasing numbers (coordinates), which is to the right in Fig. 2-1. The opposite is the negative direction.
For example, a particle might be located at i = 5 m, which means it is 5 m in the positive direction from the origin. If it were at x = — 5 m, it would be just as far from the origin but in the opposite direction. On the axis, a coordinate of — 5 m is less than a coordinate of — 1 m, and both coordinates are less than a coordinate of +5 m. A plus sign for a coordinate need not be shown, but a minus sign must always be shown.
A change from position x1 to position x2 is called a displacement Ax, where
Ax = x2 — x2.
(2-1)
(The symbol A, the Greek uppercase delta, represents a change in a quantity, and it means the final value of that quantity minus the initial value.) When numbers are inserted for the position values x1 and x2 in Eq. 2-1, a displacement in the positive direction (to the right in Fig. 2-1) always comes out positive, and a displacement in the opposite direction (left in the figure) always comes out negative. For example, if the particle moves from x1 = 5 m to x2 = 12 m, then the displacement is Ax = (12 m) — (5 m) = +7 m. The positive result indicates that the motion is in the positive direction. If, instead, the particle moves from r1 = 5mtox2 = lm, then Ax = (1 m) — (5 m) = —4 m. The negative result in¬ dicates that the motion is in the negative direction.
The actual number of meters covered for a trip is irrelevant; displacement in¬ volves only the original and final positions. For example, if the particle moves from x = 5 m out to x = 200 m and then back to x = 5 m, the displacement from start to finish is Ax = (5 m) — (5 m) = 0.
Signs. A plus sign for a displacement need not be shown, but a minus sign must always be shown. If we ignore the sign (and thus the direction) of a displace¬ ment, we are left with the magnitude (or absolute value) of the displacement. For example, a displacement of Ax = —4m has a magnitude of 4 m.
2-1 POSITION, DISPLACEMENT, AND AVERAGE VELOCITY
15
Figure 2-2 The graph of x(t) for an armadillo that is stationary at x — —2 m. The value of x is —2m for all times t.
This is a graph
t (s)
Displacement is an example of a vector quantity, which is a quantity that has both a direction and a magnitude. We explore vectors more fully in Chapter 3, but here all we need is the idea that displacement has two features: (f) Its magnitude is the distance (such as the number of meters) between the original and final po¬ sitions. (2) Its direction , from an original position to a final position, can be repre¬ sented by a plus sign or a minus sign if the motion is along a single axis.
Here is the first of many checkpoints where you can check your understanding with a bit of reasoning. The answers are in the back of the book.
Checkpoint 1
Here are three pairs of initial and final positions, respectively, along an x axis. Which pairs give a negative displacement: (a) —3 m, +5 m; (b) —3 m, —7 m; (c) 7 m, —3 m?
Average Velocity and Average Speed
A compact way to describe position is with a graph of position x plotted as a func¬ tion of time t — a graph of x(t). (The notation x(t) represents a function x of t, not the product x times t .) As a simple example, Fig. 2-2 shows the position function x(t) for a stationary armadillo (which we treat as a particle) over a 7 s time inter¬ val. The animal's position stays atx = —2 m.
Figure 2-3 is more interesting, because it involves motion. The armadillo is apparently first noticed at t = 0 when it is at the position x = — 5 m. It moves
Os 3s
Figure 2-3 The graph of x(t) for a moving armadillo. The path associated with the graph is also shown, at three times.
16
CHAPTER 2 MOTION ALONG A STRAIGHT LINE
toward x = 0, passes through that point at / = 3 s, and then moves on to increas¬ ingly larger positive values of x. Figure 2-3 also depicts the straight-line motion of the armadillo (at three times) and is something like what you would see. The graph in Fig. 2-3 is more abstract, but it reveals how fast the armadillo moves.
Actually, several quantities are associated with the phrase “how fast.” One of them is the average velocity vavg, which is the ratio of the displacement Ax that occurs during a particular time interval At to that interval:
Ax x2 — x1
Vavg = ~aT= t2-tl-
(2-2)
The notation means that the position is x1 at time t, and then x2 at time t2. A com¬ mon unit for vavg is the meter per second (m/s). You may see other units in the problems, but they are always in the form of length/time.
Graphs. On a graph of x versus t, vavg is the slope of the straight line that connects two particular points on the x{t) curve: one is the point that corresponds to x2 and f2, and the other is the point that corresponds to x2 and f,. Like displace¬ ment, vavg has both magnitude and direction (it is another vector quantity). Its magnitude is the magnitude of the line’s slope. A positive vavg (and slope) tells us that the line slants upward to the right; a negative vavg (and slope) tells us that the line slants downward to the right. The average velocity vavg always has the same sign as the displacement Ax because At in Eq. 2-2 is always positive.
Figure 2-4 shows how to find vavg in Fig. 2-3 for the time interval t = 1 s to t = 4 s. We draw the straight line that connects the point on the position curve at the begin¬ ning of the interval and the point on the curve at the end of the interval. Then we find the slope Ax/A t of the straight line. For the given time interval, the average velocity is
Average speed ,vav„ is a different way of describing “how fast” a particle moves. Whereas the average velocity involves the particle’s displacement Ax, the average speed involves the total distance covered (for example, the number of meters moved), independent of direction; that is,
Tivg
total distance At
(2-3)
Because average speed does not include direction, it lacks any algebraic sign. Sometimes savg is the same (except for the absence of a sign) as vavg. However, the two can be quite different.
Figure 2-4 Calculation of the average velocity between t— Is and t = 4 s as the slope of the line that connects the points on the x{t) curve representing those times. The swirling icon indicates that a figure is available in Wiley PLUS as an animation with voiceover.
x (m)
This is a graph of position x versus time f.
To find average velocity, first draw a straight line, start to end, and then find the slope of the line.
Start of interval
2-1 POSITION, DISPLACEMENT, AND AVERAGE VELOCITY
1 7
Sample Problem 2.01 Average velocity, beat-up pickup truck
You drive a beat-up pickup truck along a straight road for
8.4 km at 70 km/h, at which point the truck runs out of gaso¬ line and stops. Over the next 30 min, you walk another 2.0 km farther along the road to a gasoline station.
(a) What is your overall displacement from the beginning of your drive to your arrival at the station?
KEY IDEA
Assume, for convenience, that you move in the positive di¬ rection of an x axis, from a first position of x1 = 0 to a second position of x2 at the station. That second position must be at x2 = 8.4 km + 2.0 km = 10.4 km. Then your displacement Ax along the x axis is the second position minus the first position.
Calculation: From Eq. 2-1, we have
Ax = x2 - Xi = 10.4 km - 0 = 10.4 km. (Answer)
Thus, your overall displacement is 10.4 km in the positive direction of the x axis.
(b) What is the time interval At from the beginning of your drive to your arrival at the station?
KEY IDEA
We already know the walking time interval Atwlk (= 0.50 h), but we lack the driving time interval Afdr. Flowever, we know that for the drive the displacement A.rdr is 8.4 km and the average velocity vavgdr is 70 km/h. Thus, this average velocity is the ratio of the displacement for the drive to the time interval for the drive.
Calculations: We first write
= Axdl.
Vavg,dr » , ■
Rearranging and substituting data then give us
Atdr
A-vdr
Fivg.dr
8.4 km 70 km/h
0.12 h.
So, At — Atdr + A/wlk
= 0.12 h + 0.50 h = 0.62 h. (Answer)
(c) What is your average velocity vavg from the beginning of your drive to your arrival at the station? Find it both numer¬ ically and graphically.
KEY IDEA
From Eq. 2-2 we know that vavg for the entire trip is the ratio of the displacement of 10.4 km for the entire trip to the time interval of 0.62 h for the entire trip.
Calculation: Here we find
Ax _ 10.4 km
Vavg = "XT = 0.62 h
= 16.8 km/h « 17 km/h. (Answer)
To find vavg graphically, first we graph the function x(t') as shown in Fig. 2-5, where the beginning and arrival points on the graph are the origin and the point labeled as “Station.” Your average velocity is the slope of the straight line connecting those points; that is, vavg is the ratio of the rise (Ax = 10.4 km) to the run (At = 0.62 h), which gives us vavg = 16.8 km/h.
(d) Suppose that to pump the gasoline, pay for it, and walk back to the truck takes you another 45 min. What is your average speed from the beginning of your drive to your return to the truck with the gasoline?
KEY IDEA
Your average speed is the ratio of the total distance you move to the total time interval you take to make that move.
Calculation: The total distance is 8.4 km + 2.0 km + 2.0 km = 12.4 km. The total time interval is 0.12 h + 0.50 h + 0.75 h = 1.37 h.Thus, Eq. 2-3 gives us
12.4 km
■vavg = 1 37h = 9.1 km/h. (Answer)
Figure 2-5 The lines marked "Driving” and “Walking” are the position -time plots for the driving and walking stages. (The plot for the walking stage assumes a constant rate of walking.) The slope of the straight line joining the origin and the point labeled “Station” is the average velocity for the trip, from the beginning to the station.
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18
CHAPTER 2 MOTION ALONG A STRAIGHT LINE
2-2 INSTANTANEOUS VELOCITY AND SPEED
Learning Objectives _
After reading this module, you should be able to . . .
2.07 Given a particle's position as a function of time,
calculate the instantaneous velocity for any particular time.
Key Ideas _
• The instantaneous velocity (or simply velocity) v of a moving partide is dx
v = lim — — = — ,
a? — > o A t dt
where Ax = x2 — x1 and At = t2 — q.
2.08 Given a graph of a particle's position versus time, deter¬ mine the instantaneous velocity for any particular time. 2.09 Identify speed as the magnitude of the instantaneous velocity.
• The instantaneous velocity (at a particular time) may be found as the slope (at that particular time) of the graph of x versus t.
• Speed is the magnitude of instantaneous velocity.
Instantaneous Velocity and Speed
You have now seen two ways to describe how fast something moves: average velocity and average speed, both of which are measured over a time interval At. However, the phrase “how fast” more commonly refers to how fast a particle is moving at a given instant — its instantaneous velocity (or simply velocity) v.
The velocity at any instant is obtained from the average velocity by shrinking the time interval At closer and closer to 0. As At dwindles, the average velocity approaches a limiting value, which is the velocity at that instant:
v
lim Ai
Af — > 0 At
dx
dt
(2-4)
Note that v is the rate at which position x is changing with time at a given instant; that is, v is the derivative of x with respect to t. Also note that v at any instant is the slope of the position -time curve at the point representing that instant. Velocity is another vector quantity and thus has an associated direction.
Speed is the magnitude of velocity; that is, speed is velocity that has been stripped of any indication of direction, either in words or via an algebraic sign. ( Caution : Speed and average speed can be quite different.) A velocity of +5 m/s and one of —5 m/s both have an associated speed of 5 m/s. The speedometer in a car measures speed, not velocity (it cannot determine the direction).
Checkpoint 2
The following equations give the position x(t) of a particle in four situations (in each equa tion, x is in meters, /is in seconds, and f > 0):(l)x = 3 1 — 2;(2)x = —At2 — 2;
(3) x = 2/t2; and (4) x = — 2. (a) In which situation is the velocity v of the particle con¬ stant? (b) In which is v in the negative x direction?
n>
Sample Problem 2.02 Velocity and slope of x versus t, elevator cab
0
Figure 2-6 a is an x(t) plot for an elevator cab that is initially stationary, then moves upward (which we take to be the pos¬ itive direction of x), and then stops. Plot v(t).
KEY IDEA
Calculations: The slope of x(t). and so also the velocity, is zero in the intervals from 0 to 1 s and from 9 s on, so then the cab is stationary. During the interval be, the slope is con¬ stant and nonzero, so then the cab moves with constant ve¬ locity. We calculate the slope of x(f) then as
Ax 24 m — 4.0 m
At
We can find the velocity at any time from the slope of the x(t) curve at that time.
v
8.0 s — 3.0 s
= +4.0 m/s.
(2-5)
2-2 INSTANTANEOUS VELOCITY AND SPEED
19
Figure 2-6 (a) The x(t) curve for an elevator cab that moves upward along an x axis. ( b ) The v(t) curve for the cab. Note that it is the derivative of the x(t) curve (v = dxldt ). (c) The a(t) curve for the cab. It is the derivative of the v(t) curve ( a = dvldt ). The stick figures along the bottom suggest how a passenger’s body might feel dur¬ ing the accelerations.
25
20
o
0* 10
A
a
CM II
11 II
m
.0 S'""
c
d
i
i
i
x{t)
I Ax
1
1
X =
at
4.0 n = 3.0
L
1
1
i
a
b
At
y-
0 1 2 3 4 5 6
Time (s)
(a)
Slopes on the a- versus t graph Slope are the values on the v versus t graph.
of x(t)
i
/
v(«)
/
1
\
\
/
\
a
/
\
-i
0 12 3
4 5 6 7 8 9
Time (s)
Slopes on the v versus t graph are the values on the a versus t graph.
(b)
Acce
erati<
3n /
s
/
a
b
a(t)
c
d
1
(
8
9
D<
:celer
ation
What you would feel.
(c)
The plus sign indicates that the cab is moving in the posi¬ tive x direction. These intervals (where v = 0 and v = 4 m/s) are plotted in Fig. 2-6 b. In addition, as the cab ini¬ tially begins to move and then later slows to a stop, v varies as indicated in the intervals 1 s to 3 s and 8 s to 9 s. Thus, Fig. 2-6 b is the required plot. (Figure 2-6c is consid¬ ered in Module 2-3.)
Given a v(f) graph such as Fig. 2-6 b, we could “work backward” to produce the shape of the associated x(t) graph (Fig. 2-6u). Flowever, we would not know the actual values for x at various times, because the v(f) graph indicates only changes in x. To find such a change in x during any in¬
terval, we must, in the language of calculus, calculate the area “under the curve” on the v(f) graph for that interval. For example, during the interval 3 s to 8 s in which the cab has a velocity of 4.0 m/s, the change in x is
Ax = (4.0 m/s)(8.0 s - 3.0 s) = +20 m. (2-6)
(This area is positive because the v(t) curve is above the t axis.) Figure 2-6a shows that x does indeed increase by 20 m in that interval. Flowever, Fig. 2-6 b does not tell us the values of x at the beginning and end of the interval. For that, we need additional information, such as the value of x at some instant.
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20
CHAPTER 2 MOTION ALONG A STRAIGHT LINE
2-3 ACCELERATION
Learning Objectives _
After reading this module, you should be able to .. .
2.1 0 Apply the relationship between a particle's average acceleration, its change in velocity, and the time interval for that change.
2.1 1 Given a particle's velocity as a function of time, calcu¬ late the instantaneous acceleration for any particular time.
Key Ideas _
• Average acceleration is the ratio of a change in velocity Av
to the time interval At in which the change occurs:
- AL
flavg- Af-
The algebraic sign indicates the direction of navg.
2.1 2 Given a graph of a particle's velocity versus time, deter¬ mine the instantaneous acceleration for any particular time and the average acceleration between any two particular times.
• Instantaneous acceleration (or simply acceleration) a is the first time derivative of velocity v(t) and the second time deriv¬ ative of position x(t ):
dv d2x
dt dt2
• On a graph of v versus t, the acceleration a at any time t is the slope of the curve at the point that represents t.
Acceleration
When a particle’s velocity changes, the particle is said to undergo acceleration (or to accelerate). For motion along an axis, the average acceleration aavg over a time interval Af is
^avg
V2 ~ Vl
h ~ h
Av At ’
(2-7)
where the particle has velocity v1 at time q and then velocity v2 at time f2. The
instantaneous acceleration (or simply acceleration) is
a
dv dt '
(2-8)
In words, the acceleration of a particle at any instant is the rate at which its velocity is changing at that instant. Graphically, the acceleration at any point is the slope of the curve of v(f) at that point. We can combine Eq. 2-8 with Eq. 2-4 to write
dv d ( dx\ d2x
dt dt\ dt ) dt 2
(2-9)
In words, the acceleration of a particle at any instant is the second derivative of its position x(t) with respect to time.
A common unit of acceleration is the meter per second per second: m/(s • s) or m/s2. Other units are in the form of length/(time ■ time) or length/time2. Acceleration has both magnitude and direction (it is yet another vector quan¬ tity). Its algebraic sign represents its direction on an axis just as for displacement and velocity; that is, acceleration with a positive value is in the positive direction of an axis, and acceleration with a negative value is in the negative direction.
Figure 2-6 gives plots of the position, velocity, and acceleration of an ele¬ vator moving up a shaft. Compare the a(t ) curve with the v(f) curve — each point on the a(t ) curve shows the derivative (slope) of the v(f) curve at the corresponding time. When v is constant (at either 0 or 4 m/s), the derivative is zero and so also is the acceleration. When the cab first begins to move, the v(f)
2-3 ACCELERATION
21
curve has a positive derivative (the slope is positive), which means that a{t ) is positive. When the cab slows to a stop, the derivative and slope of the v(f) curve are negative; that is, a(t) is negative.
Next compare the slopes of the v(f) curve during the two acceleration peri¬ ods. The slope associated with the cab's slowing down (commonly called “decel¬ eration”) is steeper because the cab stops in half the time it took to get up to speed. The steeper slope means that the magnitude of the deceleration is larger than that of the acceleration, as indicated in Fig. 2-6c.
Sensations. The sensations you would feel while riding in the cab of Fig. 2-6 are indicated by the sketched figures at the bottom. When the cab first accelerates, you feel as though you are pressed downward; when later the cab is braked to a stop, you seem to be stretched upward. In between, you feel nothing special. In other words, your body reacts to accelerations (it is an accelerometer) but not to velocities (it is not a speedometer). When you are in a car traveling at 90 km/h or an airplane traveling at 900 km/h, you have no bodily awareness of the motion. However, if the car or plane quickly changes velocity, you may be¬ come keenly aware of the change, perhaps even frightened by it. Part of the thrill of an amusement park ride is due to the quick changes of velocity that you un¬ dergo (you pay for the accelerations, not for the speed). A more extreme example is shown in the photographs of Fig. 2-7, which were taken while a rocket sled was rapidly accelerated along a track and then rapidly braked to a stop.
g Units. Large accelerations are sometimes expressed in terms of g units, with
lg = 9.8 m/s2 (g unit). (2-10)
(As we shall discuss in Module 2-5, g is the magnitude of the acceleration of a falling object near Earth’s surface.) On a roller coaster, you may experience brief accelerations up to 3g, which is (3) (9.8 m/s2), or about 29 m/s2, more than enough to justify the cost of the ride.
Signs. In common language, the sign of an acceleration has a nonscientific meaning: positive acceleration means that the speed of an object is increasing, and negative acceleration means that the speed is decreasing (the object is decelerat¬ ing). In this book, however, the sign of an acceleration indicates a direction, not
Figure 2-7
Colonel J. P. Stapp in a rocket sled as it is brought up to high speed (acceleration out of the page) and then very rapidly braked (acceleration into the page).
Courtesy U.S. Air Force
22
CHAPTER 2 MOTION ALONG A STRAIGHT LINE
whether an object’s speed is increasing or decreasing. For example, if a car with an initial velocity v = —25 m/s is braked to a stop in 5.0 s, then aavg = +5.0 m/s1 2. The acceleration is positive, but the car’s speed has decreased. The reason is the differ¬ ence in signs: the direction of the acceleration is opposite that of the velocity.
Here then is the proper way to interpret the signs:
o If the signs of the velocity and acceleration of a particle are the same, the speed of the particle increases. If the signs are opposite, the speed decreases.
Checkpoint 3
A wombat moves along an x axis. What is the sign of its acceleration if it is moving (a) in the positive direction with increasing speed, (b) in the positive direction with decreasing speed, (c) in the negative direction with increasing speed, and (d) in the negative direction with decreasing speed?
Sample Problem 2.03 Acceleration and dv/dt
A particle’s position on the x axis of Fig. 2-1 is given by x = 4 — 27 1 + t3, with x in meters and t in seconds.
(a) Because position x depends on time t, the particle must be moving. Find the particle’s velocity function v(t) and ac¬ celeration function a(t).
KEY IDEAS
(1) To get the velocity function v(f), we differentiate the po¬ sition function x(t) with respect to time. (2) To get the accel¬ eration function a(t), we differentiate the velocity function v(t) with respect to time.
Calculations: Differentiating the position function, we find v = -27 + 3 12, (Answer)
with v in meters per second. Differentiating the velocity function then gives us
a = +6t, (Answer)
with a in meters per second squared.
(b) Is there ever a time when v = 0?
Calculation: Setting v(t) = 0 yields
0 = -27 + 3f2,
which has the solution
t = ±3 s. (Answer)
Thus, the velocity is zero both 3 s before and 3 s after the clock reads 0.
(c) Describe the particle’s motion for t > 0.
Reasoning: We need to examine the expressions for x(t), v(t), and a(t).
At t = 0, the particle is at x(0) = +4 m and is moving with a velocity of v(0) = — 27 m/s — that is, in the negative direction of the x axis. Its acceleration is a( 0) = 0 because just then the particle’s velocity is not changing (Fig. 2-8 a).
For 0 < t < 3 s, the particle still has a negative velocity, so it continues to move in the negative direction. However, its acceleration is no longer 0 but is increasing and positive. Because the signs of the velocity and the acceleration are opposite, the particle must be slowing (Fig. 2-8 b).
Indeed, we already know that it stops momentarily at t = 3 s. Just then the particle is as far to the left of the origin in Fig. 2-1 as it will ever get. Substituting t = 3 s into the expression for x(t), we find that the particle’s position just then is x = — 50 m (Fig. 2-8c). Its acceleration is still positive.
For t > 3 s, the particle moves to the right on the axis. Its acceleration remains positive and grows progressively larger in magnitude. The velocity is now positive, and it too grows progressively larger in magnitude (Fig. 2-8 d).
( b ) (a)
Figure 2-8 Four stages of the particle’s motion.
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2-4 CONSTANT ACCELERATION
23
2-4 CONSTANT ACCELERATION
Learning Objectives _
After reading this module, you should be able to . . . 2.14 Calculate a particle's change in velocity by integrating
2.13 For constant acceleration, apply the relationships be- 'ts acceleration function with respect to time.
tween position, displacement, velocity, acceleration, and J Calculate a particle s change in position by integrating
elapsed time (Table 2-1 ) its velocity function with respect to time.
Key Ideas _
• The following five equations describe the motion of a particle with constant acceleration:
1 ,
v = v0 + at, x — x0 = v0t + yor,
1 1
V2 = vl + 2 a(x - x0), x Xq — y(v0 + v)t, x - x0 = vt - —at2.
These are not valid when the acceleration is not constant.
Constant Acceleration: A Special Case
In many types of motion, the acceleration is either constant or approximately so. For example, you might accelerate a car at an approximately constant rate when a traffic light turns from red to green. Then graphs of your position, velocity, and acceleration would resemble those in Fig. 2-9. (Note that a(t) in Fig. 2-9c is constant, which requires that v(f) in Fig. 2-9 b have a constant slope.) Later when you brake the car to a stop, the acceleration (or deceleration in common language) might also be approximately constant.
Such cases are so common that a special set of equations has been derived for dealing with them. One approach to the derivation of these equations is given in this section. A second approach is given in the next section. Throughout both sections and later when you work on the homework problems, keep in mind that these equations are valid only for constant acceleration (or situations in which you can approximate the acceleration as being constant).
First Basic Equation. When the acceleration is constant, the average accel¬ eration and instantaneous acceleration are equal and we can write Eq. 2-7, with some changes in notation, as
= v ~ Vp
a flavg t _ 0 ■
Flere v0 is the velocity at time t = 0 and v is the velocity at any later time t. We can recast this equation as
X
Slopes of the position graph are plotted on the velocity graph.
V
Slope of the velocity graph is plotted on the acceleration graph.
v = v0 + at.
(2-11)
As a check, note that this equation reduces to v = v0 for t = 0, as it must. As a fur¬ ther check, take the derivative of Eq. 2-11. Doing so yields dv/dt = a , which is the definition of a. Figure 2-9 b shows a plot of Eq. 2-11, the v(t) function; the function is linear and thus the plot is a straight line.
Second Basic Equation. In a similar manner, we can rewrite Eq. 2-2 (with a few changes in notation) as
X — Xq
Vavg = t-0
_ q(t)
Slope = 0
- 1
Figure 2-9 («) The position x(t) of a particle moving with constant acceleration, (b) Its velocity v(f), given at each point by the slope of the curve of x(t). (c) Its (constant) acceleration, equal to the (constant) slope of the curve of v(f).
(e)
24
CHAPTER 2 MOTION ALONG A STRAIGHT LINE
and then as
x = x0 + vavgi, (2-12)
in which x0 is the position of the particle at t = 0 and vavg is the average velocity between t = 0 and a later time t.
For the linear velocity function in Eq. 2-11, the average velocity over any time interval (say, from t = 0 to a later time t) is the average of the velocity at the be¬ ginning of the interval (= v0) and the velocity at the end of the interval (= v). For the interval from t = 0 to the later time t then, the average velocity is
Vavg = \{VQ + V). (2-13)
Substituting the right side of Eq. 2-11 for v yields, after a little rearrangement,
Vavg = V0 + 2 at. (2-14)
Finally, substituting Eq. 2-14 into Eq. 2-12 yields
jc — x0 = v0t + \at2. (2-15)
As a check, note that putting t = 0 yields x = jc0, as it must. As a further check, taking the derivative of Eq. 2-15 yields Eq. 2-11, again as it must. Figure 2-9 a shows a plot of Eq. 2-15; the function is quadratic and thus the plot is curved.
Three Other Equations. Equations 2-11 and 2-15 are the basic equations for constant acceleration; they can be used to solve any constant acceleration prob¬ lem in this book. However, we can derive other equations that might prove useful in certain specific situations. First, note that as many as five quantities can possi¬ bly be involved in any problem about constant acceleration — namely, x — x0, v, t, a, and v0. Usually, one of these quantities is not involved in the problem, either as a given or as an unknown. We are then presented with three of the remaining quantities and asked to find the fourth.
Equations 2-11 and 2-15 each contain four of these quantities, but not the same four. In Eq. 2-11, the “missing ingredient” is the displacement x — x0. In Eq. 2-15, it is the velocity v. These two equations can also be combined in three ways to yield three additional equations, each of which involves a different “missing variable.” First, we can eliminate t to obtain
v2 = Vq + 2 a(x — x0). (2-16)
This equation is useful if we do not know t and are not required to find it. Second, we can eliminate the acceleration a between Eqs. 2-11 and 2-15 to produce an equation in which a does not appear:
Table 2-1 Equations for Motion with Constant Acceleration"
Equation
Number
Equation
Missing
Quantity
2-11
v — v0 + at
X — x0
2-15
x — x0 = vQt + \at2
V
2-16
v2 = Vq + 2 a(x — x0)
t
2-17
X - x0 = \(v0 + v)t
a
2-18
X
1
*
o
II
1
tOlK-*
a
"7c
Vo
“Make sure that the acceleration is indeed constant before using the equations in this table.
X - x0 = |(v0 + v)t.
(2-17)
Finally, we can eliminate v0, obtaining
x — xQ = vt — \at 2.
(2-18)
Note the subtle difference between this equation and Eq. 2-15. One involves the initial velocity v0; the other involves the velocity v at time t.
Table 2-1 lists the basic constant acceleration equations (Eqs. 2-11 and 2-15) as well as the specialized equations that we have derived. To solve a simple constant ac¬ celeration problem, you can usually use an equation from this list (if you have the list with you). Choose an equation for which the only unknown variable is the vari¬ able requested in the problem. A simpler plan is to remember only Eqs. 2-11 and 2-15, and then solve them as simultaneous equations whenever needed.
2-4 CONSTANT ACCELERATION
25
Checkpoint 4
The following equations give the position x(t) of a particle in four situations: (1) x = 3f — 4; (2) x = —5 13 + At2 + 6; (3) x = 2/t2 — 4/t; (4) x = 5 12 — 3. To which of these situations do the equations of Table 2-1 apply?
Sample Problem 2.04 Drag race of car and motorcycle
A popular web video shows a jet airplane, a car, and a mo¬ torcycle racing from rest along a runway (Fig. 2-10). Initially the motorcycle takes the lead, but then the jet takes the lead, and finally the car blows past the motorcycle. Flere let’s focus on the car and motorcycle and assign some reasonable values to the motion. The motorcycle first takes the lead because its (constant) acceleration am = 8.40 m/s2 is greater than the car’s (constant) acceleration ac = 5.60 m/s2, but it soon loses to the car because it reaches its greatest speed vm = 58.8 m/s before the car reaches its greatest speed vc = 106 m/s. Flow long does the car take to reach the motorcycle?
KEY IDEAS
We can apply the equations of constant acceleration to both vehicles, but for the motorcycle we must consider the mo¬ tion in two stages: (1) First it travels through distance xm\ with zero initial velocity and acceleration am = 8.40 m/s2, reaching speed vm = 58.8 m/s. (2) Then it travels through dis¬ tance xm2 with constant velocity vm = 58.8 m/s and zero ac¬ celeration (that, too, is a constant acceleration). (Note that we symbolized the distances even though we do not know their values. Symbolizing unknown quantities is often help¬ ful in solving physics problems, but introducing such un¬ knowns sometimes takes physics courage.)
Calculations: So that we can draw figures and do calcula¬ tions, let’s assume that the vehicles race along the positive di¬ rection of an x axis, starting from x = 0 at time t = 0. (We can
choose any initial numbers because we are looking for the elapsed time, not a particular time in, say, the afternoon, but let’s stick with these easy numbers.) We want the car to pass the motorcycle, but what does that mean mathematically?
It means that at some time t, the side-by-side vehicles are at the same coordinate: xc for the car and the sum xml + xm2 for the motorcycle. We can write this statement mathe¬ matically as
= xml + xm2. (2-19)
(Writing this first step is the hardest part of the problem. That is true of most physics problems. Flow do you go from the problem statement (in words) to a mathematical expres¬ sion? One purpose of this book is for you to build up that ability of writing the first step — it takes lots of practice just as in learning, say, tae-kwon-do.)
Now let’s fill out both sides of Eq. 2-19, left side first. To reach the passing point at xc, the car accelerates from rest. From Eq. 2-15 (x — x0 = vQt + \at2), with xQ and v0 = 0, we have
xc = \act 2. (2-20)
To write an expression for xml for the motorcycle, we first find the time tm it takes to reach its maximum speed vm, using Eq. 2-11 (v = v0 + at). Substituting v0 = 0, v = vm = 58.8 m/s, and a = am = 8.40 m/s2, that time is
58.8 m/s 8.40 m/s2
7.00 s.
(2-21)
To get the distance xml traveled by the motorcycle during the first stage, we again use Eq. 2-15 with x0 = 0 and v0 = 0, but we also substitute from Eq. 2-21 for the time. We find
(2-22)
For the remaining time of t — tm, the motorcycle travels at its maximum speed with zero acceleration. To get the distance, we use Eq. 2-15 for this second stage of the motion, but now the initial velocity is v0 = vm (the speed at the end of the first stage) and the acceleration is a = 0. So, the dis¬ tance traveled during the second stage is
Xml 2®mtm
2
Figure 2-10 A jet airplane, a car, and a motorcycle just after accelerating from rest.
Xm2 = vjt - tm) = vjt - 7.00 s).
(2-23)
26
CHAPTER 2 MOTION ALONG A STRAIGHT LINE
To finish the calculation, we substitute Eqs. 2-20, 2-22, and 2-23 into Eq. 2-19, obtaining
1 v2
\act2 = — — + vm(t - 7.00 s). (2-24)
l am
This is a quadratic equation. Substituting in the given data, we solve the equation (by using the usual quadratic-equa¬ tion formula or a polynomial solver on a calculator), finding t = 4.44 s and t = 16.6 s.
But what do we do with two answers? Does the car pass the motorcycle twice? No, of course not, as we can see in the video. So, one of the answers is mathematically correct but not physically meaningful. Because we know that the car passes the motorcycle after the motorcycle reaches its maxi¬ mum speed at t = 7.00 s, we discard the solution with t < 7.00 s as being the unphysical answer and conclude that the passing occurs at
t = 16.6 s. (Answer)
Figure 2-11 is a graph of the position versus time for the two vehicles, with the passing point marked. Notice
that at t = 7.00 s the plot for the motorcycle switches from being curved (because the speed had been increasing) to be¬ ing straight (because the speed is thereafter constant).
0 5 10 15 20
t( s)
Figure 2-11 Graph of position versus time for car and motorcycle.
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Another Look at Constant Acceleration*
The first two equations in Table 2-1 are the basic equations from which the others are derived. Those two can be obtained by integration of the acceleration with the condition that a is constant. To find Eq. 2-11, we rewrite the definition of ac¬ celeration (Eq. 2-8) as
dv — a dt.
We next write the indefinite integral (or antiderivative) of both sides:
Since acceleration a is a constant, it can be taken outside the integration. We obtain
or v = at + C. (2-25)
To evaluate the constant of integration C, we let / = 0, at which time v = v0. Substituting these values into Eq. 2-25 (which must hold for all values of t, including t = 0) yields
v0 = (n)(0) + C = C.
Substituting this into Eq. 2-25 gives us Eq. 2-11.
To derive Eq. 2-15, we rewrite the definition of velocity (Eq. 2-4) as
dx = v dt
and then take the indefinite integral of both sides to obtain
*This section is intended for students who have had integral calculus.
2-5 FREE-FALL ACCELERATION
27
Next, we substitute for v with Eq. 2-11:
^ dx = J (vq + at) dt.
Since v0 is a constant, as is the acceleration a , this can be rewritten as
dx = v0 \ dt + a f dt.
Integration now yields
x = v0t + \at 2 + C',
(2-26)
where C' is another constant of integration. At time f = 0, we have x = x0. Substituting these values in Eq. 2-26 yields x0= C' . Replacing C' with x0 in Eq. 2-26 gives us Eq. 2-15.
2-5 FREE-FALL ACCELERATION
Learning Objectives _
After reading this module, you should be able to . . .
2.16 Identify that if a particle is in free flight (whether upward or downward) and if we can neglect the effects of air on its motion, the particle has a constant
Key Ideas _
• An important example of straight-line motion with constant acceleration is that of an object rising or falling freely near Earth's surface. The constant acceleration equations de¬ scribe this motion, but we make two changes in notation:
downward acceleration with a magnitude g that we take to be 9.8 m/s2.
2.17 Apply the constant-acceleration equations (Table 2-1) to free-fall motion.
(1 ) we refer the motion to the vertical y axis with +y vertically up; (2) we replace a with —g, where g is the magnitude of the free-fall acceleration. Near Earth’s surface,
g = 9.8 m/s2 = 32 ft/s2.
Free-Fall Acceleration
If you tossed an object either up or down and could somehow eliminate the effects of air on its flight, you would find that the object accelerates downward at a certain constant rate. That rate is called the free-fall acceleration, and its magni¬ tude is represented by g. The acceleration is independent of the object’s charac¬ teristics, such as mass, density, or shape; it is the same for all objects.
Two examples of free-fall acceleration are shown in Fig. 2-12, which is a series of stroboscopic photos of a feather and an apple. As these objects fall, they accelerate downward — both at the same rate g. Thus, their speeds increase at the same rate, and they fall together.
The value of g varies slightly with latitude and with elevation. At sea level in Earth’s midlatitudes the value is 9.8 m/s2 (or 32 ft/s2), which is what you should use as an exact number for the problems in this book unless otherwise noted.
The equations of motion in Table 2-1 for constant acceleration also apply to free fall near Earth's surface; that is, they apply to an object in vertical flight, either up or down, when the effects of the air can be neglected. However, note that for free fall: (1) The directions of motion are now along a vertical y axis instead of the x axis, with the positive direction of y upward. (This is important for later chapters when combined horizontal and vertical motions are examined.) (2) The free-fall acceleration is negative — that is, downward on the y axis, toward Earth’s center — and so it has the value — g in the equations.
© Jim Sugar/CORBIS
Figure 2-12 A feather and an apple free fall in vacuum at the same magnitude of acceleration g. The acceleration increases the distance between successive images. In the absence of air, the feather and apple fall together.
28
CHAPTER 2 MOTION ALONG A STRAIGHT LINE
© The free-fall acceleration near Earth’s surface is a = — g = —9.8 m/s2, and the magnitude of the acceleration is g = 9.8 m/s2. Do not substitute —9.8 m/s2 for g.
Suppose you toss a tomato directly upward with an initial (positive) velocity v0 and then catch it when it returns to the release level. During its free-fall fligh t (from just after its release to just before it is caught), the equations of Table 2-1 apply to its motion. The acceleration is always a = —g = —9.8 m/s2, negative and thus down¬ ward. The velocity, however, changes, as indicated by Eqs. 2-11 and 2-16: during the ascent, the magnitude of the positive velocity decreases, until it momentarily be¬ comes zero. Because the tomato has then stopped, it is at its maximum height. During the descent, the magnitude of the (now negative) velocity increases.
Checkpoint 5
(a) If you toss a ball straight up, what is the sign of the ball’s displacement for the ascent, from the release point to the highest point? (b) What is it for the descent, from the high¬ est point back to the release point? (c) What is the ball’s acceleration at its highest point?
Sample Problem 2.05 Time for full up-down flight, baseball toss
In Fig. 2-13, a pitcher tosses a baseball up along a y axis, with an initial speed of 12 m/s.
(a) How long does the ball take to reach its maximum height?
v = 0 at highest point
y
KEY IDEAS
(1) Once the ball leaves the pitcher and before it returns to his hand, its acceleration is the free-fall acceleration a = —g. Because this is constant. Table 2-1 applies to the motion.
(2) The velocity v at the maximum height must be 0.
Calculation: Knowing v, a, and the initial velocity v0 = 12 m/s, and seeking t, we solve Eq. 2-11, which contains those four variables. This yields
t =
v ~ Vq
a
0 — 12 m/s -9.8 m/s2
(Answer)
(b) What is the ball's maximum height above its release point?
Figure 2-13 A pitcher tosses a baseball straight up into the air. The equations of free fall apply for rising as well as for falling objects, provided any effects from the air can be neglected.
During ascent, a = ~g>
speed decreases, and velocity becomes less positive
During
descent,
a = ~g>
speed
increases,
and velocity
becomes
more
negative
Calculation: We can take the ball’s release point to be y0 = 0. We can then write Eq. 2-16 in y notation, set y — y0 = y and v = 0 (at the maximum height) , and solve for y. We get
V2 - Vq = 0 ~ (12 m/s)2 2 a 2(— 9.8 m/s2)
(Answer)
(c) How long does the ball take to reach a point 5.0 m above its release point?
or 5.0 m = (12 m/s)f - (|)(9.8 m/s2)t2.
If we temporarily omit the units (having noted that they are consistent), we can rewrite this as
4.9 f2 - 12 1 + 5.0 = 0.
Solving this quadratic equation for t yields
Calculations: We know v0, a = —g, and displacement y — y0 = 5.0 m, and we want /, so we choose Eq. 2-15. Rewriting it for y and setting y0 = 0 give us
\st\
t = 0.53 s and t = 1.9 s.
(Answer)
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y = vo{ 2
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There are two such times! This is not really surprising because the ball passes twice through y = 5.0 m, once on the way up and once on the way down.
2-6 GRAPHICAL INTEGRATION IN MOTION ANALYSIS
29
2-6 GRAPHICAL INTEGRATION IN MOTION ANALYSIS
Learning Objectives _
After reading this module, you should be able to . . .
2.18 Determine a particle's change in velocity by graphical integration on a graph of acceleration versus time.
Key Ideas _
• On a graph of acceleration a versus time f, the change in the velocity is given by
P1
Vi — v0 = I a dt.
0
The integral amounts to finding an area on the graph:
/ area between acceleration curve y and time axis, from t0 to f.
2.19 Determine a particle’s change in position by graphical integration on a graph of velocity versus time.
• On a graph of velocity v versus time t, the change in the position is given by
P1
x1 — x0 = I v dt,
Jt0
where the integral can be taken from the graph as
( area between velocity curve y and time axis, from f0 to f.
Graphical Integration in Motion Analysis
Integrating Acceleration. When we have a graph of an object’s acceleration a ver¬ sus time t, we can integrate on the graph to find the velocity at any given time. Because a is defined as a = dvldt, the Fundamental Theorem of Calculus tells us that
Vl
v0 = I a dt.
Jt„
(2-27)
The right side of the equation is a definite integral (it gives a numerical result rather than a function), v0 is the velocity at time t0, and v, is the velocity at later time q.The def¬ inite integral can be evaluated from an a(t ) graph, such as in Fig. 2-1 4a. In particular,
H dt I area between acceleration curve) y and time axis, from t0 to q )'
(2-28)
If a unit of acceleration is 1 m/s2 and a unit of time is 1 s, then the correspon¬ ding unit of area on the graph is
(1 m/s2)(l s) = 1 m/s,
which is (properly) a unit of velocity. When the acceleration curve is above the time axis, the area is positive; when the curve is below the time axis, the area is negative.
Integrating Velocity. Similarly, because velocity v is defined in terms of the posi¬ tion x as v = dxldt , then
x1 — x0 = I v dt, (2-29)
Jk i
where x0 is the position at time t0 and x ^ is the position at time t1. The definite integral on the right side of Eq. 2-29 can be evaluated from a v(f) graph, like that shown in Fig. 2-14 b. In particular.
1 dt =
( area between velocity curve ] l and time axis, from t0 to q /’
(2-30)
If the unit of velocity is 1 m/s and the unit of time is 1 s, then the corre¬ sponding unit of area on the graph is
(1 m/s)(l s) = 1 m,
which is (properly) a unit of position and displacement. Whether this area is posi¬ tive or negative is determined as described for the a{t) curve of Fig. 2-14a.
Area
(b)
This area gives the change in velocity.
This area gives the change in position.
Figure 2-14 The area between a plotted curve and the horizontal time axis, from time t0 to time r, , is indicated for (a) a graph of acceleration a versus t and (b) a graph of velocity v versus t.
30
CHAPTER 2 MOTION ALONG A STRAIGHT LINE
Sample Problem 2.06 Graphical integration a versus t, whiplash injury
“Whiplash injury” commonly occurs in a rear-end collision where a front car is hit from behind by a second car. In the 1970s, researchers concluded that the injury was due to the occupant’s head being whipped back over the top of the seat as the car was slammed forward. As a result of this finding, head restraints were built into cars, yet neck injuries in rear- end collisions continued to occur.
In a recent test to study neck injury in rear-end collisions, a volunteer was strapped to a seat that was then moved abruptly to simulate a collision by a rear car moving at 10.5 km/h. Figure 2- 1 5a gives the accelerations of the volun¬ teer’s torso and head during the collision, which began at time t = 0. The torso acceleration was delayed by 40 ms because during that time interval the seat back had to compress against the volunteer. The head acceleration was delayed by an additional 70 ms. What was the torso speed when the head began to accelerate? tsS0P
KEY IDEA
We can calculate the torso speed at any time by finding an area on the torso a(t) graph.
Calculations: We know that the initial torso speed is v0 = 0 at time t0 = 0, at the start of the “collision.” We want the torso speed v1 at time q = 110 ms, which is when the head begins to accelerate.
Combining Eqs. 2-27 and 2-28, we can write
/ area between acceleration curve \
Vl v° y and time axis, from f0 to q /' '
For convenience, let us separate the area into three regions (Fig. 2-15 b). From 0 to 40 ms, region A has no area:
area^ = 0.
From 40 ms to 100 ms, region B has the shape of a triangle, with area
areaB = ^(0.060 s)(50 m/s2) = 1.5 m/s.
From 100 ms to 110 ms, region C has the shape of a rectan¬ gle, with area
areac = (0.010 s)(50 m/s2) = 0.50 m/s. Substituting these values and v0 = 0 into Eq. 2-31 gives us Vi — 0 = 0 + 1.5 m/s + 0.50 m/s, or = 2.0 m/s = 7.2 km/h. (Answer)
Comments: When the head is just starting to move forward, the torso already has a speed of 7.2 km/h. Researchers argue that it is this difference in speeds during the early stage of a rear-end collision that injures the neck. The backward whip¬ ping of the head happens later and could, especially if there is no head restraint, increase the injury.
(«)
XwiLEYO
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40
80
t (ms)
120
160
(*)
The total area gives the change in velocity.
Figure 2-15 (a) The a(t) curve of the torso and head of a volunteer in a simulation of a rear-end collision. ( b ) Breaking up the region between the plotted curve and the time axis to calculate the area.
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eview & Summary
Position The position x of a particle on an x axis locates the par¬ ticle with respect to the origin, or zero point, of the axis. The position is either positive or negative, according to which side of the origin the particle is on, or zero if the particle is at the origin. The positive direction on an axis is the direction of increasing positive numbers; the opposite direction is the negative direction on the axis.
Average Velocity When a particle has moved from position x1 to position x2 during a time interval A t — t2~ q, its average velocity during that interval is
Ax _ x2 — x1
Vavg = ~Ki~ = t2 - q ■
(2-2)
Displacement The displacement Ax of a particle is the change in its position:
Ax = x2 — X\. (2-1)
Displacement is a vector quantity. It is positive if the particle has moved in the positive direction of the x axis and negative if the particle has moved in the negative direction.
The algebraic sign of vavg indicates the direction of motion (vavg is a vector quantity). Average velocity does not depend on the actual distance a particle moves, but instead depends on its original and final positions.
On a graph of x versus ?, the average velocity for a time interval At is the slope of the straight line connecting the points on the curve that represent the two ends of the interval.
QUESTIONS
31
Average Speed The average speed savg of a particle during a time interval A t depends on the total distance the particle moves in that time interval:
total distance
*avg = - 77 - • (2-3)
Instantaneous Velocity The instantaneous velocity (or sim¬ ply velocity) v of a moving particle is
v
lim^L
Af — » 0 A t
dx ~dt ’
(2-4)
where Ax and At are defined by Eq. 2-2. The instantaneous velocity (at a particular time) may be found as the slope (at that particular time) of the graph of x versus t. Speed is the magnitude of instanta¬ neous velocity.
and the second time derivative of position x(t):
dv d2x dt dt2 "
(2-8, 2-9)
On a graph of v versus t, the acceleration a at any time t is the slope of the curve at the point that represents t.
Constant Acceleration The five equations in Table 2-1 describe the motion of a particle with constant acceleration:
v = v0 + at.
(2-11)
+
II
O
*
1
*
(2-15)
v2 = Vq + 2 a(x — x0),
(2-16)
* - *0 = \(va + v)f,
(2-17)
x x0 — vt \at2.
(2-18)
Average Acceleration Average acceleration is the ratio of a change in velocity Av to the time interval At in which the change occurs:
Av
aavg — ^ ' (2-7)
The algebraic sign indicates the direction of aavg.
Instantaneous Acceleration Instantaneous acceleration (or simply acceleration) a is the first time derivative of velocity v(t)
uestions
1 Figure 2-16 gives the velocity of a particle moving on an x axis. What are (a) the initial and (b) the final di¬ rections of travel? (c) Does the parti¬ cle stop momentarily? (d) Is the ac¬ celeration positive or negative? (e) Is it constant or varying?
2 Figure 2-17 gives the accelera¬ tion a(t) of a Chihuahua as it chases Figure 2-16 Question 1 a German shepherd along an axis. In
which of the time periods indicated does the Chihuahua move at constant speed?
Figure 2-17 Question 2.
a
3 Figure 2-18 shows four paths along which objects move from a starting point to a final point, all in the same time interval. The paths pass over a grid of equally spaced straight lines. Rank the paths according to (a) the av¬ erage velocity of the objects and (b) the average speed of the objects, great¬ est first.
Figure 2-18 Question 3.
4 Figure 2-19 is a graph of a parti¬ cle’s position along an x axis versus time, (a) At time t — 0, what
These are not valid when the acceleration is not constant.
Free-Fall Acceleration An important example of straight- line motion with constant acceleration is that of an object rising or falling freely near Earth’s surface. The constant acceleration equa¬ tions describe this motion, but we make two changes in notation:
(1) we refer the motion to the vertical y axis with +y vertically up;
(2) we replace a with — g, where g is the magnitude of the free-fall acceleration. Near Earth's surface,g = 9.8 m/s2 (= 32 ft/s2).
is the sign of the particle’s position? Is the particle’s velocity positive, negative, or 0 at (b) t = 1 s, (c) t = 2 s, and (d) t = 3 s? (e) How many times does the particle go through the point x = 0?
5 Figure 2-20 gives the velocity of a particle moving along an axis. Point 1 is at the highest point on the curve; point 4 is at the lowest point; and points 2 and 6 are at the same height. What is the direction of travel at (a) time t = 0 and (b) point 4? (c) At which of the six numbered points does the particle reverse its direction of travel? (d) Rank the six points according to the magnitude of the acceleration, greatest first.
Figure 2-19 Question 4.
6 At t = 0, a particle moving along an
x axis is at position x0 = — 20 m. The signs of the particle’s initial velocity v0 (at time f0) and constant acceleration a are, respectively, for four situations: (1) +, +; (2) +, — ; (3) +; (4) -. In
which situations will the particle (a) stop momentarily, (b) pass through the origin, and (c) never pass through the origin?
7 Hanging over the railing of a bridge, you drop an egg (no initial ve¬ locity) as you throw a second egg downward. Which curves in Fig. 2-21
v
32
CHAPTER 2 MOTION ALONG A STRAIGHT LINE
give the velocity v(f) for (a) the dropped egg and (b) the thrown egg? (Curves A and B are parallel; so are C, I), and E\ so are F and G .)
8 The following equations give the velocity v(t) of a particle in four situations; (a) v = 3; (b) v = 4r2 + 2t — 6; (c) v = 3i — 4; (d) v = 5 12 — 3.
To which of these situations do the equations of Table 2-1 apply?
9 In Fig. 2-22, a cream tangerine is thrown di¬
rectly upward past three evenly spaced windows of equal heights. Rank the windows according to (a) the average speed of the cream tangerine while passing them, (b) the time the cream tan¬ gerine takes to pass them, (c) the magnitude of the acceleration of the cream tangerine while passing them, and (d) the change Av in the speed of the cream tangerine during the pas¬ sage, greatest first. Figure 2-22
Question 9.
10 Suppose that a passenger intent on lunch
during his first ride in a hot-air balloon accidently drops an apple over the side during the balloon’s liftoff. At the moment of the
, 1
1
, 2
. 3
apple’s release, the balloon is accelerating upward with a magni¬ tude of 4.0 m/s2 and has an upward velocity of magnitude 2 m/s. What are the (a) magnitude and (b) direction of the acceleration of the apple just after it is released? (c) Just then, is the apple moving upward or downward, or is it stationary? (d) What is the magni¬ tude of its velocity just then? (e) In the next few moments, does the speed of the apple increase, decrease, or remain constant?
1 1 Figure 2-23 shows that a particle moving along an x axis un¬ dergoes three periods of acceleration. Without written computa¬ tion, rank the acceleration periods according to the increases they produce in the particle’s velocity, greatest first.
(3)
(1
)
(5
!)
Time t
Figure 2-23 Question 11.
roblems
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign SSM Worked-out solution available in Student Solutions Manual WWW Worked-out solution is at
• - ••• Number of dots indicates level of problem difficulty ILW Interactive solution is at
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
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Module 2-1 Position, Displacement, and Average Velocity
•1 While driving a car at 90 km/h, how far do you move while your eyes shut for 0.50 s during a hard sneeze?
•2 Compute your average velocity in the following two cases: (a) You walk 73.2 m at a speed of 1.22 m/s and then run 73.2 m at a speed of 3.05 m/s along a straight track, (b) You walk for 1.00 min at a speed of 1.22 m/s and then run for 1.00 min at 3.05 m/s along a straight track, (c) Graph x versus t for both cases and indicate how the average velocity is found on the graph.
•3 SSM WWW An automobile travels on a straight road for 40 km at 30 km/h. It then continues in the same direction for an¬ other 40 km at 60 km/h. (a) What is the average velocity of the car during the full 80 km trip? (Assume that it moves in the positive x direction.) (b) What is the average speed? (c) Graph x versus t and indicate how the average velocity is found on the graph.
•4 A car moves uphill at 40 km/h and then back downhill at 60 km/h. What is the average speed for the round trip?
•5 SSM The position of an object moving along an x axis is given by x = 3 1 — 4 12 + r3, where x is in meters and 1 in seconds. Find the position of the object at the following values of f: (a) 1 s, (b) 2 s, (c) 3 s, and (d) 4 s. (e) What is the object’s displacement between t — 0 and t = 4 s? (f) What is its average velocity for the time interval from t = 2 s to t = 4 s? (g) Graph x versus t for 0 < t < 4 s and indi¬ cate how the answer for (f) can be found on the graph.
•6 The 1992 world speed record for a bicycle (human-powered vehicle) was set by Chris Huber. His time through the measured 200 m stretch was a sizzling 6.509 s, at which he commented,
“Cogito ergo zoom!” (I think, therefore I go fast!). In 2001, Sam Whittingham beat Huber’s record by 19.0 km/h. What was Whittingham's time through the 200 m?
••7 Two trains, each having a speed of 30 km/h, are headed at each other on the same straight track. A bird that can fly 60 km/h flies off the front of one train when they are 60 km apart and heads directly for the other train. On reaching the other train, the (crazy) bird flies directly back to the first train, and so forth. What is the to¬ tal distance the bird travels before the trains collide?
"8 £3 Panic escape. Figure 2-24 shows a general situation in
which a stream of people attempt to escape through an exit door that turns out to be locked. The people move toward the door at speed vs = 3.50 m/s, are each d = 0.25 m in depth, and are sepa¬ rated by L = 1.75 m. The arrangement in Fig. 2-24 occurs at time 1=0. (a) At what average rate does the layer of people at the door increase? (b) At what time does the layer’s depth reach 5.0 m? (The answers reveal how quickly such a situation becomes dangerous.)
-
"C
Locked - door
Figure 2-24 Problem 8.
••9 ILW In 1 km races, runner 1 on track 1 (with time 2 min, 27.95 s) appears to be faster than runner 2 on track 2 (2 min, 28.15 s). However, length L2 of track 2 might be slightly greater than length Ll of track 1. How large can L2 — Lx be for us still to conclude that runner 1 is faster?
PROBLEMS
33
••10 -z?* To set a speed record in a measured (straight-line) distance d, a race car must be driven first in one direction (in time t{) and then in the opposite direction (in time t2). (a) To eliminate the ef¬ fects of the wind and obtain the car’s speed vc in a windless situation, should we find the average of d/ti and d/t2 (method 1) or should we di¬ vide d by the average of and t{l (b) What is the fractional difference in the two methods when a steady wind blows along the car’s route and the ratio of the wind speed vw to the car’s speed vc is 0.0240?
•*1 1 O You are to drive 300 km to an interview. The interview is at 11:15 a.m. You plan to drive at 100 km/h, so you leave at 8:00 a.m. to allow some extra time. You drive at that speed for the first 100 km, but then construction work forces you to slow to 40 km/h for 40 km. What would be the least speed needed for the rest of the trip to arrive in time for the interview?
•••12 Traffic shock wave. An abrupt slowdown in concen¬
trated traffic can travel as a pulse, termed a shock wave, along the line of cars, either downstream (in the traffic direction) or up¬ stream, or it can be stationary. Figure 2-25 shows a uniformly spaced line of cars moving at speed v = 25.0 m/s toward a uni¬ formly spaced line of slow cars moving at speed vs = 5.00 m/s. Assume that each faster car adds length L = 12.0 m (car length plus buffer zone) to the line of slow cars when it joins the line, and as¬ sume it slows abruptly at the last instant, (a) For what separation dis¬ tance d between the faster cars does the shock wave remain stationary? If the separation is twice that amount, what are the (b) speed and (c) direction (upstream or downstream) of the shock wave?
rm
-L—\-
-SpR
'PT "POTT
Car Buffer
Figure 2-25 Problem 12.
•••13 ilw You drive on Interstate 10 from San Antonio to Houston, half the time at 55 km/h and the other half at 90 km/h. On the way back you travel half the distance at 55 km/h and the other half at 90 km/h. What is your average speed (a) from San Antonio to Houston, (b) from Houston back to San Antonio, and (c) for the entire trip? (d) What is your average velocity for the entire trip? (e) Sketch x versus t for (a), assuming the motion is all in the positive x direc¬ tion. Indicate how the average velocity can be found on the sketch.
Module 2-2 Instantaneous Velocity and Speed
•14 ^ An electron moving along the x axis has a position given by x = 16te~‘ m, where t is in seconds. How far is the electron from the origin when it momentarily stops?
•15 ® (a) If a particle’s position is given by x = 4 — 12r + 3 12 (where t is in seconds and x is in meters), what is its velocity at t = Is? (b) Is it moving in the positive or negative direction of x just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculation.) (e) Is there ever an instant when the velocity is zero? If so, give the time f; if not, answer no. (f) Is there a time after t — 3 s when the particle is moving in the negative direction of xl If so, give the time l: if not, answer no.
•1 6 The position function x(t) of a particle moving along an x axis is x = 4.0 — 6.0t2, with x in meters and t in seconds, (a) At what time and (b) where does the particle (momentarily) stop? At what (c) negative time and (d) positive time does the particle pass through the origin? (e) Graph x versus t for the range —5 s to +5 s. (f) To shift the curve rightward on the graph, should we include the
term +20r or the term — 20r in x(t)1 (g) Does that inclusion increase or decrease the value of x at which the particle momentarily stops?
••17 The position of a particle moving along the x axis is given in centimeters by x = 9.75 + 1.50f3, where t is in seconds. Calculate (a) the average velocity during the time interval t = 2.00 s to f = 3.00 s; (b) the instantaneous velocity at t — 2.00 s; (c) the instantaneous ve¬ locity at t = 3.00 s; (d) the instantaneous velocity at t = 2.50 s; and (e) the instantaneous velocity when the particle is midway between its positions at t = 2.00 s and t = 3.00 s. (f) Graph x versus t and in¬ dicate your answers graphically.
Module 2-3 Acceleration
•1 8 The position of a particle moving along an x axis is given by x = 12t2 — 2 r3, where x is in meters and t is in seconds. Determine (a) the position, (b) the velocity, and (c) the acceleration of the particle at t = 3.0 s. (d) What is the maximum positive coordinate reached by the particle and (e) at what lime is it reached? (f) What is the maxi¬ mum positive velocity reached by the particle and (g) at what time is it reached? (h) What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)? (i) Determine the av¬ erage velocity of the particle between t = 0 and t — 3 s.
•19 SSM At a certain time a particle had a speed of 18 m/s in the positive x direction, and 2.4 s later its speed was 30 m/s in the opposite direction. What is the average acceleration of the particle during this 2.4 s interval?
•20 (a) If the position of a particle is given by x — 20 1 — 5 f3, where x is in meters and t is in seconds, when, if ever, is the parti¬ cle’s velocity zero? (b) When is its acceleration a zero? (c) For what time range (positive or negative) is a negative? (d) Positive? (e) Graph x(t), v(t), and aft).
••21 From t = 0 to t = 5.00 min, a man stands still, and from t = 5.00 min to f = 10.0 min, he walks briskly in a straight line at a constant speed of 2.20 m/s. What are (a) his average velocity vavg and (b) his average acceleration «avg in the time interval 2.00 min to 8.00 min? What are (c) vavg and (d) aavg in the time interval 3.00 min to 9.00 min? (e) Sketch x versus t and v versus t, and indicate how the answers to (a) through (d) can be obtained from the graphs.
••22 The position of a particle moving along the x axis depends on the time according to the equation x = ct2 — bt 3, where x is in me¬ ters and r in seconds. What are the units of (a) constant c and (b) con¬ stant bl Let their numerical values be 3.0 and 2.0, respectively, (c) At what time does the particle reach its maximum positive x position? From t = 0.0 s to t = 4.0 s, (d) what distance does the particle move and (e) what is its displacement? Find its velocity at times (f) 1.0 s, (g) 2.0 s, (h) 3.0 s, and (i) 4.0 s. Find its acceleration at times (j) 1.0 s, (k) 2.0 s, (1) 3.0 s, and (m) 4.0 s.
Module 2-4 Constant Acceleration
•23 SSM An electron with an initial velocity v0 = 1.50 X 105 m/s
enters a region of length L = 1.00 cm where it is electrically acceler¬ ated (Fig. 2-26). It emerges with v = 5.70 X 106 m/s. What is its ac¬ celeration, assumed constant?
•24 -^58? Catapulting mush¬
rooms. Certain mushrooms launch their spores by a catapult mecha¬ nism. As water condenses from the air onto a spore that is attached to
Nonaccelerating Accelerating region region
Figure 2-26 Problem 23.
34
CHAPTER 2 MOTION ALONG A STRAIGHT LINE
the mushroom, a drop grows on one side of the spore and a film grows on the other side. The spore is bent over by the drop’s weight, but when the film reaches the drop, the drop's water suddenly spreads into the film and the spore springs upward so rapidly that it is slung off into the air. Typically, the spore reaches a speed of 1.6 m/s in a 5.0 /r m launch; its speed is then reduced to zero in 1.0 mm by the air. Using those data and assuming constant accelerations, find the acceleration in terms of g during (a) the launch and (b) the speed reduction.
•25 An electric vehicle starts from rest and accelerates at a rate of 2.0 m/s2 in a straight line until it reaches a speed of 20 m/s. The vehicle then slows at a constant rate of 1.0 m/s2 until it stops, (a) How much time elapses from start to stop? (b) How far does the vehicle travel from start to stop?
•26 A muon (an elementary particle) enters a region with a speed of 5.00 X 106 m/s and then is slowed at the rate of 1.25 X 1014 m/s2.
(a) How far does the muon take to stop? (b) Graph x versus t and v versus t for the muon.
•27 An electron has a constant acceleration of +3.2 m/s2. At a certain instant its velocity is +9.6 m/s. What is its velocity (a) 2.5 s earlier and (b) 2.5 s later?
•28 On a dry road, a car with good tires may be able to brake with a constant deceleration of 4.92 m/s2, (a) How long does such a car, initially traveling at 24.6 m/s, take to stop? (b) How far does it travel in this time? (c) Graph x versus t and v versus t for the deceleration.
•29 ILW A certain elevator cab has a total run of 190 m and a max¬ imum speed of 305 m/min, and it accelerates from rest and then back to rest at 1.22 m/s2, (a) How far does the cab move while ac¬ celerating to full speed from rest? (b) How long does it take to make the nonstop 190 m run, starting and ending at rest?
•30 The brakes on your car can slow you at a rate of 5.2 m/s2, (a) If you are going 137 km/h and suddenly see a state trooper, what is the minimum time in which you can get your car under the 90 km/h speed limit? (The answer reveals the futility of braking to keep your high speed from being detected with a radar or laser gun.)
(b) Graph x versus t and v versus t for such a slowing.
•31 SSM Suppose a rocket ship in deep space moves with con¬ stant acceleration equal to 9.8 m/s2, which gives the illusion of nor¬ mal gravity during the flight, (a) If it starts from rest, how long will it take to acquire a speed one-tenth that of light, which travels at 3.0 X 10s m/s? (b) How far will it travel in so doing?
•32 --yT- A world’s land speed record was set by Colonel John P. Stapp when in March 1954 he rode a rocket-propelled sled that moved along a track at 1020 km/h. He and the sled were brought to a stop in 1.4 s. (See Fig. 2-7.) In terms of g, what acceleration did he experience while stopping?
•33 SSM ILW A car traveling 56.0 km/h is 24.0 m from a barrier when the driver slams on the brakes. The car hits the barrier 2.00 s later, (a) What is the magnitude of the car’s constant acceleration before impact? (b) How fast is the car traveling at impact?
••34 © In Fig. 2-27, a red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an x axis. At time t = 0, the red car is at xr = 0 and the green car is at xg = 220 m. If the red car has a constant velocity of 20 km/h, the cars pass each other at x = 44.5 m, and if it has a constant velocity of 40 km/h, they pass each other at x = 76.6 m. What are (a) the initial velocity and (b) the constant acceleration of the green car?
xr
1
^ Green
Red i car u
Xg
Figure 2-27 Problems 34 and 35.
••35 Figure 2-27 shows a red car and a green car that move toward each other. Figure 2-28 is a graph of their motion, showing the positions xg0 = 270 m and xr0 = —35.0 m at time t = 0. The green car has a con¬ stant speed of 20.0 m/s and the red car begins from rest. What is the ac¬ celeration magnitude of the red car?
Figure 2-28 Problem 35.
••36 A car moves along an x axis through a distance of 900 m, starting at rest (at x = 0) and ending at rest (at x = 900 m). Through the first ^ of that distance, its acceleration is +2.25 m/s2. Through the rest of that distance, its acceleration is —0.750 m/s2. What are (a) its travel time through the 900 m and (b) its maxi¬ mum speed? (c) Graph position x, velocity v, and acceleration a versus time t for the trip.
t( s)
••37 Figure 2-29 depicts the motion x of a particle moving along an x axis with a constant acceleration. The fig¬ ure’s vertical scaling is set by xs = 6.0 m.
What are the (a) magnitude and (b) di¬ rection of the particle’s acceleration?
••38 (a) If the maximum acceleration
that is tolerable for passengers in a subway train is 1.34 m/s2 and subway stations are located 806 m apart, what is the maximum speed a subway train can attain between stations? (b) What is the travel time between stations? (c) If a subway train stops for 20 s at each station, what is the maximum average speed of the train, from one start-up to the next? (d) Graph x, v, and a versus t for the interval from one start-up to the next.
Figure 2-29 Problem 37.
••39 Cars A and B move in the same direction in adjacent lanes. The position x of car A is given in Fig. 2-30, from time t — 0 to t = 7.0 s. The figure’s vertical scaling is set by xs =
32.0 m. At t = 0, car B is at x —
0, with a velocity of 12 m/s and a negative constant accelera¬ tion aB. (a) What must aB be such that the cars are (momen¬ tarily) side by side (momentarily at the same value of x) at t = 4.0 s?
(b) For that value of aB , how many times are the cars side by side?
(c) Sketch the position x of car B versus time t on Fig. 2-30. How many times will the cars be side by side if the magnitude of accelera¬ tion aB is (d) more than and (e) less than the answer to part (a)?
1 2 3 4 5
t (s)
Figure 2-30 Problem 39.
••40 You are driving toward a traffic signal when it turns yel¬
low. Your speed is the legal speed limit of v0 = 55 km/h; your best deceleration rate has the magnitude a = 5.18 m/s2. Your best reaction time to begin braking is T = 0.75 s. To avoid having the front of your car enter the intersection after the light turns red, should you brake to a stop or continue to move at 55 km/h if the distance to
PROBLEMS
35
the intersection and the duration of the yellow light are (a) 40 m and 2.8 s, and (b) 32 m and 1.8 s? Give an answer of brake, continue, either (if either strategy works), or neither (if neither strategy works and the yellow duration is inappropriate).
t (s)
••41 © As two trains move along a track, their conductors suddenly notice that they are headed toward each other.
Figure 2-31 gives their velocities v as functions of time t as the conductors slow the trains. The figure’s vertical scaling is set by vs = 40.0 m/s. The slowing
processes begin when the trains are 200 m apart. What is their separa¬ tion when both trains have stopped?
Figure 2-31 Problem 41.
...42 © You are arguing over a cell phone while trailing an unmarked police car by 25 m; both your car and the police car are traveling at 110 km/h. Your argument diverts your attention from the police car for 2.0 s (long enough for you to look at the phone and yell, “I won’t do that!”). At the beginning of that 2.0 s, the po¬ lice officer begins braking suddenly at 5.0 m/s2, (a) What is the sep¬ aration between the two cars when your attention finally returns? Suppose that you take another 0.40 s to realize your danger and begin braking, (b) If you too brake at 5.0 m/s2, what is your speed when you hit the police car?
•••43 © When a high-speed passenger train traveling at
161 km/h rounds a bend, the engineer is shocked to see that a locomotive has improperly entered onto the track from a siding and is a distance D = 676 m ahead (Fig. 2-32). The locomotive is moving at 29.0 km/h. The engineer of the high-speed train imme¬ diately applies the brakes, (a) What must be the magnitude of the resulting constant deceleration if a collision is to be just avoided? (b) Assume that the engineer is at x = 0 when, at t = 0, he first spots the locomotive. Sketch x(t) curves for the locomotive and high-speed train for the cases in which a collision is just avoided and is not quite avoided.
Module 2-5 Free-Fall Acceleration
•44 When startled, an armadillo will leap upward. Suppose it rises 0.544 m in the first 0.200 s. (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of 0.544 m? (c) How much higher does it go?
•45 SSM WWW (a) With what speed must a ball be thrown verti¬ cally from ground level to rise to a maximum height of 50 m? (b) How long will it be in the air? (c) Sketch graphs of y, v, and a versus t for the ball. On the first two graphs, indicate the time at which 50 m is reached.
•46 Raindrops fall 1700 m from a cloud to the ground, (a) If they were not slowed by air resistance, how fast would the drops be moving when they struck the ground? (b) Would it be safe to walk outside during a rainstorm?
•47 SSM At a construction site a pipe wrench struck the ground with a speed of 24 m/s. (a) From what height was it inadvertently dropped? (b) How long was it falling? (c) Sketch graphs of y, v, and a versus t for the wrench.
•48 A hoodlum throws a stone vertically downward with an ini¬ tial speed of 12.0 m/s from the roof of a building, 30.0 m above the ground, (a) How long does it take the stone to reach the ground? (b) What is the speed of the stone at impact?
•49 SSM A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side, (a) How long does the package take to reach the ground? (b) With what speed does it hit the ground?
••50 At time t = 0, apple 1 is dropped from a bridge onto a road¬ way beneath the bridge; somewhat later, apple 2 is thrown down from the same height. Figure 2-33 gives the vertical positions y of the apples versus t during the falling, until both apples have hit the roadway. The scaling is set by ts = 2.0 s. With approximately what speed is apple 2 thrown down?
Figure 2-33 Problem 50.
••51 As a runaway scientific bal¬ loon ascends at 19.6 m/s, one of its instrument packages breaks free of a harness and free-falls. Figure 2-34 gives the vertical velocity of the package versus time, from before it breaks free to when it reaches the ground, (a) What maximum height above the break-free point does it rise? (b) How high is the break-free point above the ground?
Figure 2-34 Problem 51.
••52 © A bolt is dropped from a bridge under construction, falling 90 m to the valley below the bridge, (a) In how much time does it pass through the last 20% of its fall? What is its speed (b) when it begins that last 20% of its fall and (c) when it reaches the valley beneath the bridge?
••53 SSM ILW A key falls from a bridge that is 45 m above the water. It falls directly into a model boat, moving with constant velocity, that is 12 m from the point of impact when the key is re¬ leased. What is the speed of the boat?
••54 © A stone is dropped into a river from a bridge 43.9 m above the water. Another stone is thrown vertically down 1.00 s after the first is dropped. The stones strike the water at the same time, (a) What is the initial speed of the second stone? (b) Plot velocity versus time on a graph for each stone, taking zero time as the instant the first stone is released.
36
CHAPTER 2 MOTION ALONG A STRAIGHT LINE
••55 SSM A ball of moist clay falls 15.0 m to the ground. It is in contact with the ground for 20.0 ms before stopping, (a) What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground? (Treat the ball as a particle.) (b) Is the average acceleration up or down?
••56 ® Figure 2-35
shows the speed v versus height y of a ball tossed directly upward, along a y axis. Distance d is 0.40 m.
The speed at height yA is v^.The speed at height yB is \vA. What is speed vA?
••57 To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. ft re¬ bounds to a height of 2.00 m. If the ball is in contact with the floor for 12.0 ms, (a) what is the magnitude of its average acceleration during that contact and (b) is the average acceleration up or down?
••58 An object falls a distance h from rest. If it travels 0.50/; in the last 1.00 s, find (a) the time and (b) the height of its fall, (c) Explain the physically unacceptable solution of the quadratic equation in t that you obtain.
••59 Water drips from the nozzle of a shower onto the floor 200 cm below. The drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall. When the first drop strikes the floor, how far below the nozzle are the (a) second and (b) third drops?
••60 ® A rock is thrown vertically upward from ground level at time t = 0. At t = 1.5 s it passes the top of a tall tower, and 1.0 s later it reaches its maximum height. What is the height of the tower?
•••61 ® A steel ball is dropped from a building’s roof and passes a window, taking 0.125 s to fall from the top to the bottom of the window, a distance of 1.20 m. It then falls to a sidewalk and bounces back past the window, moving from bottom to top in 0.125 s. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is 2.00 s. How tall is the building?
•••62 A basketball player grabbing a rebound jumps
76.0 cm vertically. How much total time (ascent and descent) does the player spend (a) in the top 15.0 cm of this jump and (b) in the bottom 15.0 cm? (The player seems to hang in the air at the top.)
•••63 ® A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of 0.50 s, and the top-to-bottom height of the window is 2.00 m. How high above the window top does the flowerpot go?
•••64 A ball is shot vertically up¬ ward from the surface of another planet. A plot of y versus t for the ball is shown in Fig. 2-36, where y is the height of the ball above its start¬ ing point and t = 0 at the instant the ball is shot. The figure’s vertical scal¬ ing is set by ys = 30.0 m. What are the magnitudes of (a) the free-fall accel¬ eration on the planet and (b) the ini¬ tial velocity of the ball?
t (s)
Figure 2-36 Problem 64.
v
Figure 2-35 Problem 56.
Module 2-6 Graphical Integration in Motion Analysis
•65 Figure 2-15 a gives the acceleration of a volunteer’s
head and torso during a rear-end collision. At maximum head ac¬ celeration, what is the speed of (a) the head and (b) the torso?
••66 —^jSr In a forward punch in karate, the fist begins at rest at the waist and is brought rapidly forward until the arm is fully ex¬ tended. The speed v(r) of the fist is given in Fig. 2-37 for someone skilled in karate. The vertical scaling is set by vs = 8.0 m/s. How far has the fist moved at (a) time t = 50 ms and (b) when the speed of the fist is maximum?
0 50 100 140
t (ms)
Figure 2-37 Problem 66.
••67 When a soccer ball is kicked to¬ ward a player and the player deflects the ball by “head¬ ing” it, the accelera¬ tion of the head dur¬ ing the collision can be significant. Figure 2-38 gives the meas¬ ured acceleration
a{t) of a soccer player’s head for a bare head and a helmeted head, starting from rest. The scaling on the vertical axis is set by as = 200 m/s2. At time t = 7.0 ms, what is the difference in the speed acquired by the bare head and the speed acquired by the helmeted head?
t (ms)
Figure 2-38 Problem 67.
••68 A salamander of the genus Hydromantes captures
prey by launching its tongue as a projectile: The skeletal part of the tongue is shot for¬ ward, unfolding the rest of T the tongue, until the outer c portion lands on the prey, ' sticking to it. Figure 2-39 shows the acceleration mag¬ nitude a versus time t for the acceleration phase of the launch in a typical situation.
The indicated accelerations are a2 = 400 m/s2 and a1 = 100 m/s2. What is the outward speed of the tongue at the end of the acceleration phase?
••69 1LW How far does the run¬ ner whose velocity -time graph is shown in Fig. 2-40 travel in 16 s? The figure’s vertical scaling is set by Vj = 8.0 m/s.
t (s)
Figure 2-40 Problem 69.
PROBLEMS
37
•••70 Two particles move along an x axis. The position of particle 1 is given by x = 6.00 12 + 3.00f + 2.00 (in meters and seconds); the ac¬ celeration of particle 2 is given by a — — 8.00f (in meters per second squared and seconds) and, at f = 0, its velocity is 20 m/s. When the velocities of the particles match, what is their velocity?
Additional Problems
71 In an arcade video game, a spot is programmed to move across the screen according to x — 9.00 1 — 0.750 13, where x is dis¬ tance in centimeters measured from the left edge of the screen and t is time in seconds. When the spot reaches a screen edge, at either x = 0 or x = 1 5.0 cm, t is reset to 0 and the spot starts moving again according to x(t). (a) At what time after starting is the spot instan¬ taneously at rest? (b) At what value of x does this occur? (c) What is the spot’s acceleration (including sign) when this occurs? (d) Is it moving right or left just prior to coming to rest? (e) Just after? (f) At what time t > 0 does it first reach an edge of the screen?
72 A rock is shot vertically upward from the edge of the top of a tall building. The rock reaches its maximum height above the top of the building 1.60 s after being shot. Then, after barely missing the edge of the building as it falls downward, the rock strikes the ground 6.00 s after it is launched. In SI units: (a) with what upward velocity is the rock shot, (b) what maximum height above the top of the building is reached by the rock, and (c) how tall is the building?
73 © At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.2 m/s2. At the same instant a truck, traveling with a constant speed of 9.5 m/s, overtakes and passes the automobile, (a) How far beyond the traffic signal will the automobile overtake the truck? (b) How fast will the automo¬ bile be traveling at that instant?
74 A pilot flies horizontally at 1300 km/h, at height /? = 35 m above initially level ground. However, at time t = 0, the pilot be¬ gins to fly over ground sloping upward at angle 0 = 4.3° (Fig. 2-41). If the pilot does not change the airplane’s heading, at what time t does the plane strike the ground?
75 © To stop a car, first you require a certain reaction time to be¬ gin braking; then the car slows at a constant rate. Suppose that the total distance moved by your car during these two phases is 56.7 m when its initial speed is 80.5 km/h, and 24.4 m when its initial speed is 48.3 km/h. What are (a) your reaction time and (b) the magni¬ tude of the acceleration?
76 © Figure 2-42 shows part of a street where traffic flow
is to be controlled to allow a platoon of cars to move smoothly along the street. Suppose that the platoon leaders have just
1 2 3
Y - A 2 - A3 -*j
reached intersection 2, where the green appeared when they were distance d from the intersection. They continue to travel at a cer¬ tain speed vp (the speed limit) to reach intersection 3, where the green appears when they are distance d from it. The intersections are separated by distances H23 and Z)12. (a) What should be the time delay of the onset of green at intersection 3 relative to that at intersection 2 to keep the platoon moving smoothly?
Suppose, instead, that the platoon had been stopped by a red light at intersection 1. When the green comes on there, the leaders require a certain time tr to respond to the change and an additional time to accelerate at some rate a to the cruising speed vp. (b) If the green at intersection 2 is to appear when the leaders are distance d from that intersection, how long after the light at intersection 1 turns green should the light at intersection 2 turn green?
77 SSM A hot rod can accelerate from 0 to 60 km/h in 5.4 s. (a) What is its average acceleration, in m/s2, during this time? (b) How far will it travel during the 5.4 s, assuming its acceleration is con¬ stant? (c) From rest, how much time would it require to go a distance of 0.25 km if its acceleration could be maintained at the value in (a)?
78 © A red train traveling at 72 km/h and a green train traveling at 144 km/h are headed toward each other along a straight, level track. When they are 950 m apart, each engineer sees the other’s train and applies the brakes. The brakes slow each train at the rate of 1.0 m/s2. Is there a collision? If so, answer yes and give the speed of the red train and the speed of the green train at impact, respec¬ tively. If not, answer no and give the separation between the trains when they stop.
79 © At time t = 0, a rock climber accidentally allows a piton to fall freely from a high point on the rock wall to the valley below him. Then, after a short delay, his climbing part¬ ner, who is 10 m higher on the wall, throws a piton down¬ ward. The positions y of the pitons versus t during the falling are given in Fig. 2-43.
With what speed is the second piton thrown?
80 A train started from rest and moved with constant accelera¬ tion. At one time it was traveling 30 m/s, and 160 m farther on it was traveling 50 m/s. Calculate (a) the acceleration, (b) the time re¬ quired to travel the 160 m mentioned, (c) the time required to at¬ tain the speed of 30 m/s, and (d) the distance moved from rest to the time the train had a speed of 30 m/s. (e) Graph x versus t and v versus t for the train, from rest.
81 SSM A particle’s acceleration along an x axis is « = 5.0r, with t in seconds and a in meters per
second squared. At t = 2.0 s, its velocity is +17 m/s. What is its velocity at t = 4.0 s?
82 Figure 2-44 gives the ac¬ celeration a versus time t for a particle moving along an x axis. The a-axis scale is set by as = 12.0 m/s2. At t = —2.0 s, the particle’s velocity is 7.0 m/s. What is its velocity at t = 6.0 s?
a (m/s2)
2
0
1
t (s)
Figure 2-42 Problem 76.
Figure 2-44 Problem 82.
38
CHAPTER 2 MOTION ALONG A STRAIGHT LINE
83 Figure 2-45 shows a simple device for measuring your reaction time. It consists of a cardboard strip marked with a scale and two large dots. A friend holds the strip vertically, with thumb and forefinger at the dot on the right in Fig. 2-45. You then posi¬ tion your thumb and forefinger at the other dot (on the left in Fig. 2-45), being careful not to touch the strip. Your friend re¬ leases the strip, and you try to pinch it as soon as possible after you see it begin to fall. The mark at the place where you pinch the strip gives your reaction time, (a) Flow far from the lower dot should you place the 50.0 ms mark? Flow much higher should you place the marks for (b) 100, (c) 150, (d) 200, and (e) 250 ms? (For example, should the 100 ms marker be 2 times as far from the dot as the 50 ms marker? If so, give an answer of 2 times. Can you find any pattern in the answers?)
Reaction time (ms)
O OX b- ‘ b- ‘ NO NO
O O CJT o Ox
© O O O
Figure 2-45 Problem 83.
84 A rocket-driven sled running on a straight, level track is used to investigate the effects of large accelerations on humans. One such sled can attain a speed of 1600 km/h in 1.8 s, starting from rest. Find (a) the acceleration (assumed constant) in terms of g and (b) the distance traveled.
85 A mining cart is pulled up a hill at 20 km/h and then pulled back down the hill at 35 km/h through its original level. (The time required for the cart’s reversal at the top of its climb is negligible.) What is the average speed of the cart for its round trip, from its original level back to its original level?
86 A motorcyclist who is moving along an x axis directed to¬ ward the east has an acceleration given by a = (6.1 — 1.2 1) m/s2 for 0 < t < 6.0 s. At t = 0, the velocity and position of the cyclist are 2.7 m/s and 7.3 m. (a) What is the maximum speed achieved by the cyclist? (b) What total distance does the cyclist travel be¬ tween t = 0 and 6.0 s?
87 SSM When the legal speed limit for the New York Thruway was increased from 55 mi/h to 65 mi/h, how much time was saved by a motorist who drove the 700 km between the Buffalo entrance and the New York City exit at the legal speed limit?
88 A car moving with constant acceleration covered the distance between two points 60.0 m apart in 6.00 s. Its speed as it passed the second point was 15.0 m/s. (a) What was the speed at the first point? (b) What was the magnitude of the acceleration? (c) At what prior distance from the first point was the car at rest? (d) Graph x versus t and v versus t for the car, from rest (t = 0).
89 SSM — yS- A certain juggler usually tosses balls vertically to a height H. To what height must they be tossed if they are to spend twice as much time in the air?
90 A particle starts from the ori¬ gin at t = 0 and moves along the positive x axis. A graph of the veloc¬ ity of the particle as a function of the time is shown in Fig. 2-46; the v-axis scale is set by = 4.0 m/s. (a) What is the coordinate of the particle at t = 5.0 s? (b) What is the velocity of the particle at t — 5.0 s? (c) What is
the acceleration of the particle at t = 5.0 s? (d) What is the average ve¬ locity of the particle between t = 1.0 s and t = 5.0 s? (e) What is the average acceleration of the particle between t = 1.0 s and t — 5.0 s?
91 A rock is dropped from a 100-m-high cliff. Flow long does it take to fall (a) the first 50 m and (b) the second 50 m?
92 Two subway stops are separated by 1100 m. If a subway train accelerates at +1.2 m/s2 from rest through the first half of the dis¬ tance and decelerates at —1.2 m/s2 through the second half, what are (a) its travel time and (b) its maximum speed? (c) Graph x, v, and a versus t for the trip.
93 A stone is thrown vertically upward. On its way up it passes point A with speed v, and point B, 3.00 m higher than A, with speed \ v. Calculate (a) the speed v and (b) the maximum height reached by the stone above point B.
94 A rock is dropped (from rest) from the top of a 60-m-tall building. Flow far above the ground is the rock 1.2 s before it reaches the ground?
95 SSM An iceboat has a constant velocity toward the east when a sudden gust of wind causes the iceboat to have a constant accel¬ eration toward the east for a period of 3.0 s. A plot of x versus t is shown in Fig. 2-47, where t = 0 is taken to be the instant the wind starts to blow and the positive x axis is toward the east, (a) What is the acceleration of the iceboat during the 3.0 s interval? (b) What is the velocity of the iceboat at the end of the 3.0 s interval? (c) If the acceleration remains constant for an additional 3.0 s, how far does the iceboat travel during this second 3.0 s interval?
t (s)
Figure 2-47 Problem 95.
96 A lead ball is dropped in a lake from a diving board 5.20 m above the water. It hits the water with a certain velocity and then sinks to the bottom with this same constant velocity. It reaches the bottom 4.80 s after it is dropped, (a) Flow deep is the lake? What are the (b) magnitude and (c) direction (up or down) of the aver¬ age velocity of the ball for the entire fall? Suppose that all the wa¬ ter is drained from the lake. The ball is now thrown from the diving board so that it again reaches the bottom in 4.80 s. What are the (d) magnitude and (e) direction of the initial velocity of the ball?
97 The single cable supporting an unoccupied construction ele¬ vator breaks when the elevator is at rest at the top of a 120-m-high building, (a) With what speed does the elevator strike the ground? (b) How long is it falling? (c) What is its speed when it passes the halfway point on the way down? (d) How long has it been falling when it passes the halfway point?
98 Two diamonds begin a free fall from rest from the same height, 1.0 s apart. How long after the first diamond begins to fall will the two diamonds be 10 m apart?
99 A ball is thrown vertically downward from the top of a 36.6- m-tall building. The ball passes the top of a window that is 12.2 m above the ground 2.00 s after being thrown. What is the speed of the ball as it passes the top of the window?
Figure 2-46 Problem 90.
PROBLEMS
39
100 A parachutist bails out and freely falls 50 m. Then the para¬ chute opens, and thereafter she decelerates at 2.0 m/s2. She reaches the ground with a speed of 3.0 m/s. (a) How long is the parachutist in the air? (b) At what height does the fall begin?
101 A ball is thrown down vertically with an initial speed of v0 from a height of h. (a) What is its speed just before it strikes the ground? (b) How long does the ball take to reach the ground? What would be the answers to (c) part a and (d) part b if the ball were thrown upward from the same height and with the same ini¬ tial speed? Before solving any equations, decide whether the an¬ swers to (c) and (d) should be greater than, less than, or the same as in (a) and (b).
102 The sport with the fastest moving ball is jai alai, where measured speeds have reached 303 km/h. If a professional jai alai player faces a ball at that speed and involuntarily blinks, he blacks out the scene for 100 ms. How far does the ball move dur¬ ing the blackout?
103 If a baseball pitcher throws a fastball at a horizontal speed of 160 km/h, how long does the ball take to reach home plate 18.4 m away?
104 A proton moves along the x axis according to the equation x = 50 1 + 10f2, where x is in meters and t is in seconds. Calculate (a) the average velocity of the proton during the first 3.0 s of its motion, (b) the instantaneous velocity of the proton at t = 3.0 s, and (c) the instantaneous acceleration of the proton at t = 3.0 s. (d) Graph x versus t and indicate how the answer to (a) can be obtained from the plot, (e) Indicate the answer to (b) on the graph, (f) Plot v versus t and indicate on it the answer to (c).
105 A motorcycle is moving at 30 m/s when the rider applies the brakes, giving the motorcycle a constant deceleration. During the 3.0 s interval immediately after braking begins, the speed decreases to 15 m/s. What distance does the motorcycle travel from the instant braking begins until the motorcycle stops?
106 A shuffleboard disk is accelerated at a constant rate from rest to a speed of 6.0 m/s over a 1.8 m distance by a player using a cue. At this point the disk loses contact with the cue and slows at a constant rate of 2.5 m/s2 until it stops, (a) How much time elapses from when the disk begins to accelerate until it stops? (b) What total distance does the disk travel?
1 07 The head of a rattlesnake can accelerate at 50 m/s2 in striking a victim. If a car could do as well, how long would it take to reach a speed of 100 km/h from rest?
108 A jumbo jet must reach a speed of 360 km/h on the runway for takeoff. What is the lowest constant acceleration needed for takeoff from a 1.80 km runway?
1 09 An automobile driver increases the speed at a constant rate from 25 km/h to 55 km/h in 0.50 min. A bicycle rider speeds up at a constant rate from rest to 30 km/h in 0.50 min. What are the magni¬ tudes of (a) the driver’s acceleration and (b) the rider's acceleration?
110 On average, an eye blink lasts about 100 ms. How far does a MiG-25 “Foxbat” fighter travel during a pilot’s blink if the plane’s average velocity is 3400 km/h?
111 A certain sprinter has a top speed of 11.0 m/s. If the sprinter starts from rest and accelerates at a constant rate, he is able to reach his top speed in a distance of 12.0 m. He is then able to main¬ tain this top speed for the remainder of a 100 m race, (a) What is his time for the 100 m race? (b) In order to improve his time, the sprinter tries to decrease the distance required for him to reach his
top speed. What must this distance be if he is to achieve a time of 10.0 s for the race?
112 The speed of a bullet is measured to be 640 m/s as the bullet emerges from a barrel of length 1.20 m. Assuming constant accelera¬ tion, find the time that the bullet spends in the barrel after it is fired.
113 The Zero Gravity Research Facility at the NASA Glenn Research Center includes a 145 m drop tower. This is an evacuated ver¬ tical tower through which, among other possibilities, a 1-m-diameter sphere containing an experimental package can be dropped, (a) How long is the sphere in free fall? (b) What is its speed just as it reaches a catching device at the bottom of the tower? (c) When caught, the sphere experiences an average deceleration of 25 g as its speed is reduced to zero. Through what distance does it travel during the deceleration?
114 -sjg? A car can be braked to a stop from the autobahn-like speed of 200 km/h in 170 m. Assuming the acceleration is constant, find its magnitude in (a) SI units and (b) in terms of g. (c) How much time Tb is required for the braking? Your reaction time Tr is the time you require to perceive an emergency, move your foot to the brake, and begin the braking. If Tr — 400 ms, then (d) what is Tb in terms of Tr, and (e) is most of the full time required to stop spent in reacting or braking? Dark sunglasses delay the visual signals sent from the eyes to the visual cortex in the brain, increasing Tr. (f) In the extreme case in which Tr is increased by 100 ms, how much farther does the car travel during your reaction time?
115 In 1889, at Jubbulpore, India, a tug-of-war was finally won af¬ ter 2 h 41 min, with the winning team displacing the center of the rope 3.7 m. In centimeters per minute, what was the magnitude of the average velocity of that center point during the contest?
116 Most important in an investigation of an airplane crash by the U.S. National Transportation Safety Board is the data stored on the airplane’s flight-data recorder, commonly called the “black box” in spite of its orange coloring and reflective tape. The recorder is engi¬ neered to withstand a crash with an average deceleration of magni¬ tude 3400g during a time interval of 6.50 ms. In such a crash, if the recorder and airplane have zero speed at the end of that time inter¬ val, what is their speed at the beginning of the interval?
117 From January 26, 1977, to September 18, 1983, George Meegan of Great Britain walked from Ushuaia, at the southern tip of South America, to Prudhoe Bay in Alaska, covering 30 600 km. In meters per second, what was the magnitude of his average velocity during that time period?
118 The wings on a stonefly do not flap, and thus the insect cannot fly. However, when the insect is on a water surface, it can sail across the surface by lifting its wings into a breeze. Suppose that you time stoneflies as they move at constant speed along a straight path of a certain length. On average, the trips each take 7.1 s with the wings set as sails and 25.0 s with the wings tucked in. (a) What is the ratio of the sailing speed vs to the nonsailing speed vnp. (b) In terms of vs, what is the difference in the times the insects take to travel the first 2.0 m along the path with and without sailing?
119 The position of a particle as it moves along a y axis is given by
y = (2.0 cm) sin (-7rf/4),
with t in seconds and y in centimeters, (a) What is the average veloc¬ ity of the particle between t — 0 and t = 2.0 s? (b) What is the instan¬ taneous velocity of the particle at t = 0, 1.0, and 2.0 s? (c) What is the average acceleration of the particle between t = 0 and t = 2.0 s? (d) What is the instantaneous acceleration of the particle at t = 0, 1.0, and 2.0 s?
c
3
Vectors
3-1 VECTORS AND THEIR COMPONENTS
Learning Objectives _
After reading this module, you should be able to . . .
3.01 Add vectors by drawing them in head-to-tail arrange- 3.04 Given the components of a vector, draw the vector
ments, applying the commutative and associative laws. and determine its magnitude and orientation.
3.02 Subtract a vector from a second one. 3.05 Convert angle measures between degrees and radians.
3.03 Calculate the components of a vector on a given coordi¬ nate system, showing them in a drawing.
Key Ideas _
• Scalars, such as temperature, have magnitude only. They are specified by a number with a unit (1 0°C) and obey the rules of arithmetic and ordinary algebra. Vectors, such as dis¬ placement, have both magnitude and direction (5 m, north) and obey the rules of vector algebra.
• Two vectors It and b may be added geometrically by draw¬ ing them to a common scale and placing them head to tail.
The vector connecting the tail of the first to the head of the second is the vector sum ~s. To subtract b from a, reverse the direction of b to get —ft; then add — ft to It. Vector addition is commutative and obeys the associative law.
a = \/al + a? and tan ft = —
" y n
• The (scalar) components ax and ay of any two-dimensional vector ~a along the coordinate axes are found by dropping perpendicular lines from the ends of It onto the coordinate axes. The components are given by
ax = a cos 0 and ay = a sin ft,
where ft is the angle between the positive direction of the x axis and the direction of It. The algebraic sign of a component indicates its direction along the associated axis. Given its components, we can find the magnitude and orientation of the vector ~a with
What Is Physics?
Physics deals with a great many quantities that have both size and direction, and it needs a special mathematical language — the language of vectors — to describe those quantities. This language is also used in engineering, the other sciences, and even in common speech. If you have ever given directions such as “Go five blocks down this street and then hang a left,” you have used the language of vectors. In fact, navigation of any sort is based on vectors, but physics and engineering also need vectors in special ways to explain phenomena involving rotation and mag¬ netic forces, which we get to in later chapters. In this chapter, we focus on the basic language of vectors.
Vectors and Scalars
A particle moving along a straight line can move in only two directions. We can take its motion to be positive in one of these directions and negative in the other. For a particle moving in three dimensions, however, a plus sign or minus sign is no longer enough to indicate a direction. Instead, we must use a vector.
40
3-1 VECTORS AND THEIR COMPONENTS
41
A vector has magnitude as well as direction, and vectors follow certain (vector) rules of combination, which we examine in this chapter. A vector quantity is a quantity that has both a magnitude and a direction and thus can be represented with a vector. Some physical quantities that are vector quantities are displacement, velocity, and acceleration. You will see many more throughout this book, so learning the rules of vector combination now will help you greatly in later chapters.
Not all physical quantities involve a direction. Temperature, pressure, energy, mass, and time, for example, do not “point” in the spatial sense. We call such quantities scalars, and we deal with them by the rules of ordinary algebra. A sin¬ gle value, with a sign (as in a temperature of — 40°F), specifies a scalar.
The simplest vector quantity is displacement, or change of position. A vec¬ tor that represents a displacement is called, reasonably, a displacement vector. (Similarly, we have velocity vectors and acceleration vectors.) If a particle changes its position by moving from A to B in Fig. 3-la, we say that it undergoes a displace¬ ment from A to B. which we represent with an arrow pointing from AtoB. The ar¬ row specifies the vector graphically. To distinguish vector symbols from other kinds of arrows in this book, we use the outline of a triangle as the arrowhead.
In Fig. 3-la, the arrows from A to B, from A' to B' , and from A" to B" have the same magnitude and direction. Thus, they specify identical displacement vec¬ tors and represent the same change of position for the particle. A vector can be shifted without changing its value if its length and direction are not changed.
The displacement vector tells us nothing about the actual path that the parti¬ cle takes. In Fig. 3-1 b, for example, all three paths connecting points A and B cor¬ respond to the same displacement vector, that of Fig. 3-la. Displacement vectors represent only the overall effect of the motion, not the motion itself.
(b)
Figure 3-1 (a) All three arrows have the same magnitude and direction and thus represent the same displacement, (b) All three paths connecting the two points cor¬ respond to the same displacement vector.
Adding Vectors Geometrically
Suppose that, as in the vector diagram of Fig. 3-2 a, a particle moves from A to B and then later from B to C. We can represent its overall displacement (no matter what its actual path) with two successive displacement vectors, AB and BC. The net displacement of these two displacements is a single displacement from A to C. We call AC the vector sum (or resultant) of the vectors AB and BC. This sum is not the usual algebraic sum.
In Fig. 3-2 b, we redraw the vectors of Fig. 3-2a and relabel them in the way that we shall use from now on, namely, with an arrow over an italic symbol, as in o’. If we want to indicate only the magnitude of the vector (a quantity that lacks a sign or direction), we shall use the italic symbol, as in a, b, and s. (You can use just a handwritten symbol.) A symbol with an overhead arrow always implies both properties of a vector, magnitude and direction.
We can represent the relation among the three vectors in Fig. 3-2 b with the vector equation
~s = ~a + b. (3-1)
which says that the vector ~s is the vector sum of vectors a and h.The symbol + in Eq. 3-1 and the words “sum” and “add” have different meanings for vectors than they do in the usual algebra because they involve both magnitude and direction.
Figure 3-2 suggests a procedure for adding two-dimensional vectors a and b geometrically. (1) On paper, sketch vector o’ to some convenient scale and at the proper angle. (2) Sketch vector b to the same scale, with its tail at the head of vec¬ tor ~a, again at the proper angle. (3) The vector sum ~s is the vector that extends from the tail of 7? to the head of b.
Properties. Vector addition, defined in this way, has two important proper¬ ties. First, the order of addition does not matter. Adding a to b gives the same
B
To add a*and b , draw them head to tail.
This is the resulting vector, from tail of a to head of b.
Figure 3-2 (a) AC is the vector sum of the vectors AB and BC. ( b ) The same vectors relabeled.
42
CHAPTER 3 VECTORS
You get the same vector result for either order of adding vectors.
Figure 3-3 The two vectors 77 and ft can be added in either order; see Eq. 3-2.
Figure 3-5 The vectors b and —b have the same magnitude and opposite directions.
(A)
Figure 3-6 (a) Vectors 77, ft, and — ft.
(ft) To subtract vector ft from vector 77, add vector — ft to vector 77.
result as adding ft to 7? (Fig. 3-3); that is,
7? + ft = ft + 7? (commutative law). (3-2)
Second, when there are more than two vectors, we can group them in any order as we add them. Thus, if we want to add vectors 7z, ft, and ~c, we can add 7? and ft first and then add their vector sum to ~c. We can also add ft and 7 ; first and then add that sum to 7z. We get the same result either way, as shown in Fig. 3-4. That is,
(7? + ft) + ~C. = 7z + (ft + 7?) (associative law). (3-3)
Figure 3-4 The three vectors 77, ft, and ~c can be grouped in any way as they are added; see Eq. 3-3.
The vector —ft is a vector with the same magnitude as ft but the opposite direction (see Fig. 3-5). Adding the two vectors in Fig. 3-5 would yield
ft + (-ft) = 0.
Thus, adding — ft has the effect of subtracting ft . We use this property to define the difference between two vectors; let d = ~a — ft. Then
d — 7? — ft — 7? + ( — ft) (vector subtraction); (3-4)
that is, we find the difference vector d by adding the vector — ft to the vector 7? . Figure 3-6 shows how this is done geometrically.
As in the usual algebra, we can move a term that includes a vector symbol from one side of a vector equation to the other, but we must change its sign. For example, if we are given Eq. 3-4 and need to solve for 77. we can rearrange the equation as
d + ft = 7? or a = d + b.
Remember that, although we have used displacement vectors here, the rules for addition and subtraction hold for vectors of all kinds, whether they represent velocities, accelerations, or any other vector quantity. However, we can add only vectors of the same kind. For example, we can add two displacements, or two velocities, but adding a displacement and a velocity makes no sense. In the arith¬ metic of scalars, that would be like trying to add 21 s and 12 m.
0
Checkpoint 1
The magnitudes of displacements 77 and ft are 3 m and 4 m, respectively, and 7? = 7? + ft . Considering various orientations of 77 and ft, what are (a) the maximum possible magnitude for ~c and (b) the minimum possible magnitude?
Components of Vectors
Adding vectors geometrically can be tedious. A neater and easier technique involves algebra but requires that the vectors be placed on a rectangular coordi¬ nate system. The x and y axes are usually drawn in the plane of the page, as shown
43
3-1 VECTORS AND THEIR COMPONENTS
in Fig. 3-7 a. The z axis comes directly out of the page at the origin; we ignore it for now and deal only with two-dimensional vectors.
A component of a vector is the projection of the vector on an axis. In Fig. 3-7 a. for example, ax is the component of vector ~a on (or along) the x axis and ay is the component along the y axis. To find the projection of a vector along an axis, we draw perpendicular lines from the two ends of the vector to the axis, as shown. The projection of a vector on an x axis is its x component , and similarly the projection on the y axis is the y component. The process of finding the components of a vector is called resolving the vector.
A component of a vector has the same direction (along an axis) as the vector. In Fig. 3-7, ax and ay are both positive because ~a extends in the positive direction of both axes. (Note the small arrowheads on the components, to indicate their di¬ rection.) If we were to reverse vector a, then both components would be negative and their arrowheads would point toward negative x and y. Resolving vector b in Fig. 3-8 yields a positive component bx and a negative component bv.
In general, a vector has three components, although for the case of Fig. 3-la the component along the z axis is zero. As Figs. 3-la and b show, if you shift a vec¬ tor without changing its direction, its components do not change.
Finding the Components. We can find the components of ~a in Fig. 3-la geo¬ metrically from the right triangle there:
ax = a cos 9 and ay = a sin 9 , (3-5)
where 9 is the angle that the vector ~a makes with the positive direction of the x axis, and a is the magnitude of ~a. Figure 3-1 c shows that ~a and its x and y com¬ ponents form a right triangle. It also shows how we can reconstruct a vector from its components: we arrange those components head to tail. Then we complete a right triangle with the vector forming the hypotenuse, from the tail of one com¬ ponent to the head of the other component.
Once a vector has been resolved into its components along a set of axes, the components themselves can be used in place of the vector. For example, ~a in Fig. 3-la is given (completely determined) by a and 6. It can also be given by its components ax and ay. Both pairs of values contain the same information. If we know a vector in component notation (ax and ay) and want it in magnitude-angle notation ( a and 9), we can use the equations
a = Va* + aj and tan 9 = — (3-6)
This is the y component of the vector.
of the vector.
(c)
Figure 3-7 ( a ) The components ax and ay of vector ~a. ( b ) The components are unchanged if the vector is shifted, as long as the magnitude and orientation are maintained, (c) The com¬ ponents form the legs of a right triangle whose hypotenuse is the magnitude of the vector.
The components and the vector form a right triangle.
to transform it. Figure 3-8 The component of b on the
In the more general three-dimensional case, we need a magnitude and two x axis is positive, and that on the y axis is angles (say, a, 9, and (/>) or three components (ax, a and a,) to specify a vector. negative.
0
Checkpoint 2
In the figure, which of the indicated methods for combining the x and y components of vector ~a are proper to determine that vector?
(a)
(6)
(c)
(d)
(/)
44
CHAPTER 3 VECTORS
Sample Problem 3.01 Adding vectors in a drawing, orienteering
In an orienteering class, you have the goal of moving as far (straight-line distance) from base camp as possible by making three straight-line moves. You may use the follow¬ ing displacements in any order: (a) ~a , 2.0 km due east (directly toward the east); (b) b, 2.0 km 30° north of east (at an angle of 30° toward the north from due east); (c) c, 1.0 km due west. Alternatively, you may substitute either — b for b or — ~c for ~c. What is the greatest distance you can be from base camp at the end of the third displace¬ ment? (We are not concerned about the direction.)
Reasoning: Using a convenient scale, we draw vectors ~a , b,~c,—b, and — "c as in Fig. 3-9a. We then mentally slide the vectors over the page, connecting three of them at a time in head-to-tail arrangements to find their vector sum d. The tail of the first vector represents base camp. The head of the third vector represents the point at which you stop. The vector sum d extends from the tail of the first vector to the head of the third vector. Its magnitude d is your dis¬ tance from base camp. Our goal here is to maximize that base-camp distance.
We find that distance d is greatest for a head-to-tail arrangement of vectors ~a, b, and —~c. They can be in any
-This is the vector result for adding those three vectors in any order.
(a) (*)
Figure 3-9 (a) Displacement vectors; three are to be used, (b) Your distance from base camp is greatest if you undergo displacements a, b, and —~c, in any order.
order, because their vector sum is the same for any order. (Recall from Eq. 3-2 that vectors commute.) The order shown in Fig. 3-9 b is for the vector sum
d = b + 7z + (— ' ~c).
Using the scale given in Fig. 3-9 a, we measure the length d of this vector sum, finding
d = 4.8 m. (Answer)
Sample Problem 3.02 Finding components, airplane flight
A small airplane leaves an airport on an overcast day and is later sighted 215 km away, in a direction making an angle of 22° east of due north. This means that the direction is not due north (directly toward the north) but is rotated 22° to¬ ward the east from due north. How far east and north is the airplane from the airport when sighted?
y
Figure 3-10 A plane takes off from an airport at the origin and is later sighted at P.
KEY IDEA
We are given the magnitude (215 km) and the angle (22° east of due north) of a vector and need to find the components of the vector.
Calculations: We draw an xy coordinate system with the positive direction of x due east and that of y due north (Fig. 3-10). For convenience, the origin is placed at the airport. (We don’t have to do this. We could shift and misalign the coordinate system but, given a choice, why make the prob¬ lem more difficult?) The airplane’s displacement d points from the origin to where the airplane is sighted.
To find the components of d , we use Eq. 3-5 with 9 = 68° (= 90° - 22°):
dx = d cos 9 = (215 km)(cos 68°)
= 81 km (Answer)
dy = d sin 9 = (215 km)(sin 68°)
= 199 km ~ 2.0 X 102 km. (Answer)
Thus, the airplane is 81 km east and 2.0 X 102 km north of the airport.
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3-1 VECTORS AND THEIR COMPONENTS
45
Problem-Solving Tactics Angles, trig functions, and inverse trig functions
Tactic 1: Angles— Degrees and Radians Angles that are measured relative to the positive direction of the x axis are positive if they are measured in the counterclockwise direc¬ tion and negative if measured clockwise. For example, 210° and —150° are the same angle.
Angles may be measured in degrees or radians (rad). To relate the two measures, recall that a full circle is 360° and 2 77 rad. To convert, say, 40° to radians, write
40°
277 rad 360°
0.70 rad.
Quadrants
IV I II
III IV
+1
X"
x
-90° Jf
90° 180° 270° 360°
(a)
Tactic 2: Trig Functions You need to know the definitions of the common trigonometric functions — sine, cosine, and tangent — because they are part of the language of science and engineering. They are given in Fig. 3-11 in a form that does not depend on how the triangle is labeled.
You should also be able to sketch how the trig functions vary with angle, as in Fig. 3-12, in order to be able to judge whether a calculator result is reasonable. Even knowing the signs of the functions in the various quadrants can be of help.
Tactic 3: Inverse Trig Functions When the inverse trig functions sin"1, cos"1, and tan"1 are taken on a calculator, you must consider the reasonableness of the answer you get, because there is usually another possible answer that the calculator does not give. The range of operation for a calculator in taking each inverse trig function is indicated in Fig. 3-12. As an example, sin"1 0.5 has associated angles of 30° (which is displayed by the calculator, since 30° falls within its range of operation) and 150°. To see both values, draw a horizontal line through 0.5 in Fig. 3-l2a and note where it cuts the sine curve. How do you distinguish a cor¬ rect answer? It is the one that seems more reasonable for the given situation.
Tactic 4: Measuring Vector Angles The equations for cos 6 and sin 6 in Eq. 3-5 and for tan 0 in Eq. 3-6 are valid only if the angle is measured from the positive direction of
_
T'l
N\ c
*
1
-9o° o
9(K 18
f V
0° 270° 36
0°
(b)
XT
11 |
-9
0°
9
)° 18
0° 27
0° 36
0°
(c)
Figure 3-12 Three useful curves to remember. A calculator’s range of operation for taking inverse trig functions is indicated by the darker portions of the colored curves.
sin 6 = k;S°PPOS'tce hypotenuse
„ leg adjacent to 0
cos 6 = — \ J , -
hypotenuse
tane= legoppositeS leg adjacent to 6
Leg adjacent to 6
Figure 3-11 A triangle used to define the trigonometric functions. See also Appendix E.
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the x axis. If it is measured relative to some other direc¬ tion, then the trig functions in Eq. 3-5 may have to be in¬ terchanged and the ratio in Eq. 3-6 may have to be inverted. A safer method is to convert the angle to one measured from the positive direction of the x axis. In Wiley PLUS, the system expects you to report an angle of direction like this (and positive if counterclockwise and negative if clockwise).
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46
CHAPTER 3 VECTORS
3-2 UNIT VECTORS, ADDING VECTORS BY COMPONENTS
Learning Objectives _
After reading this module, you should be able to . . .
3.06 Convert a vector between magnitude-angle and unit- vector notations.
3.07 Add and subtract vectors in magnitude-angle notation and in unit-vector notation.
Key Ideas _
• Unit vectors i, j, and k have magnitudes of unity and are directed in the positive directions of the x, y, and z axes, respectively, in a right-handed coordinate system. We can write a vector ~a in terms of unit vectors as
~a = ax i + ay j + az k,
3.08 Identify that, for a given vector, rotating the coordinate system about the origin can change the vector's compo¬ nents but not the vector itself.
in which ax i, av j, and az k are the vector components of ~a and axl ayi and az are its scalar components.
• To add vectors in component form, we use the rules
rx = ax + bx ry = ay + by rz = az + bz.
Here ~a and b are the vectors to be added, and ~r is the vector sum. Note that we add components axis by axis.
Unit Vectors
The unit vectors point along axes.
y
Figure 3.13 Unit vectors i, j, and k define the directions of a right-handed coordinate system.
A unit vector is a vector that has a magnitude of exactly 1 and points in a particu¬ lar direction. It lacks both dimension and unit. Its sole purpose is to point — that is, to specify a direction. The unit vectors in the positive directions of the x, y, and z axes are labeled i, j , and k, where the hat " is used instead of an overhead arrow as for other vectors (Fig. 3-13). The arrangement of axes in Fig. 3-13 is said to be a right-handed coordinate system. The system remains right-handed if it is rotated rigidly. We use such coordinate systems exclusively in this book.
Unit vectors are very useful for expressing other vectors; for example, we can express ~a and b of Figs. 3-7 and 3-8 as
~ci = ax i + ay j (3-7)
and b = bx i + by j. (3-8)
These two equations are illustrated in Fig. 3-14. The quantities ax i and a j are vec¬ tors, called the vector components of ~a. The quantities a, and ay are scalars, called the scalar components of 7f (or, as before, simply its components).
Figure 3-14 (a) The vector components of vector ~a. ( b ) The vector components of vector b.
Adding Vectors by Components
We can add vectors geometrically on a sketch or directly on a vector-capable calculator. A third way is to combine their components axis by axis.
3-2 UNIT VECTORS, ADDING VECTORS BY COMPONENTS
47
To start, consider the statement
7 = ~a + b, (3-9)
which says that the vector 7 is the same as the vector (a + b). Thus, each component of 7 must be the same as the corresponding component of {a + b):
rx = ax + bx (3-10)
ry = ay + by (3-H)
rz = 7- + bz. (3-12)
In other words, two vectors must be equal if their corresponding components are equal. Equations 3-9 to 3-12 tell us that to add vectors ~a and b, we must (1) re¬ solve the vectors into their scalar components; (2) combine these scalar compo¬ nents, axis by axis, to get the components of the sum 7; and (3) combine the components of 7 to get 7 itself. We have a choice in step 3. We can express 7 in unit-vector notation or in magnitude-angle notation.
This procedure for adding vectors by components also applies to vector subtractions. Recall that a subtraction such as d = ~a — b can be rewritten as an addition d = ~a + ( — b). To subtract, we add 7 and — b by components, to get
dx = ax — bx, dy = ay — by, and dz = az — bz, where d = dx i + dv j + dz k. (3-13)
Checkpoint 3
(a) In the figure here, what are the signs of the x components of d1 and d{! (b) What are the signs of the y components of dt and rf2? (c) What are the signs of the x and y components of d1 + d{>.
y
Vectors and the Laws of Physics
So far, in every figure that includes a coordinate system, the x and y axes are par¬ allel to the edges of the book page. Thus, when a vector ~a is included, its compo¬ nents ax and ay are also parallel to the edges (as in Fig. 3-15a). The only reason for that orientation of the axes is that it looks “proper”; there is no deeper reason. We could, instead, rotate the axes (but not the vector ~a) through an angle cf> as in Fig. 3-15b, in which case the components would have new values, call them a'x and a'y. Since there are an infinite number of choices of <b. there are an infinite num¬ ber of different pairs of components for ~a.
Which then is the “right” pair of components? The answer is that they are all equally valid because each pair (with its axes) just gives us a different way of de¬ scribing the same vector Ti\ all produce the same magnitude and direction for the vector. In Fig. 3-15 we have
a = V a l + a 2 = Va(2 + a'y2 (3-14)
and
6= 6' + 4>. (3-15)
The point is that we have great freedom in choosing a coordinate system, be¬ cause the relations among vectors do not depend on the location of the origin or on the orientation of the axes. This is also true of the relations of physics; they are all independent of the choice of coordinate system. Add to that the simplicity and richness of the language of vectors and you can see why the laws of physics are almost always presented in that language: one equation, like Eq. 3-9, can repre¬ sent three (or even more) relations, like Eqs. 3-10, 3-11, and 3-12.
y
°
(a)
Rotating the axes changes the components but not the vector.
y
( b )
Figure 3-15 (a) The vector ~a and its components. ( b ) The same vector, with the axes of the coordinate system rotated through an angle </>.
48
CHAPTER 3 VECTORS
Sample Problem 3.03 Searching through a hedge maze
A hedge maze is a maze formed by tall rows of hedge. After entering, you search for the center point and then for the exit. Figure 3-16a shows the entrance to such a maze and the first two choices we make at the junctions we encounter in moving from point i to point c. We un¬ dergo three displacements as indicated in the overhead view of Fig. 3-16h:
d3 = 6.00 m 0\ = 40°
d2 = 8.00 m 02 = 30°
d3 = 5.00 m 03 = 0°,
where the last segment is parallel to the superimposed x axis. When we reach point c, what are the magnitude and angle of our net displacement dnet from point it
KEY IDEAS
(1) To find the net displacement dDet, we need to sum the three individual displacement vectors:
dnet = di + d2 + d3.
(2) To do this, we first evaluate this sum for the x compo¬ nents alone,
dnet ,* d|A T d2x + d3x, (3-16)
and then the y components alone,
dnet,_y d | v, d2y T d3y. (3-17)
(3) Finally, we construct c/net from its x and y components.
Calculations: To evaluate Eqs. 3-16 and 3-17, we find the x and y components of each displacement. As an example, the com¬ ponents for the first displacement are shown in Fig. 3-16c. We draw similar diagrams for the other two displacements and then we apply the x part of Eq. 3-5 to each displacement, using angles relative to the positive direction of the x axis:
dLr = (6.00 m) cos 40° = 4.60 m d2x = (8.00 m) cos ( — 60°) = 4.00 m d3x = (5.00 m) cos 0° = 5.00 m.
Equation 3-16 then gives us
dnet,* = +4.60 m + 4.00 m + 5.00 m = 13.60 m.
Similarly, to evaluate Eq. 3-17, we apply the y part of Eq. 3-5 to each displacement:
dlv = (6.00 m) sin 40° = 3.86 m d2 y = (8.00 m) sin (—60°) = —6.93 m d3y = (5.00 m) sin 0° = 0 m.
Equation 3-17 then gives us
dnei y = +3.86 m — 6.93 m + 0 m = -3.07 m.
Next we use these components of dnet to construct the vec¬ tor as shown in Fig. 3-16 d: the components are in a head-to- tail arrangement and form the legs of a right triangle, and
Figure 3-1 6 (a) Three displacements through a hedge maze, (b) The displacement vectors, (c) The first displacement vector and its components. ( d ) The net displacement vector and its components.
3-2 UNIT VECTORS, ADDING VECTORS BY COMPONENTS
49
the vector forms the hypotenuse. We find the magnitude and angle of dnet with Eq. 3-6. The magnitude is
dnet = Vd2ne u + d2neUy (3-18)
= V(13.60 m)2 + (—3.07 m)2 = 13.9 m. (Answer)
To find the angle (measured from the positive direction of x ), we take an inverse tangent:
9 = tan"1
\ ^net,x
= tan”1 ( 7!'°! m ) = _12.7°. (Answer) \ 13.60 m / v ’
The angle is negative because it is measured clockwise from
positive x. We must always be alert when we take an inverse
tangent on a calculator. The answer it displays is mathe¬ matically correct but it may not be the correct answer for the physical situation. In those cases, we have to add 180° to the displayed answer, to reverse the vector. To check, we always need to draw the vector and its components as we did in Fig. 3-16d. In our physical situation, the figure shows us that 0 = —12.7° is a reasonable answer, whereas -12.7° + 180° = 167° is clearly not.
We can see all this on the graph of tangent versus angle in Fig. 3-12c. In our maze problem, the argument of the in¬ verse tangent is -3.07/13.60, or -0.226. On the graph draw a horizontal line through that value on the vertical axis. The line cuts through the darker plotted branch at —12.7° and also through the lighter branch at 167°. The first cut is what a calculator displays.
(3-19)
Sample Problem 3.04 Adding vectors, unit-vector components
Figure 3- 1 la shows the following three vectors:
~a = (4.2 m)i — (1.5 m)j, b = (—1.6 m)i + (2.9 m)j, and ~c = (—3.7 m)j.
What is their vector sum 7 which is also shown?
y
Figure 3-17 Vector 7 is the vector sum of the other three vectors.
KEY IDEA
We can add the three vectors by components, axis by axis, and then combine the components to write the vector sum 7.
Calculations: For the x axis, we add the x components of 7, b, and 7, to get the x component of the vector sum 7:
rx = ax + bx + cx
= 4.2 m — 1.6 m + 0 = 2.6 m.
Similarly, for the y axis,
ry = ay + by + cy
= — 1.5 m + 2.9 m — 3.7 m = —2.3 m.
We then combine these components of 7 to write the vector in unit-vector notation:
7 = (2.6 m)i — (2.3 m)j, (Answer)
where (2.6 m)i is the vector component of 7 along the x axis and —(2.3 m)j is that along the y axis. Figure 3-176 shows one way to arrange these vector components to form 7. (Can you sketch the other way?)
We can also answer the question by giving the magnitude and an angle for 7. From Eq. 3-6, the magnitude is
r = V(2.6m)2 + (-2.3 m)2 ~ 3.5 m (Answer)
and the angle (measured from the +x direction) is
0 = tan ^ 26 = (Answer)
where the minus sign means clockwise.
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50
CHAPTER 3 VECTORS
3-3 MULTIPLYING VECTORS
Learning Objectives _
After reading this module, you should be able to . . .
3.09 Multiply vectors by scalars.
3.1 0 Identify that multiplying a vector by a scalar gives a vec¬ tor, taking the dot (or scalar) product of two vectors gives a scalar, and taking the cross (or vector) product gives a new vector that is perpendicular to the original two.
3.1 1 Find the dot product of two vectors in magnitude-angle notation and in unit-vector notation.
3.12 Find the angle between two vectors by taking their dot prod¬ uct in both magnitude-angle notation and unit-vector notation.
Key Ideas _
• The product of a scalar s and a vector "visa new vector whose magnitude is sv and whose direction is the same as that of 7 if s is positive, and opposite that of 7 if s is negative. To divide 7 by s, multiply 7 by 1/s.
• The scalar (or dot) product of two vectors ~d and b is writ¬ ten ~a • b and is the scalar quantity given by
a - b = ab cos (/>,
in which (f> is the angle between the directions of ~a and b.
A scalar product is the product of the magnitude of one vec¬ tor and the scalar component of the second vector along the direction of the first vector. In unit-vector notation,
~a • b = ( axi + av j + azk)-(bxi + by j + bzk),
which may be expanded according to the distributive law.
Note that ~a • b = b • ~a.
3.1 3 Given two vectors, use a dot product to find how much of one vector lies along the other vector.
3.14 Find the cross product of two vectors in magnitude- angle and unit-vector notations.
3.1 5 Use the right-hand rule to find the direction of the vector that results from a cross product.
3.16 In nested products, where one product is buried inside another, follow the normal algebraic procedure by starting with the innermost product and working outward.
• The vector (or cross) product of two vectors ~a and b is written ~a X b and is a vector~c whose magnitude c is given by
c = ab sin </>,
in which cf) is the smaller of the angles between the directions of ~a and b. The direction of 7 is perpendicular to the plane defined by ~a and b and is given by a right-hand rule, as shown in Fig. 3-19. Note that 7z X b = —(b X a). In unit-vector notation,
~a X b = (ax i + avj + a zk) X (bx i + by j + bzk ),
which we may expand with the distributive law.
• In nested products, where one product is buried inside an¬ other, follow the normal algebraic procedure by starting with the innermost product and working outward.
Multiplying Vectors*
There are three ways in which vectors can be multiplied, but none is exactly like the usual algebraic multiplication. As you read this material, keep in mind that a vector-capable calculator will help you multiply vectors only if you understand the basic rules of that multiplication.
Multiplying a Vector by a Scalar
If we multiply a vector a by a scalar .v. we get a new vector. Its magnitude is the product of the magnitude of ~a and the absolute value of s. Its direction is the direction of 7z if s is positive but the opposite direction if .v is negative. To divide ~a by s, we multiply ~a by 1/s.
Multiplying a Vector by a Vector
There are two ways to multiply a vector by a vector: one way produces a scalar (called the scalar product ), and the other produces a new vector (called the vector product). (Students commonly confuse the two ways.)
*This material will not be employed until later (Chapter 7 for scalar products and Chapter 11 for vec¬ tor products), and so your instructor may wish to postpone it.
3-3 MULTIPLYING VECTORS
51
The Scalar Product
The scalar product of the vectors ~a and b in Fig. 3-18n is written as ~a • b and defined to be
~a • b = ab cos cj), (3-20)
where a is the magnitude of ~a, b is the magnitude of b , and </> is the angle between ~a and b (or, more properly, between the directions of o’ and b). There are actually two such angles: cj) and 360° — 4>. Either can be used in Eq. 3-20, because their cosines are the same.
Note that there are only scalars on the right side of Eq. 3-20 (including the value of cos c/q. Thus ~a • b on the left side represents a scalar quantity. Because of the notation, ~a • b is also known as the dot product and is spoken as “a dot b.”
A dot product can be regarded as the product of two quantities: (1) the mag¬ nitude of one of the vectors and (2) the scalar component of the second vector along the direction of the first vector. For example, in Fig. 3- 1 Hb. ~a has a scalar component a cos cj) along the direction of b \ note that a perpendicular dropped from the head of ~a onto b determines that component. Similarly, b has a scalar component b cos <j> along the direction of ~a .
© If the angle <f> between two vectors is 0°, the component of one vector along the other is maximum, and so also is the dot product of the vectors. If, instead, <f> is 90°, the component of one vector along the other is zero, and so is the dot product.
Equation 3-20 can be rewritten as follows to emphasize the components:
~a -b = (a cos cf>)(b) = ( a)(b cos cf>). (3-21)
The commutative law applies to a scalar product, so we can write
~a ‘b = b • ~a.
When two vectors are in unit-vector notation, we write their dot product as
~d‘b = (ax i + ay j + az k) • (bx\ + by j + bz k), (3-22)
which we can expand according to the distributive law: Each vector component of the first vector is to be dotted with each vector component of the second vec¬ tor. By doing so, we can show that
~a ‘b = axbx + ayby + azbz. (3-23)
Figure 3-18 (a) Two vectors a and b, with an angle <f> between them. ( b ) Each vector has a component along the direction of the other vector.
Component of b along direction of
a is b cos (j)
Multiplying these gives the dot product.
Or multiplying these gives the dot product
Component of a along direction of b is a cos 0
52
CHAPTER 3 VECTORS
0
Checkpoint 4
Vectors C and D have magnitudes of 3 units and 4 units, respectively. What is the angle between the directions of C and D if C • D equals (a) zero, (b) 12 units, and (c) —12 units?
The Vector Product
The vector product of 7? and b, written 7? X b, produces a third vector ~c whose magnitude is
c = ab sin <£, (3-24)
where </> is the smaller of the two angles between 7? and b. (You must use the smaller of the two angles between the vectors because sin cf> and sin(360° — cf>) differ in algebraic sign.) Because of the notation,?? X b is also known as the cross product, and in speech it is “a cross b.”
o If 7? and b are parallel or antiparallel, ~a X b = 0. The magnitude of a X b , which can be written as | a X b | , is maximum when ~a and b are perpendicular to each other.
The direction of ~c is perpendicular to the plane that contains 7? and b. Figure 3-19a shows how to determine the direction of ~c = 7? X b with what is known as a right-hand rule. Place the vectors 7? and b tail to tail without altering their orientations, and imagine a line that is perpendicular to their plane where they meet. Pretend to place your right hand around that line in such a way that your fingers would sweep 7? into b through the smaller angle between them. Your outstretched thumb points in the direction of ~c.
The order of the vector multiplication is important. In Fig. 3-19b, we are determining the direction of ~c'= b X 7?, so the fingers are placed to sweep b into 7? through the smaller angle. The thumb ends up in the opposite direction from previously, and so it must be that?:' = —7?; that is,
b X 7? = —(7? X b ). (3-25)
In other words, the commutative law does not apply to a vector product.
In unit-vector notation, we write
7 J X b = (axi + ay j + n-k) X {bx i + by j + b-k), (3-26)
which can be expanded according to the distributive law; that is, each component of the first vector is to be crossed with each component of the second vector. The cross products of unit vectors are given in Appendix E (see “Products of Vectors”). For example, in the expansion of Eq. 3-26, we have
ax i X bxi = axbx(i X i) = 0,
because the two unit vectors i and i are parallel and thus have a zero cross prod¬ uct. Similarly, we have
ax i X bvj = axby{\ X j) = axby k.
In the last step we used Eq. 3-24 to evaluate the magnitude of i X j as unity. (These vectors i and j each have a magnitude of unity, and the angle between them is 90°. ) Also, we used the right-hand rule to get the direction of i X j as being in the positive direction of the z axis (thus in the direction of k).
3-3 MULTIPLYING VECTORS
53
Continuing to expand Eq. 3-26, you can show that
~a X b = (avbz — byaz) i + ( azbx — b.ax )j + ( axby — bxay) k. (3-27)
A determinant (Appendix E) or a vector-capable calculator can also be used.
To check whether any xyz coordinate system is a right-handed coordinate system, use the right-hand rule for the cross product i X j = k with that system. If your fingers sweep i (positive direction of x) into j (positive direction of y) with the outstretched thumb pointing in the positive direction of z (not the negative direction), then the system is right-handed.
Checkpoint 5
Vectors C and D have magnitudes of 3 units and 4 units, respectively. What is the an¬ gle between the directions of C and D if the magnitude of the vector product C X D is (a) zero and (b) 12 units?
0
Figure 3-19 Illustration of the right-hand rule for vector products, (a) Sweep vector ~a into vector b with the fingers of your right hand. Your outstretched thumb shows the direction of vector ~c = ~a X b. (b) Showing that b X ~a is the reverse of ~a X b.
54
CHAPTER 3 VECTORS
Sample Problem 3.05 Angle between two vectors using dot products
What is the angle (f> between a = 3.0i — 4.0j and b = — 2.0i + 3.0k? ( Caution : Although many of the following steps can be bypassed with a vector-capable calculator, you will learn more about scalar products if, at least here, you use these steps.)
KEY IDEA
The angle between the directions of two vectors is included in the definition of their scalar product (Eq. 3-20):
~a • b = ab cos </>. (3-28)
Calculations: In Eq. 3-28, a is the magnitude of a, or
a =V3.02 + (— 4.0)2 = 5.00, (3-29)
and b is the magnitude of b, or
b = V(-2.0)2 + 3.02 = 3.61. (3-30)
We can separately evaluate the left side of Eq. 3-28 by writing the vectors in unit-vector notation and using the distributive law:
a-b = (3.0i - 4.0j) •(— 2.0t + 3.0k)
= (3.0t) •(— 2.0t) + (3.0i) • (3.0k)
+ (-4.0j)-(-2.0i) + (-4.0j)-(3.0k).
We next apply Eq. 3-20 to each term in this last expression. The angle between the unit vectors in the first term (i and i) is 0°, and in the other terms it is 90°. We then have
a-b = -(6.0)(1) + (9.0)(0) + (8.0)(0) - (12)(0)
= -6.0.
Substituting this result and the results of Eqs. 3-29 and 3-30 into Eq. 3-28 yields
-6.0 = (5.00)(3.61) cos <£,
S° (5. 00X3.61) - 109""110" Answer)
Sample Problem 3.06 Cross product, right-hand rule
In Fig. 3-20, vector a lies in the xy plane, has a magnitude of 18 units, and points in a direction 250° from the positive di¬ rection of the x axis. Also, vector b has a magnitude of 12 units and points in the positive direction of the z axis. What is the vector product ~c = a X bl
KEY IDEA
When we have two vectors in magnitude-angle notation, we find the magnitude of their cross product with Eq. 3-24 and the direction of their cross product with the right-hand rule of Fig. 3-19.
Calculations: For the magnitude we write
c = ab sin 4> = (18)(12)(sin 90°) = 216. (Answer)
To determine the direction in Fig. 3-20, imagine placing the fingers of your right hand around a line perpendicular to the plane of ~a and b (the line on which ~c is shown) such that your fingers sweep ~a into b. Your outstretched thumb then
Figure 3-20 Vector ~c (in the xy plane) is the vector (or cross) product of vectors ~a and b.
gives the direction of ~c. Thus, as shown in the figure, ~c lies in the xy plane. Because its direction is perpendicular to the direction of ~ci (a cross product always gives a perpendicular vector), it is at an angle of
250° - 90° = 160° (Answer)
from the positive direction of the x axis.
Sample Problem 3.07 Cross product, unit-vector notation
If ~a = 3i — 4j and b = — 2i + 3k, what is ~c = 7z X bl
KEY IDEA
Calculations: Here we write ~c — (3t — 4j) X (— 2t + 3k)
= 3t X (— 2t) + 3i X 3k + (—4]) X (— 2t)
+ (-4f) X 3k.
When two vectors are in unit-vector notation, we can find their cross product by using the distributive law.
REVIEW & SUMMARY
55
We next evaluate each term with Eq. 3-24, finding the direction with the right-hand rule. For the first term here, the angle </> between the two vectors being crossed is 0. For the other terms, cf> is 90°. We find
c = -6(0) + 9(— j) + 8(— k) - 12i = — 12i — 9j — 8k. (Answer)
This vector ~c is perpendicular to both ~a and b, a fact you can check by showing that ~c • ~a = 0 and ~c • b = 0; that is, there is no component of"? along the direction of either ~a or b.
In general: A cross product gives a perpendicular vector, two perpendicular vectors have a zero dot prod¬ uct, and two vectors along the same axis have a zero cross product.
^WILEY©
PLUS
Additional examples, video, and practice available at WiteyPLUS
eview & Summary
Scalars and Vectors Scalars, such as temperature, have magni¬ tude only. They are specified by a number with a unit (10°C) and obey the rules of arithmetic and ordinary algebra. Vectors, such as displacement, have both magnitude and direction (5 m, north) and obey the rules of vector algebra.
Adding Vectors Geometrically Two vectors a and b may
be added geometrically by drawing them to a common scale and placing them head to tail. The vector connecting the tail of the first to the head of the second is the vector sum ~s. To subtract b from 7f, reverse the direction of b to get —b; then add —ft to a. Vector addition is commutative
~a + ft = ft + ~a
(3-2)
and obeys the associative law
(a + ft) + ? = ~a + (ft + ?).
(3-3)
Components of a Vector The (scalar) components ax and ay of any two-dimensional vector ~a along the coordinate axes are found by dropping perpendicular lines from the ends of ? onto the coor¬ dinate axes. The components are given by
ax — a cos 8 and ay = a sin 8, (3-5)
where 8 is the angle between the positive direction of the x axis and the direction of ~a. The algebraic sign of a component indi¬ cates its direction along the associated axis. Given its compo¬ nents, we can find the magnitude and orientation (direction) of the vector ~a by using
a = s/ax + ay and tan 9 = — (3-6)
^ X
Unit-Vector Notation Unit vectors i, j, and k have magnitudes of unity and are directed in the positive directions of the x, y, and z axes, respectively, in a right-handed coordinate system (as defined by the vector products of the unit vectors). We can write a vector ~a in terms of unit vectors as
~a = ax\ + ay) +fl-k, (3-7)
in which aAi, ay j , and a7k are the vector components of 7! and a,. ay,
and a, are its scalar components.
Adding Vectors in Component Form To add vectors in com¬ ponent form, we use the rules
rx = ax + bx ry = ay + by rz — az + bz. (3-10 to 3-12)
Here ~a and ft are the vectors to be added, and 7 is the vector sum. Note that we add components axis by axis. We can then express the sum in unit-vector notation or magnitude-angle notation.
Product of a Scalar and a Vector The product of a scalar s and a vector 7 is a new vector whose magnitude is sv and whose direc¬ tion is the same as that of 7 if s is positive, and opposite that of 7 if s is negative. (The negative sign reverses the vector.) To divide 7 by s, multiply 7 by 1 Is.
The Scalar Product The scalar (or dot) product of two vectors ~a and ft is written ~a • ft and is the scalar quantity given by
~a- ft = ab cos <f> , (3-20)
in which 4> is the angle between the directions of ~a and ft. A scalar product is the product of the magnitude of one vector and the scalar component of the second vector along the direction of the first vector. Note that 7t-b =b-~a, which means that the scalar product obeys the commutative law.
In unit-vector notation,
Tf-ft = (ax\ + ay] + azk)-(bxi + ftvj + ftzk), (3-22) which may be expanded according to the distributive law.
The Vector Product The vector (or cross) product of two vectors ~a and ft is written ~a X ft and is a vector 7 whose magnitude c is given by
c = ab sin <J> , (3-24)
in which </> is the smaller of the angles between the directions of ~a and ft. The direction of ~c is perpendicular to the plane defined by ~a and ft and is given by a right-hand rule, as shown in Fig. 3-19. Note that ~a X ft = — (ft X ~a), which means that the vec¬ tor product does not obey the commutative law.
In unit-vector notation,
~a X ft = (ax i + ayj + a.k) X (bx i + ftvj + ftzk), (3-26) which we may expand with the distributive law.
56
CHAPTER 3 VECTORS
questions
1 Can the sum of the magnitudes of two vectors ever be equal to the magnitude of the sum of the same two vectors? If no, why not? If yes, when?
2 The two vectors shown in Fig. 3-21 lie in an xy plane. What are the signs of the x and y components, respec¬ tively, of (a) dx + tl2, (b) d1 — d2, and (c) d2 ~ dil
3 Being part of the “Gators,” the University of Florida golfing team must play on a putting green with an alligator pit. Figure 3-22 shows an overhead view of one putting chal¬ lenge of the team; an xy coordinate system is superimposed. Team mem¬ bers must putt from the origin to the hole, which is at xy coordinates (8 m,
12 m), but they can putt the golf ball using only one or more of the fol¬ lowing displacements, one or more times:
d1 = (8 m)i + (6 m)j, d2 = (6 m)j, d3 = (8 m)i.
The pit is at coordinates (8 m, 6 m). If a team member putts the ball into or through the pit, the member is automatically trans¬ ferred to Florida State University, the arch rival. What sequence of displacements should a team member use to avoid the pit and the school transfer?
4 Equation 3-2 shows that the addition of two vectors ~a and b is commutative. Does that mean subtraction is commutative, so that ~a — b = b —~al
5 Which of the arrangements of axes in Fig. 3-23 can be labeled “right-handed coordinate system”? As usual, each axis label indi¬ cates the positive side of the axis.
oHole
D
Gator
pit
Figure 3-22 Question 3.
6 Describe two vectors ~a and b such that
(a) ~a + b = ~c and a + b = c;
(b) 7f + b = ~a — b\
(c) ~a + b = ~c and a2 + b2 = c2.
7 If c? = a + b + (—"?), does (a) ~a + ( — d) = ~c + (~b), (b) ~a = (— b) + d + T, and (c) ~c + (— d) = ~a + bl
8 If ~a • b = ~a • ~c , must b equal "c?
9 If F = q(v X B) and ~v is perpendicular to B, then what is the direction of B in the three situations shown in Fig. 3-24 when con¬ stant q is (a) positive and (b) negative?
Figure 3-24 Question 9.
10 Figure 3-25 shows vector A and four other vectors that have the same magnitude but differ in orientation.
(a) Which of those other four vectors have the same dot product with A? (b)
Which have a negative dot product with A?
11 In a game held within a three- dimensional maze, you must move Figure 3-25 Question 10. your game piece from start , at xyz co¬ ordinates (0, 0, 0), to finish, at coordinates (—2 cm, 4 cm, —4 cm). The game piece can undergo only the displacements (in centime¬ ters) given below. If, along the way, the game piece lands at coordi¬ nates (—5 cm, —1 cm, —1 cm) or (5 cm, 2 cm, —1 cm), you lose the game. Which displacements and in what sequence will get your game piece to finish!
p — — 7i + 2j — 3k ~r = 2i — 3j + 2k q = 2i — j + 4k ~s = 3i + 5j — 3k.
12 The x and y components of four vectors ~a ,b ,~c, and d are given below. For which vectors will your calculator give you the correct an¬ gle 6 when you use it to find 6 with Eq. 3-6? Answer first by examin¬ ing Fig. 3-12, and then check your answers with your calculator.
ax = 3 a = 3 cx— —3
bx = —3 by = 3 dx = !>
-3
-3.
13 Which of the following are correct (meaningful) vector expressions? What is wrong with any incorrect expression?
(a) A • (B -C)
(b) A X (B ■ C)
(c) A-(BX C)
(d) A X (B X C)
(e) A + (B ■ C)
(f) A + (B X C)
(g) 5+A
(h) 5 + 0 ■ C)
(i) 5 + (B X C)
0) (A • B) + (B X C)
PROBLEMS
57
roblems
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign SSM Worked-out solution available in Student Solutions Manual WWW Worked-out solution is at
• - ••• Number of dots indicates level of problem difficulty ILW Interactive solution is at
T^yM***' Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
http://www.wiley.com/college/halliday
Module 3-1 Vectors and Their Components
•1 SSM What are (a) the x component and (b) the y component of a vector 7 in the xy plane if its direction is 250° counterclockwise from the positive direction of the x axis and its magnitude is 7.3 m?
•2 A displacement vector 7 in the xy plane is 15 m long and directed at angle 0 = 30° in Fig. 3-26. Determine (a) the x component and (b) the y component of the vector.
•3 SSM The x component of vector A is —25.0 m and the y component is +40.0 m. (a) What is the magni¬ tude of A? (b) What is the angle between the direction of A and the positive direction of x?
•4 Express the following angles in radians: (a) 20.0°, (b) 50.0°, (c) 100°. Convert the following angles to degrees: (d) 0.330 rad, (e) 2.10 rad, (f)7.70 rad.
•5 A ship sets out to sail to a point 120 km due north. An unex¬ pected storm blows the ship to a point 100 km due east of its starting point, (a) How far and (b) in what direction must it now sail to reach its original destination?
•6 In Fig. 3-27, a heavy piece of machinery is raised by sliding it a distance d — 12.5 m along a plank oriented at angle 6 — 20.0° to the horizontal. How far is it moved (a) vertically and (b) horizontally?
•7 Consider two displacements, one of magnitude 3 m and another Figure 3-27 Problem 6. of magnitude 4 m. Show how the
displacement vectors may be combined to get a resultant displace¬ ment of magnitude (a) 7 m, (b) 1 m, and (c) 5 m.
Module 3-2 Unit Vectors, Adding Vectors by Components
•8 A person walks in the following pattern: 3.1 km north, then 2.4 km west, and finally 5.2 km south, (a) Sketch the vector dia¬ gram that represents this motion, (b) How far and (c) in what di¬ rection would a bird fly in a straight line from the same starting point to the same final point?
•9 Two vectors are given by
7 = (4.0 m)i — (3.0 m)j + (1.0 m)k
and b = (-l.Om)i + (1.0m)j + (4.0 m)k.
In unit-vector notation, find (a) 7 + b, (b) 7 — b, and (c) a third vector 7 such that 7 — b + 7 — 0.
•10 Find the (a) x, (b) y, and (c) z components of the sum 7 of the displacements 7 and d whose components in meters are cx — 7A,cy = — 3.8,cz = —6.1 \dx = 4.4, 7^, = —2.0 ,dz = 3.3.
•11 SSM (a) In unit- vector notation, what is the sum 7 + b if 7 = (4.0 m)i + (3.0 m)j and b = (— 13.0m)i + (7.0 m)j? What are the (b) magnitude and (c) direction of 7 + bl
Figure 3-26
Problem 2.
•12 A car is driven east for a distance of 50 km, then north for 30 km, and then in a direction 30° east of north for 25 km. Sketch the vector diagram and determine (a) the magnitude and (b) the angle of the car’s total displacement from its starting point.
•13 A person desires to reach a point that is 3.40 km from her present location and in a direction that is 35.0° north of east. However, she must travel along streets that are oriented either north-south or east-west. What is the minimum distance she could travel to reach her destination?
•14 You are to make four straight-line moves over a flat desert floor, starting at the origin of an xy coordinate system and ending at the xy coordinates (—140 m, 30 m). The x component and y component of your moves are the following, respectively, in me¬ ters: (20 and 60), then ( bx and —70), then (—20 and cy ), then (—60 and —70). What are (a) component bx and (b) component cy7 What are (c) the magnitude and (d) the angle (relative to the pos¬ itive direction of the x axis) of the overall displacement?
•15 SSM ILW WWW The two vec¬ tors 7 and b in Fig. 3-28 have equal magnitudes of 10.0 m and the angles are d1 = 30° and d2 = 105°. Find the (a) x and (b) y components of their vector sum 7, (c) the magnitude of 7, and (d) the angle 7 makes with the positive direction of the x axis.
•16 For the displacement vectors 7 = (3.0 m)i + (4.0 m)j and b =
(5.0 m)i + (-2.0 m)j, give ~a + b in (a) unit-vector notation, and as (b) a magnitude and (c) an angle (rela¬ tive to i). Now give b — 7 in (d) unit-vector notation, and as (e) a magnitude and (f) an angle.
Figure 3-28 Problem 15.
•17 © ILW Three vectors 7, h. and 7 each have a magnitude of 50 m and lie in an xy plane. Their directions relative to the positive direction of the x axis are 30°, 195°, and 315°, respectively. What are (a) the magnitude and (b) the angle of the vector 7 + b + 7, and (c) the magnitude and (d) the angle of 7 — b +7? What are the (e) magnitude and (f) angle of a fourth vector d such that (7 + b) — (7 + d) = 0?
•1 8 In the sum A + B = C, vector A has a magnitude of 12.0 m and is angled 40.0° counterclockwise from the +x direction, and vec¬ tor C has a magnitude of 15.0 m and is angled 20.0° counterclock¬ wise from the — x direction. What are (a) the magnitude and (b) the angle (relative to +x ) of B ?
•19 In a game of lawn chess, where pieces are moved between the centers of squares that are each 1.00 m on edge, a knight is moved in the following way: (1) two squares forward, one square rightward; (2) two squares leftward, one square forward; (3) two squares forward, one square leftward. What are (a) the magnitude and (b) the angle (relative to “forward”) of the knight’s overall dis¬ placement for the series of three moves?
58
CHAPTER 3 VECTORS
••20 — ^Tr~ An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) while returning to base camp. He was supposed to travel due north for 5.6 km, but when the snow clears, he discovers that he actually traveled 7.8 km at 50° north of due east, (a) How far and (b) in what direction must he now travel to reach base camp?
••21 © An ant, crazed by the Sun on a hot Texas afternoon, darts over an xy plane scratched in the dirt. The x and y components of four consecutive darts are the following, all in centimeters: (30.0, 40.0), (bx, -70.0), (-20.0, cy), (-80.0, -70.0). The overall displace¬ ment of the four darts has the xy components (—140, —20.0). What are (a) bx and (b) cyl What are the (c) magnitude and (d) angle (relative to the positive direction of the x axis) of the overall displacement?
••22 (a) What is the sum of the following four vectors in unit-
vector notation? For that sum, what are (b) the magnitude, (c) the angle in degrees, and (d) the angle in radians?
E: 6.00 m at +0.900 rad F: 5.00 m at —75.0°
G: 4.00 m at +1.20 rad H: 6.00 m at -210°
••23 If B is added to C = 3.0i + 4.0j , the result is a vector in the positive direction of the y axis, with a magnitude equal to that of C. What is the magnitude of B?
••24 Q Vector A, which is directed along an x axis, is to be added to vector B , which has a magnitude of 7.0 m. The sum is a third vec¬ tor that is directed along the y axis, with a magnitude that is 3.0 times that of A. What is that magnitude of A?
••25 Q Oasis B is 25 km due east of oasis A. Starting from oasis A, a camel walks 24 km in a direction 15° south of east and then walks 8.0 km due north. How far is the camel then from oasis B1
••26 What is the sum of the following four vectors in (a) unit- vector notation, and as (b) a magnitude and (c) an angle?
an ant’s displacement from the nest (find it in the figure) if the ant enters the trail at point A? What are the (c) magnitude and (d) angle if it enters at point B ?
••30 ® Here are two vectors:
~a = (4.0 m)i — (3.0 m)j and b = (6.0 m)i + (8.0 m)j.
What are (a) the magnitude and (b) the angle (relative to i) of 7f? What are (c) the magnitude and (d) the angle of h? What are (e) the magnitude and (f) the angle of 7? + b\ (g) the magnitude and (h) the angle of b — ~a\ and (i) the magnitude and (j) the angle of ~ci — h? (k) What is the angle between the directions of b — ~a and ~a — bl
••31 In Fig. 3-30, a vector ~a with a magnitude of 17.0 m is directed at angle 0 = 56.0° counterclockwise from the +x axis. What are the components (a) ax and (b) ay of the vector? A sec¬ ond coordinate system is inclined by angle 6' = 18.0° with respect to the first. What are the components (c) a'x and (d) a'y in this primed coordinate system?
A = (2.00 m)i + (3.00 m)j B: 4.00 m, at +65.0°
C = (-4.00 m)5 + (-6.00 m)J D : 5.00 m, at -235°
••27 ® If d1+ d2 = 5 d3, dl — d2 = 3 d3, and d3 = 2i + 4j, then what are, in unit-vector notation, (a) d1 and (b) d21
••28 Two beetles run across flat sand, starting at the same point. Beetle 1 runs 0.50 m due east, then 0.80 m at 30° north of due east. Beetle 2 also makes two runs; the first is 1.6 m at 40° east of due north. What must be (a) the magnitude and (b) the direction of its second run if it is to end up at the new location of beetle 1?
••29 -iSB? O Typical backyard ants often create a network of chemical trails for guidance. Extending outward from the nest, a trail branches ( bifurcates ) repeatedly, with 60° between the branches. If a roaming ant chances upon a trail, it can tell the way to the nest at any branch point: If it is moving away from the nest, it has two choices of path requiring a small turn in its travel direction, either 30° leftward or 30° rightward. If it is moving toward the nest, it has only one such choice. Figure 3-29 shows a typical ant trail, with lettered straight sec¬ tions of 2.0 cm length and symmetric bifurcation of 60°. Path v is parallel to the y axis. What are the (a) magnitude and (b) angle (relative to the positive direction of the superimposed x axis) of
Figure 3-30 Problem 31.
•••32 In Fig. 3-31, a cube of edge length a sits with one corner at the ori¬ gin of an xyz coordinate system. A body diagonal is a line that extends from one corner to another through the center. In unit-vector notation, what is the body diagonal that extends from the corner at (a) coordinates (0,
0, 0), (b) coordinates (a, 0, 0), (c) coor¬ dinates (0, a, 0), and (d) coordinates (a, a, 0)? (e) Determine the
z
Figure 3-31 Problem 32.
PROBLEMS
59
angles that the body diagonals make with the adjacent edges, (f) Determine the length of the body diagonals in terms of a.
Module 3-3 Multiplying Vectors
•33 For the vectors in Fig. 3-32, with a = 4, are (a) the magnitude and (b) the direction of 7 X b, (c) the magnitude and (d) the di¬ rection of aXc, and (e) the magnitude and (f) the direction of b X 7' ? (The z axis is not shown.)
•34 Two vectors are presented as 7? = 3.0i + 5.0j and b = 2.0i + 4.0j. Find (a) 7 1 X b, (b) 7 • b, (c) (7? + b)- b, and (d) the component of 7? along the direc¬ tion of b. (Hint: For (d), consider Eq. 3-20 and Fig. 3-18.)
b = 3, and c = 5, what
•35 Two vectors, 7 and 7, lie in the xy plane. Their magnitudes are 4.50 and 7.30 units, respectively, and their directions are 320° and 85.0°, respectively, as measured counterclockwise from the positive x axis. What are the values of (a) 7 • 7 and (b) 7x7?
•36 If d1 = 3i — 2j T 4k and d2 — — 5i + 2j — k, then what is (d1 + d2 ) • (d1 X 4 72)?
•37 Three vectors are given by 7 = 3.0i + 3.0j — 2.0k, b = — l.Oi — 4.0j + 2.0k, and 7 = 2.0i + 2.0j + 1.0k. Find (a) 7? • (b X 7), (b)7-(h + 7), and (c) 7 X (b + 7).
••38 ® For the following three vectors, what is 3C • (2/1 X B)1
A = 2.00i + 3.00J - 4.00k
3 = -3.00i + 4.00] + 2.00k C = 7.00i - 8.00]
••39 Vector A has a magnitude of 6.00 units, vector B has a mag¬ nitude of 7.00 units, and A • B has a value of 14.0. What is the angle between the directions of A and B1
••40 ® Displacement d2 is in the yz plane 63.0° from the positive direction of the y axis, has a positive z component, and has a mag¬ nitude of 4.50 m. Displacement d2 is in the xz plane 30.0° from the positive direction of the x axis, has ajtositive z component, and has magnitude 1.40 m. What are (a) d1 • d2, (b) d1 X d2, and (c) the an¬ gle between dl and 72?
••41 SSM ILW WWW Use the definition of scalar product, 7 • b = ab cos 6 , and the fact that 7 • b = axbx + ayby + azbz to cal¬ culate the angle between the two vectors given by 7 = 3.0i + 3.0j + 3.0k and b — 2.0i + l.Oj + 3.0k.
••42 In a meeting of mimes, mime 1 goes through a displacement d1 = (4.0 m)i + (5.0 m)j and mime 2 goes through a displacement d2 = (— 3.0m)i + (4.0 m)j. What are (a) d1 X d2, (b) dx- d2, (c) (d2 + d2) • d2, and (d) the com¬ ponent of rfj along the direction of d2l (Hint: For (d), see Eq. 3-20 and Fig. 3-18.)
••43 SSM ILW The three vectors in Fig. 3-33 have magnitudes a = 3.00 m, b = 4.00 m, and c = 10.0 m and angle 0 = 30.0°. What are (a) the x compo¬ nent and (b) the y component of 7; (c)
the x component and (d) the y com- Figure 3-33 Problem 43.
ponent of b ; and (e) the x component and (f) the y component of 7? If 7 = pa + qb, what are the values of (g)p and (h)g?
••44 ® In the product F = q~v X B, take q = 2,
7 = 2.0i + 4.0j + 6.0k and F = 4.0i - 20j + 12k.
What then is B in unit-vector notation if Bx = Byl
Additional Problems
45 Vectors A and B lie in an xy plane. A has magnitude 8.00 and angle 130°; B has components Bx = —7.72 and By — —9.20. (a) What is 5/1 • B ? What is 4A X 3 B in (b) unit-vector notation and (c) magnitude-angle notation with spherical coordinates (see Fig. 3-34)? (d) What is the angle between the directions of A and 4A X 3 B? (Hint: Think a bit before you resort to a calculation.) What is A + 3.00k in (e) unit-vector notation and (f) magnitude- angle notation with spherical coordinates?
Z
Figure 3-34 Problem 45.
46 © Vector 7 has a magnitude of 5.0 m and is directed east. Vector b has a magnitude of 4.0 m and is directed 35° west of due north. What are (a) the magnitude and (b) the direction of 7 + i>? What are (c) the magnitude and (d) the direction of b — 7? (e) Draw a vector diagram for each combination.
47 Vectors A and B lie in an xy plane. A has magnitude 8.00 and angle 130°; B has components Bx = —7.72 and By = —9.20. What are the angles between the negative direction of the y axis and (a) the direction of A, (b) the direction of the product A X B, and (c) the direction of A X (B + 3.00k)?
48 ® Two vectors 7 and b have the components, in meters, ax = 3.2, ay = 1.6, bx — 0.50, by — 4.5. (a) Find the angle between the directions of 7 and b. There are two vectors in the xy plane that are perpendicular to 7 and have a magnitude of 5.0 m. One, vector 7, has a positive x component and the other, vector rf, a negative x component. What are (b) the x component and (c) the y compo¬ nent of vector 7, and (d) the x component and (e) the y component of vector dl
49 SSM A sailboat sets out from the U.S. side of Lake Erie for a point on the Canadian side, 90.0 km due north. The sailor, how¬ ever, ends up 50.0 km due east of the starting point, (a) Flow far and (b) in what direction must the sailor now sail to reach the orig¬ inal destination?
50 Vector dx is in the negative direction of a v axis, and vector d2 is in the positive direction of an x axis. What are the directions of (a) d2/4 and (b) d1l(^—4)l What are the magnitudes of products (c) dt • d2 and (d) d2 • (d2/4)l What is the direction of the vector result¬ ing from (e) dx X d2 and (f) d2 X d{! What is the magnitude of the vector product in (g) part (e) and (h) part (f)? What are the (i) magnitude and (j) direction of d2 X (72/4)?
60
CHAPTER 3 VECTORS
51 Rock faults are ruptures along which opposite faces of rock have slid past each other. In Fig. 3-35, points A and B coincided be¬ fore the rock in the foreground slid down to the right. The net dis- placement A B is along the plane of the fault. The horizontal compo¬ nent of AB is the strike-slip AC. The component of AB that is directed down the plane of the fault is the dip-slip AD. (a) What is the magnitude of the net displacement AB if the strike-slip is 22.0 m and the dip-slip is 17.0 m? (b) If the plane of the fault is inclined at angle f = 52.0° to the horizontal, what is the vertical component of AB ?
52 Here are three displacements, each measured in meters: d3 = 4.0i + 5.0j — 6.0k, d2 = — l.Oi + 2.0j + 3.0k, and d3 — 4.0i + 3.0j + 2.0k. (a) What is 7 = dx — d2 + df! (b) What is the angle between 7 and the positive z axis? (c) What is the compo¬ nent of d3 along the direction of dfl (d) What is the component of d3 that is perpendicular to the direction of d2 and in the plane of d3 and d{! {Hint: For (c), consider Eq. 3-20 and Fig. 3-18; for (d), con¬ sider Eq. 3-24.)
53 SSM A vector 7 of magnitude 10 units and another vector b of magnitude 6.0 units differ in directions by 60°. Find (a) the scalar product of the two vectors and (b) the magnitude of the vec¬ tor product 7f X b.
54 For the vectors in Fig. 3-32, with a = 4, b = 3, and c — 5, calcu¬ late (a) 7 • b, (b) 7 • 7, and (c) b - 7.
55 A particle undergoes three successive displacements in a plane, as follows: d\, 4.00 m southwest; then d2, 5.00 m east; and finally d3, 6.00 m in a direction 60.0° north of east. Choose a coor¬ dinate system with the y axis pointing north and the x axis pointing east. What are (a) the x component and (b) the y component of d{l What are (c) the x component and (d) the y component of d2l What are (e) the x component and (f) the y component of d3? Next, consider the net displacement of the particle for the three successive displacements. What are (g) the x component, (h) the y component, (i) the magnitude, and ( j) the direction of the net dis¬ placement? If the particle is to return directly to the starting point, (k) how far and (1) in what direction should it move?
56 Find the sum of the following four vectors in (a) unit-vector notation, and as (b) a magnitude and (c) an angle relative to +x.
P: 10.0 m, at 25.0° counterclockwise from +x Q: 12.0 m, at 10.0° counterclockwise from +y R: 8.00 m, at 20.0° clockwise from — y S: 9.00 m, at 40.0° counterclockwise from — y
57 SSM If Bis added to A, the result is 6.0i + l.Oj. If B is subtracted fromA, the result is — 4.0i + 7.0j.What is the magnitude of A?
58 A vector d has a magnitude of 2.5 m and points north. What are (a) the magnitude and (b) the direction of 4.0 dl What are (c) the magnitude and (d) the direction of —3.0 dl
59 A has the magnitude 12.0 m and is angled 60.0° counterclock¬ wise from the jiositive direction of the x axis of an xy coordinate system. Also, B = (12.0 m)i + (8.00 m)j on that same coordinate system. We now rotate the system counterclockwise about the origin by 20.0° to form an x'y' system. On this new system, what are (a) A and (b) B , both in unit-vector notation?
60 If a — b = 2"? , 7 + b = 47, and 7 = 3i + 4j , then what are (a) 7 and (b) £>?
61 (a) In unit-vector notation, what is 7 = 7— b + 7 if 7 = 5.0i + 4.0j — 6.0k, b = — 2.0i + 2.0j + 3.0k, and 7 = 4.0i + 3.0j + 2.0k? (b) Calculate the angle between 7 and the positive z axis, (c) What is the component of 7 along the direction of b? (d) What is the component of 7 perpendicular to the direction of b but in the plane of 7 and b? {Hint: For (c), see Eq. 3-20 and Fig. 3-18; for (d), see Eq. 3-24.)
62 A golfer takes three putts to get the ball into the hole. The first putt displaces the ball 3.66 m north, the second 1.83 m south¬ east, and the third 0.91 m southwest. What are (a) the magnitude and (b) the direction of the displacement needed to get the ball into the hole on the first putt?
63 Here are three vectors in meters:
d3 = — 3.0i + 3.0j + 2.0k d2 = — 2.0i - 4.0] + 2.0k d3 — 2.0i + 3.0j + 1.0k.
What results from (a) d3 • {d2 + d3), (b) d3 • {d2 X d3), and (c) dx X {d2 + d3)l
64 SSM WWW A room has dimensions 3.00 m (height) X 3.70 m X 4.30 m. A fly starting at one corner flies around, ending up at the diagonally opposite corner, (a) What is the magnitude of its displacement? (b) Could the length of its path be less than this magnitude? (c) Greater? (d) Equal? (e) Choose a suitable coordi¬ nate system and express the components of the displacement vec¬ tor in that system in unit-vector notation, (f) If the fly walks, what is the length of the shortest path? {Hint: This can be answered without calculus. The room is like a box. Unfold its walls to flatten them into a plane.)
65 A protester carries his sign of protest, starting from the ori¬ gin of an xyz coordinate system, with the xy plane horizontal. He moves 40 m in the negative direction of the x axis, then 20 m along a perpendicular path to his left, and then 25 m up a water tower, (a) In unit-vector notation, what is the displacement of the sign from start to end? (b) The sign then falls to the foot of the tower. What is the magnitude of the displacement of the sign from start to this new end?
66 Consider 7 in the positive direction of x, b in the positive di¬ rection of y, and a scalar d. What is the direction of bid if d is (a) positive and (b) negative? What is the magnitude of (c) 7 • b and (d) 7 • bid ? What is the direction of the vector resulting from
(e) 7 X b and (f) b X 72 (g) What is the magnitude of the vector product in (e)? (h) What is the magnitude of the vector product in
(f) ? What are (i) the magnitude and (j) the direction of o' X bid if d is positive?
PROBLEMS
61
67 Let i be directed to the east, j be directed to the north, and k be directed upward. What are the values of products (a) i • k, (b) (— k) • (— j), and (c) j • (— j)? What are the directions (such as east or down) of products (d) k X j,(e) (— i) X (-j),and (f) (-k) X (— j)?
68 A bank in downtown Boston is robbed (see the map in Fig. 3-36). To elude police, the robbers escape by helicopter, mak¬ ing three successive flights described by the following displace¬ ments: 32 km, 45° south of east; 53 km, 26° north of west; 26 km, 18° east of south. At the end of the third flight they are captured. In what town are they apprehended?
BOSTON N
and Vicinity
5 10 km
Salem •
Woburn
Lexington *
Lynn
Arlington Medford Bank
* • -
Waltham
•
Winthrop
•
•TSOSTON
Newton ,
Massachusetts
,.r „ , Brookline
Wellesley
Bay
•
Framingham
•
Dedham
•
Quincy
Walpole
Weymouth
Figure 3-36 Problem 68.
69 A wheel with a radius of 45.0 cm rolls without slipping along a hori¬ zontal floor (Fig. 3-37). At time tv the dot P painted on the rim of the wheel is at the point of contact be¬ tween the wheel and the floor. At a later time t2, the wheel has rolled through one-half of a revolution. What are (a) the magnitude and (b) the angle (relative to the floor) of the displacement of PI
P
At time /, At time t»
Figure 3-37 Problem 69.
70 A woman walks 250 m in the direction 30° east of north, then 175 m directly east. Find (a) the magnitude and (b) the angle of her final displacement from the starting point, (c) Find the distance she walks, (d) Which is greater, that distance or the magnitude of her displacement?
71 A vector d has a magnitude 3.0 m and is directed south. What are (a) the magnitude and (b) the direction of the vector 5.0 dl What are (c) the magnitude and (d) the direction of the vector —2.0 h?
72 A fire ant, searching for hot sauce in a picnic area, goes through three displacements along level ground: dx for 0.40 m southwest (that is, at 45° from directly south and from directly west), d2 for 0.50 m due east, d3 for 0.60 m at 60° north of east. Let the positive x direction be east and the positive y direction be north. What are (a) the x component and (b) the y compo¬ nent of d{! Next, what are (c) the x component and (d) the y component of d{! Also, what are (e) the x component and (f) the y component of d{t
What are (g) the x component, (h) the y component, (i) the magnitude, and (j) the direction of the ant’s net displacement? If the ant is to return directly to the starting point, (k) how far and (1) in what direction should it move?
73 Two vectors are given by ~a = 3.0i + 5.0j and b = 2.0i + 4.0j. Find (a) ~a X b, (b) ~a • b, (c) (a + b) ■ b, and (d) the component of ~a along the direction of b.
74 Vector ~a lies in the yz plane 63.0° from the positive direction of the y axis, has a positive z component, and has magnitude 3.20 units. Vector b lies in the xz plane 48.0° from the positive direction of the x axis, has a positive z component, and has magnitude 1.40 units. Find (a) 7f • h, (b) a X b, and (c) the angle between 7? and b.
75 Find (a) “north cross west,” (b) “down dot south,” (c) “east cross up,” (d) “west dot west,” and (e) “south cross south.” Let each “vector” have unit magnitude.
76 A vector B , with a magnitude of 8.0 m, is added to a vector A, which lies along an x axis. The sum of these two vectors is a third vector that lies along the y axis and has a magnitude that is twice the magnitude of A. What is the magnitude of A?
77 A man goes for a walk, starting from the origin of an xyz coordinate system, with the xy plane horizontal and the x axis east¬ ward. Carrying a bad penny, he walks 1300 m east, 2200 m north, and then drops the penny from a cliff 410 m high, (a) In unit-vector notation, what is the displacement of the penny from start to its landing point? (b) When the man returns to the origin, what is the magnitude of his displacement for the return trip?
78 What is the magnitude of ~a X (b X ~a ) if a = 3.90, b = 2.70, and the angle between the two vectors is 63.0°?
79 In Fig. 3-38, the magnitude of ~a is 4.3, the magnitude of b is 5.4, and cf> = 46°. Find the area of the triangle contained between the two vectors and the thin diagonal line.
Figure 3-38 Problem 79.
c
Motion in Two and Three Dimensions
4-1 POSITION AND DISPLACEMENT
Learning Objectives _
After reading this module, you should be able to . . .
4.01 Draw two-dimensional and three-dimensional position vectors for a particle, indicating the components along the axes of a coordinate system.
4.02 On a coordinate system, determine the direction and
Key Ideas _
• The location of a particle relative to the origin of a coordi¬ nate system is given by a position vector 7, which in unit- vector notation is
7 = xi + y) + zk.
Here xi, y] , and zk are the vector components of position vector 7, and x, y, and z are its scalar components (as well as the coordinates of the particle).
• A position vector is described either by a magnitude and
magnitude of a particle's position vector from its compo¬ nents, and vice versa.
4.03 Apply the relationship between a particle's displace¬ ment vector and its initial and final position vectors.
one or two angles for orientation, or by its vector or scalar components.
• If a particle moves so that its position vector changes from 7, to 72, the particle's displacement A7 is
A7 = 72 - 7|.
The displacement can also be written as
A7 = (x2 - xj)i + (y2 - y,)J + (z2 - Zi)k = Axi + Ayj + Azk.
What Is Physics?
In this chapter we continue looking at the aspect of physics that analyzes motion, but now the motion can be in two or three dimensions. For example, medical researchers and aeronautical engineers might concentrate on the physics of the two- and three-dimensional turns taken by fighter pilots in dog¬ fights because a modern high-performance jet can take a tight turn so quickly that the pilot immediately loses consciousness. A sports engineer might focus on the physics of basketball. For example, in a free throw (where a player gets an uncontested shot at the basket from about 4.3 m), a player might employ the overhand push shot, in which the ball is pushed away from about shoulder height and then released. Or the player might use an underhand loop shot, in which the ball is brought upward from about the belt-line level and released. The first technique is the overwhelming choice among professional players, but the legendary Rick Barry set the record for free-throw shooting with the under¬ hand technique. -sSfi«7
Motion in three dimensions is not easy to understand. For example, you are probably good at driving a car along a freeway (one-dimensional motion) but would probably have a difficult time in landing an airplane on a runway (three- dimensional motion) without a lot of training.
In our study of two- and three-dimensional motion, we start with position and displacement.
62
4-1 POSITION AND DISPLACEMENT
63
Position and Displacement
One general way of locating a particle (or particle-like object) is with a position vector 7, which is a vector that extends from a reference point (usually the origin) to the particle. In the unit-vector notation of Module 3-2, 7 can be written
7 = xi + yj + zk, (4-1)
where xi, yj , and zk are the vector components of 7 and the coefficients x, y , and z are its scalar components.
The coefficients x, y, and z give the particle’s location along the coordinate axes and relative to the origin; that is, the particle has the rectangular coordinates (x, y, z). For instance, Fig. 4-1 shows a particle with position vector
7 = (—3 m)i + (2 m)j + (5 m)k
and rectangular coordinates (— 3 m, 2 m, 5 m). Along the x axis the particle is 3 m from the origin, in the — i direction. Along the y axis it is 2 m from the origin, in the +j direction. Along the z axis it is 5 m from the origin, in the +k direction.
As a particle moves, its position vector changes in such a way that the vector always extends to the particle from the reference point (the origin). If the posi¬ tion vector changes — say, from 7, to 72 during a certain time interval — then the particle’s displacement A7 during that time interval is
To locate the particle, this
cle is the vector sum of its vector compo¬ nents.
A7 = 72 - 7,. (4-2)
Using the unit-vector notation of Eq. 4-1, we can rewrite this displacement as A7 = (x2r + y2j + z2k) - (xp + yj + Zjk)
or as A7 = (x2 - x,)i + (y2 - y:)j + (z2 - Zi)k, (4-3)
where coordinates (xu yu Zi) correspond to position vector 7, and coordinates (x2, y2, z2) correspond to position vector 72. We can also rewrite the displacement by substituting Ax for (x2 — x,). Ay for (y2 — yi), and Az for (z2 — Zi):
A7 = Ad + Ayj + Azk. (4-4)
Sample Problem 4.01 Two-dimensional position vector, rabbit run
A rabbit runs across a parking lot on which a set of coordinate axes has, strangely enough, been drawn. The co¬ ordinates (meters) of the rabbit's position as functions of time t (seconds) are given by
x = —0.317 + lit + 28 (4-5)
and y = 0.22 12 - 9.1 1 + 30. (4-6)
(a) At t = 15 s, what is the rabbit's position vector 7 in unit- vector notation and in magnitude-angle notation?
KEY IDEA
The x and y coordinates of the rabbit’s position, as given by Eqs. 4-5 and 4-6, are the scalar components of the rabbit’s
position vector 7. Let’s evaluate those coordinates at the given time, and then we can use Eq. 3-6 to evaluate the mag¬ nitude and orientation of the position vector.
Calculations: We can write
7(f) = x(f)i + y(0j. (4-7)
(We write 7(0 rather than 7 because the components are functions of f, and thus 7 is also.)
At t = 15 s, the scalar components are
x = (— 0.31)(15)2 + (7.2)(15) + 28 = 66 m and y = (0.22)(15)2 - (9.1)(15) + 30 = -57 m, so 7 = (66 m)i — (57 m)j, (Answer)
64
CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
Figure 4-2 (a) A rabbit’s position vector 7 at time t = 15 s. The scalar compo¬ nents of 7 are shown along the axes.
( b ) The rabbit’s path and its position at six values of t.
y (m)
various times indicated.
x (m)
which is drawn in Fig. 4-2 a. To get the magnitude and angle of 7, notice that the components form the legs of a right tri¬ angle and r is the hypotenuse. So, we use Eq. 3-6:
r = V x2 + y2 = V(66 m)2 + (— 57 m)2 = 87 m, (Answer)
and 9 = tan^1 — = tan_1( — ^ m )= -41°. (Answer) x \ 66 m
Check: Although 9 = 139° has the same tangent as —41°, the components of position vector 7 indicate that the de¬ sired angle is 139° - 180° = -41°.
(b) Graph the rabbit’s path for t = 0 to t = 25 s.
Graphing: We have located the rabbit at one instant, but to see its path we need a graph. So we repeat part (a) for sev¬ eral values of t and then plot the results. Figure 4-2 b shows the plots for six values of t and the path connecting them.
PLUS Additional examples, video, and practice available at WileyPLUS
4-2 AVERAGE VELOCITY AND INSTANTANEOUS VELOCITY
Learning Objectives _
After reading this module, you should be able to .. .
4.04 Identify that velocity is a vector quantity and thus has both magnitude and direction and also has components.
4.05 Draw two-dimensional and three-dimensional velocity vectors for a particle, indicating the components along the axes of the coordinate system.
Key Ideas _
• If a particle undergoes a displacement A7 in time interval At,
its average velocity 7,vg for that time interval is
A7
• As A t is shrunk to 0, v^vg reaches a limit called either the velocity or the instantaneous velocity 7:
dT
dt
4.06 In magnitude-angle and unit-vector notations, relate a parti¬ cle's initial and final position vectors, the time interval between those positions, and the particle's average velocity vector.
4.07 Given a particle's position vector as a function of time, determine its (instantaneous) velocity vector.
which can be rewritten in unit-vector notation as
v = vxi + vy j + vzk,
where vx = dx/dt, vy = dy/dt, and vz = dz/dt.
• The instantaneous velocity 7 of a particle is always directed along the tangent to the particle's path at the particle's position.
v
4-2 AVERAGE VELOCITY AND INSTANTANEOUS VELOCITY
65
Average Velocity and Instantaneous Velocity
If a particle moves from one point to another, we might need to know how fast it moves. Just as in Chapter 2, we can define two quantities that deal with “how fast”: average velocity and instantaneous velocity. However, here we must con¬ sider these quantities as vectors and use vector notation.
If a particle moves through a displacement A 7 in a time interval At, then its average velocity 7avg is
displacement
average velocity = — - ; - — ,
time interval
or
_ _ A7
Vavg ~ A t '
(4-8)
This tells us that the direction of 7avg (the vector on the left side of Eq. 4-8) must be the same as that of the displacement A7 (the vector on the right side). Using Eq. 4-4, we can write Eq. 4-8 in vector components as
Vavg
Axi + Ayj + Azk Ax At
At
i +
Ay
At
J
For example, if a particle moves through displacement 2.0 s, then its average velocity during that move is
Vavg
At
(12 m)i + (3.0 m)k 2.0 s
= (6.0 m/s)i
+ L ^
(12 m)i + (3.0 m)k in
+ (1.5 m/s)k.
That is, the average velocity (a vector quantity) has a component of 6.0 m/s along the x axis and a component of 1.5 m/s along the z axis.
When we speak of the velocity of a particle, we usually mean the particle’s instantaneous velocity 7 at some instant. This 7 is the value that 7avg approaches in the limit as we shrink the time interval At to 0 about that instant. Using the lan¬ guage of calculus, we may write 7 as the derivative
v =
d~r
dt
(4-10)
Figure 4-3 shows the path of a particle that is restricted to the xy plane. As the particle travels to the right along the curve, its position vector sweeps to the right. During time interval At, the position vector changes from 7, to 72 and the particle’s displacement is A7.
To find the instantaneous velocity of the particle at, say, instant q (when the particle is at position 1), we shrink interval At to 0 about q. Three things happen as we do so. (1) Position vector 72 in Fig. 4-3 moves toward 7, so that A7 shrinks
Figure 4-3 The displacement A7 of a particle during a time interval At, from position 1 with position vector ~rl at time q to position 2 with position vector 72 at time t2. The tangent to the particle’s path at position 1 is shown.
As the particle moves, the position vector must change.
This is the displacement.
66
CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
toward zero. (2) The direction of ATlAt (and thus of 7avg) approaches the direction of the line tangent to the particle’s path at position 1. (3) The average velocity 7avg approaches the instantaneous velocity 7 at tl.
In the limit as At — > 0, we have 7avg — » 7 and, most important here, 7avg takes on the direction of the tangent line. Thus, 7 has that direction as well:
The direction of the instantaneous velocity 7 of a particle is always tangent to the particle’s path at the particle’s position.
The result is the same in three dimensions: 7 is always tangent to the particle’s path. To write Eq. 4-10 in unit-vector form, we substitute for 7 from Eq. 4-1:
_» d , f ~ dx t dy ~ dz ~
,.-w+r, +A),—l+— ,+— k.
This equation can be simplified somewhat by writing it as
7 = vj + Vy\ + vzk,
(4-11)
where the scalar components of 7 are
dx dy
dz
(4-12)
V’-TT- and v*'
dt '
For example, dx/dt is the scalar component of 7 along the x axis. Thus, we can find the scalar components of 7 by differentiating the scalar components of 7.
Figure 4-4 shows a velocity vector 7 and its scalar x and y components. Note that 7 is tangent to the particle’s path at the particle’s position. Caution: When a position vector is drawn, as in Figs. 4-1 through 4-3, it is an arrow that extends from one point (a “here”) to another point (a “there”). However, when a velocity vector is drawn, as in Fig. 4-4, it does not extend from one point to another. Rather, it shows the instantaneous direction of travel of a particle at the tail, and its length (representing the velocity magnitude) can be drawn to any scale.
The velocity vector is always tangent to the path.
Figure 4-4 The velocity 7 of a particle, along with the scalar components of 7.
Checkpoint 1
The figure shows a circular path taken by a particle. If the instantaneous velocity of the particle is 7 = (2 m/s)i — (2 m/s)j , through which quadrant is the par¬ ticle moving at that instant if it is traveling (a) clockwise and (b) counterclockwise around the circle? For both cases, draw 7 on the figure.
4-3 AVERAGE ACCELERATION AND INSTANTANEOUS ACCELERATION
67
Sample Problem 4.02 Two-dimensional velocity, rabbit run
For the rabbit in the preceding sample problem, find the velocity 7 at time t = 15 s.
KEY IDEA
We can find 7 by taking derivatives of the components of the rabbit’s position vector.
Calculations: Applying the vx part of Eq. 4-12 to Eq. 4-5, we find the x component of 7 to be
Vl = J^ = ^(_a31f2 + 7-2f + 28)
= — 0.62t + 7.2. (4-13)
At t = 15 s, this gives vx = —2.1 m/s. Similarly, applying the vy part of Eq. 4-12 to Eq. 4-6, we find
vy — = — T- (0.22f 2 - 9.1t + 30)
y dt dt v '
= 0.44t - 9.1. (4-14)
At t = 15 s, this gives vv = —2.5 m/s. Equation 4-11 then yields
7 = (-2.1 m/s)i + (-2.5 m/s )j, (Answer)
which is shown in Fig. 4-5, tangent to the rabbit’s path and in the direction the rabbit is running at t = 15 s.
To get the magnitude and angle of 7, either we use a vector-capable calculator or we follow Eq. 3-6 to write
v = Vv* + Vy = V(— 2.1 m/s)2 + (-2.5 m/s)2
= 3.3 m/s (Answer)
, „ vy ( “2.5 m/s \
and 6 = tan 1 — = tan 1 -
vx \ —2.1 m/s /
= tan1 1.19 = -130°. (Answer)
Check: Is the angle -130° or -130° + 180° = 50°?
y (m)
at this instant.
Figure 4-5 The rabbit’s velocity 7 at t = 15 s.
^wileTSs
PLUS Additional examples, video, and practice available at WileyPLUS
4-3 AVERAGE ACCELERATION AND INSTANTANEOUS ACCELERATION
Learning Objectives _
After reading this module, you should be able to . . .
4.08 Identify that acceleration is a vector quantity and thus has both magnitude and direction and also has components.
4.09 Draw two-dimensional and three-dimensional accelera¬ tion vectors for a particle, indicating the components.
4.1 0 Given the initial and final velocity vectors of a particle and the time interval between those velocities, determine
Key Ideas _
• If a particle's velocity changes from ~v1 to 72 in time interval
At, its average acceleration during At is
_» 72 ~v i A7
flavg = ~^At = ~at
• As At is shrunk to 0, r^vg reaches a limiting value called
the average acceleration vector in magnitude-angle and unit-vector notations.
4.1 1 Given a particle's velocity vector as a function of time, determine its (instantaneous) acceleration vector.
4.12 For each dimension of motion, apply the constant- acceleration equations (Chapter 2) to relate acceleration, velocity, position, and time.
either the acceleration or the instantaneous acceleration 7:
^ civ* a = — — . dt
• In unit-vector notation,
7 = ax i + ay j + mk,
where a, = dvjdt, ay = dvy/dt , and a, = dvz/dt.
68
CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
Average Acceleration and Instantaneous Acceleration
When a particle’s velocity changes from v) to ~v2 in a time interval Af, its average acceleration 7avg during At is
average change in velocity acceleration “ time interval ’
or
72 — vq _ A?
A t A t '
(4-15)
If we shrink At to zero about some instant, then in the limit 77a v„ approaches the
instantaneous acceleration (or acceleration) ~a at that instant; that is.
a =
dv
dt
(4-16)
If the velocity changes in either magnitude or direction (or both), the particle must have an acceleration.
We can write Eq. 4-16 in unit-vector form by substituting Eq. 4-11 for ~v to obtain
_ d . * a=~dt Vx' + Vy] +
dvr „ dvv „ dv. „
= ~r~i +~ri +-7^k. dt dt dt
We can rewrite this as
7 = ax i + nvj + mk, where the scalar components of ~a are
(4-17)
dvx
dt
dVy
n = - —
y dt '
and
dvz
dt
(4-18)
To find the scalar components of ~a, we differentiate the scalar components of 7.
Figure 4-6 shows an acceleration vector ~a and its scalar components for a particle moving in two dimensions. Caution: When an acceleration vector is drawn, as in Fig. 4-6, it does not extend from one position to another. Rather, it shows the direction of acceleration for a particle located at its tail, and its length (representing the acceleration magnitude) can be drawn to any scale.
These are the x and y components of the vector y at this instant.
Figure 4-6 The acceleration It of a particle and the scalar components of ~a.
O
Path
4-3 AVERAGE ACCELERATION AND INSTANTANEOUS ACCELERATION
69
Checkpoint 2
Here are four descriptions of the position (in meters) of a puck as it moves in an xy plane:
(1) x = -3 12 + At - 2 and y = 6 12 - 4 1 (3) ~r = 2t2l - (4 1 + 3)j
(2) x = —3 13 — At and y — —St 2 + 6 (4) ~r = (At2 — 2f)i + 3j
Are the x and v acceleration components constant? Is acceleration ~a constant?
Sample Problem 4.03 Two-dimensional acceleration, rabbit run
For the rabbit in the preceding two sample problems, find the acceleration ~a at time t = 15 s.
KEY IDEA
We can find ~a by taking derivatives of the rabbit’s velocity components.
Calculations: Applying the ax part of Eq. 4-18 to Eq. 4-13, we find the x component of ~a to be
ax = = —7 - (— 0.62t + 7.2) = —0.62 m/s2.
dt dt
Similarly, applying the ay part of Eq. 4-18 to Eq. 4-14 yields the y component as
dvv d
av = — y- = — (0.44t - 9.1) = 0.44 m/s2. y dt dt v 2
We see that the acceleration does not vary with time (it is a constant) because the time variable t does not appear in the expression for either acceleration component. Equation 4-17 then yields
~a = (-0.62m/s2)i + (0.44m/s2)j, (Answer)
which is superimposed on the rabbit’s path in Fig. 4-7.
To get the magnitude and angle of ~a, either we use a vector-capable calculator or we follow Eq. 3-6. For the mag¬ nitude we have
a = Va2 + a2y = V(— 0.62 m/s2)2 + (0.44 m/s2)2 = 0.76 m/s2. (Answer)
For the angle we have
6 = tan 1 — = tan 1 ar
0.44 m/s2 -0.62 m/s2
= -35°.
However, this angle, which is the one displayed on a calcula¬ tor, indicates that ~a is directed to the right and downward in Fig. 4-7. Yet, we know from the components that ~a must be directed to the left and upward. To find the other angle that
has the same tangent as —35° but is not displayed on a cal¬ culator, we add 180°:
-35° + 180° = 145°. (Answer)
This is consistent with the components of ~a because it gives a vector that is to the left and upward. Note that ~a has the same magnitude and direction throughout the rabbit’s run because the acceleration is constant. That means that we could draw the very same vector at any other point along the rabbit’s path (just shift the vector to put its tail at some other point on the path without changing the length or orientation).
This has been the second sample problem in which we needed to take the derivative of a vector that is written in unit -vector notation. One common error is to neglect the unit vectors themselves, with a result of only a set of numbers and symbols. Keep in mind that a derivative of a vector is always another vector.
y (m)
These are the x and y components of the vector at this instant.
Figure 4-7 The acceleration a of the rabbit at t = 15 s. The rabbit happens to have this same acceleration at all points on its path.
^WILEYO
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70
CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
4-4 PROJECTILE MOTION
Learning Objectives _
After reading this module, you should be able to . . .
4.14 Given the launch velocity in either magnitude-angle or unit-vector notation, calculate the particle's position, dis¬ placement, and velocity at a given instant during the flight.
4.15 Given data for an instant during the flight, calculate the launch velocity.
4.1 3 On a sketch of the path taken in projectile motion, explain the magnitudes and directions of the velocity and acceleration components during the flight.
Key Ideas _
• In projectile motion, a particle is launched into the air with a speed v0 and at an angle 90 (as measured from a horizontal x axis). During flight, its horizontal acceleration is zero and its vertical acceleration is — g (downward on a vertical y axis).
• The equations of motion for the particle (while in flight) can be written as
x - x0= (v0 cos 60)t,
y - yo= (v0 sin e0)t - \gt2, vy = v0sin 0O - gt, v2y = (v0 sin 0O)2 - 2 g(y - y0).
• The trajectory (path) of a particle in projectile motion is par¬ abolic and is given by
y = (tan 60)x
if x0 and y0 are zero.
2(v0cos 0O)2 ’
• The particle's horizontal range R, which is the horizontal distance from the launch point to the point at which the parti¬ cle returns to the launch height, is
D vo ■ on R = — sin 260.
g
Projectile Motion
Richard Megna/Fundamental Photographs
Figure 4-8 A stroboscopic photograph of a yellow tennis ball bouncing off a hard surface. Between impacts, the ball has projectile motion.
We next consider a special case of two-dimensional motion: A particle moves in a vertical plane with some initial velocity 70 but its acceleration is always the free- fall acceleration g, which is downward. Such a particle is called a projectile (mean¬ ing that it is projected or launched), and its motion is called projectile motion. A projectile might be a tennis ball (Fig. 4-8) or baseball in flight, but it is not a duck in flight. Many sports involve the study of the projectile motion of a ball. For ex¬ ample, the racquetball player who discovered the Z-shot in the 1970s easily won his games because of the ball’s perplexing flight to the rear of the court. -iS*- Our goal here is to analyze projectile motion using the tools for two- dimensional motion described in Module 4-1 through 4-3 and making the assumption that air has no effect on the projectile. Figure 4-9, which we shall ana¬ lyze soon, shows the path followed by a projectile when the air has no effect. The projectile is launched with an initial velocity 70 that can be written as
?0 = + VoyJ. (4-19)
The components v0x and v0v can then be found if we know the angle 80 between 70 and the positive x direction:
vox = v0 cos 00 and vQy = v0 sin 60. (4-20)
During its two-dimensional motion, the projectile’s position vector ~r and velocity vector 7 change continuously, but its acceleration vector ~a is constant and always directed vertically downward. The projectile has no horizontal acceleration.
Projectile motion, like that in Figs. 4-8 and 4-9, looks complicated, but we have the following simplifying feature (known from experiment):
In projectile motion, the horizontal motion and the vertical motion are indepen¬ dent of each other; that is, neither motion affects the other.
4-4 PROJECTILE MOTION
71
Figure 4-9 The projectile motion of an object launched into the air at the origin of a coordinate system and with launch velocity 70 at angle 60. The motion is a combination of vertical motion (constant acceleration) and horizontal motion (constant velocity), as shown by the velocity components.
72
CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
Richard Megna/Fundamental Photographs
Figure 4-10 One ball is released from rest at the same instant that another ball is shot horizontally to the right. Their vertical motions are identical.
The ball and the can fall the same distance h.
Figure 4-11 The projectile ball always hits the falling can. Each falls a distance h from where it would be were there no free-fall acceleration.
This feature allows us to break up a problem involving two-dimensional motion into two separate and easier one-dimensional problems, one for the horizontal motion (with zero acceleration) and one for the vertical motion (with constant downward acceleration). Here are two experiments that show that the horizontal motion and the vertical motion are independent.
Two Golf Balls
Figure 4-10 is a stroboscopic photograph of two golf balls, one simply released and the other shot horizontally by a spring. The golf balls have the same vertical motion, both falling through the same vertical distance in the same interval of time. The fact that one ball is moving horizontally while it is falling has no effect on its vertical mo¬ tion; that is, the horizontal and vertical motions are independent of each other.
A Great Student Rouser
In Fig. 4-11, a blowgun G using a ball as a projectile is aimed directly at a can sus¬ pended from a magnet M. Just as the ball leaves the blowgun, the can is released. If g (the magnitude of the free-fall acceleration) were zero, the ball would follow the straight-line path shown in Fig. 4-11 and the can would float in place after the magnet released it. The ball would certainly hit the can. However, g is not zero, but the ball still hits the can! As Fig. 4-11 shows, during the time of flight of the ball, both ball and can fall the same distance h from their zero -g locations. The harder the demonstrator blows, the greater is the ball’s initial speed, the shorter the flight time, and the smaller the value of h.
3 Checkpoint 3
At a certain instant, a fly ball has velocity ~v = 25i — 4.9j (the x axis is horizontal, the y axis is upward, and ~v is in meters per second). Has the ball passed its highest point?
The Horizontal Motion
Now we are ready to analyze projectile motion, horizontally and vertically. We start with the horizontal motion. Because there is no acceleration in the hori¬ zontal direction, the horizontal component vx of the projectile’s velocity remains unchanged from its initial value vo* throughout the motion, as demonstrated in Fig. 4-12. At any time t, the projectile’s horizontal displacement x — x0 from an initial position x0 is given by Eq. 2-15 with a = 0, which we write as
x - x0 = v0xt.
Because v0x = v0 cos 90, this becomes
x - xQ = (v0 cos 90)t. (4-21)
The Vertical Motion
The vertical motion is the motion we discussed in Module 2-5 for a particle in free fall. Most important is that the acceleration is constant. Thus, the equations of Table 2-1 apply, provided we substitute — g for a and switch to y notation. Then, for example, Eq. 2-15 becomes
y ~ To = V ~ 1st2
= (v0 sin Oo)t - \gt2, (4-22)
where the initial vertical velocity component v0y is replaced with the equivalent v0 sin 60. Similarly, Eqs. 2-11 and 2-16 become
vy = v0 sin d0 - gt (4-23)
and v2y = (v0 sin Of)2 - 2 g(y - y0). (4-24)
4-4 PROJECTILE MOTION
73
As is illustrated in Fig. 4-9 and Eq. 4-23, the vertical velocity component be¬ haves just as for a ball thrown vertically upward. It is directed upward initially, and its magnitude steadily decreases to zero, which marks the maximum height of the path. The vertical velocity component then reverses direction, and its magni¬ tude becomes larger with time.
The Equation of the Path
We can find the equation of the projectile’s path (its trajectory) by eliminating time t between Eqs. 4-21 and 4-22. Solving Eq. 4-21 for t and substituting into Eq. 4-22, we obtain, after a little rearrangement,
ex 2
y = (tan Qq)x - - TTf (trajectory). (4-25)
Z(V0 cos t)0)
This is the equation of the path shown in Fig. 4-9. In deriving it, for simplicity we let x0 = 0 and y0 = 0 in Eqs. 4-21 and 4-22, respectively. Because g, 0O, and v0 are constants, Eq. 4-25 is of the form y = ax + bx2, in which a and b are constants. This is the equation of a parabola, so the path is parabolic.
The Horizontal Range
The horizontal range R of the projectile is the horizontal distance the projectile has traveled when it returns to its initial height (the height at which it is launched). To find range R , let us put x — x0 = R in Eq. 4-21 and y — y0 = 0 in Eq. 4-22, obtaining
R = (v0 cos 0o)t
and 0 = (v0 sin 0o)t — \gt 2.
Eliminating t between these two equations yields
2vq .
R = - sin 0O cos 60.
g
Using the identity sin 200 = 2 sin 90 cos 00 (see Appendix E), we obtain
v2
R = — sin 290. (4-26)
g
This equation does not give the horizontal distance traveled by a projectile when the final height is not the launch height. Note that R in Eq. 4-26 has its maximum value when sin 2 60 = 1, which corresponds to 2 90 = 90° or 60 = 45°.
The horizontal range R is maximum for a launch angle of 45°.
However, when the launch and landing heights differ, as in many sports, a launch angle of 45° does not yield the maximum horizontal distance. -^35?
The Effects of the Air
We have assumed that the air through which the projectile moves has no effect on its motion. However, in many situations, the disagreement between our calcu¬ lations and the actual motion of the projectile can be large because the air resists (opposes) the motion. Figure 4-13, for example, shows two paths for a fly ball that leaves the bat at an angle of 60° with the horizontal and an initial speed of 44.7 m/s. Path I (the baseball player’s fly ball) is a calculated path that approximates normal conditions of play, in air. Path II (the physics professor’s fly ball) is the path the ball would follow in a vacuum.
■
Jamie Budge
Figure 4-12 The vertical component of this skateboarder’s velocity is changing but not the horizontal component, which matches the skateboard’s velocity. As a result, the skateboard stays underneath him, allowing him to land on it.
Air reduces
height ... ... and range.
Figure 4-13 (I) The path of a fly ball calcu¬ lated by taking air resistance into account. (II) The path the ball would follow in a vacuum, calculated by the methods of this chapter. See Table 4-1 for corresponding data. (Based on “The Trajectory of a Fly Ball,” by Peter J. Brancazio, The Physics Teacher, January 1985.)
Table 4-1 Two Fly Balls"
Path I (Air)
Path II (Vacuum)
Range
98.5 m
177 m
Maximum
height
53.0 m
76.8 m
Time
of flight
6.6 s
7.9 s
“See Fig. 4-13. The launch angle is 60° and the launch speed is 44.7 m/s.
74
CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
Checkpoint 4
A fly ball is hit to the outfield. During its flight (ignore the effects of the air), what happens to its (a) horizontal and (b) vertical components of velocity? What are the (c) horizontal and (d) vertical components of its acceleration during ascent, during de¬ scent, and at the topmost point of its flight?
Sample Problem 4.04 Projectile dropped from airplane
In Fig. 4-14, a rescue plane flies at 198 km/h (= 55.0 m/s) and constant height h = 500 m toward a point directly over a victim, where a rescue capsule is to land.
(a) What should be the angle cf> of the pilot's line of sight to the victim when the capsule release is made?
KEY IDEAS
Once released, the capsule is a projectile, so its horizontal and vertical motions can be considered separately (we need not consider the actual curved path of the capsule).
Calculations: In Fig. 4-14. we see that </> is given by
4> = tan-1 -f , (4-27)
h
where x is the horizontal coordinate of the victim (and of the capsule when it hits the water) and h = 500 m. We should be able to find x with Eq. 4-21:
x - x0 = (v0 cos 80)t. (4-28)
Here we know that x0 = 0 because the origin is placed at the point of release. Because the capsule is released and not shot from the plane, its initial velocity "v0 is equal to the plane’s velocity. Thus, we know also that the initial ve¬ locity has magnitude v0 = 55.0 m/s and angle 0O = 0° (measured relative to the positive direction of the x axis). However, we do not know the time t the capsule takes to move from the plane to the victim.
To find t, we next consider the vertical motion and specifically Eq. 4-22:
y ~ yo = (v0 sin 80)t - \gt2. (4-29)
Here the vertical displacement y — y0 of the capsule is —500 m (the negative value indicates that the capsule moves downward). So,
—500 m = (55.0 m/s)(sin 0°)t — ^(9.8 m/s 2)t2. (4-30)
Solving for t, we find t = 10.1 s. Using that value in Eq. 4-28 yields
* - 0 = (55.0 m/s)(cos 0°)(10.1 s), (4-31)
Figure 4-14 A plane drops a rescue capsule while moving at constant velocity in level flight. While falling, the capsule remains under the plane.
Then Eq. 4-27 gives us
, 555.5 m
d, = tan1 - = 48.0°. (Answer)
v 500 m v ’
(b) As the capsule reaches the water, what is its velocity v ?
KEY IDEAS
(1) The horizontal and vertical components of the capsule’s velocity are independent. (2) Component vx does not change from its initial value v0x = v0 cos 0O because there is no hori¬ zontal acceleration. (3) Component vy changes from its initial value v0y = v0 sin 0O because there is a vertical acceleration.
Calculations: When the capsule reaches the water, vx = v0 cos 0O = (55.0 m/s)(cos 0°) = 55.0 m/s.
Using Eq. 4-23 and the capsule’s time of fall t = 10.1 s, we also find that when the capsule reaches the water,
Vy = v0 sin 0O - gt
= (55.0 m/s)(sin 0°) - (9.8 m/s2)(10.1 s)
= -99.0 m/s.
Thus, at the water
V = (55.0m/s)i - (99.0m/s)j. (Answer)
From Eq. 3-6, the magnitude and the angle of F are
v = 113 m/s and 0 =-60.9°. (Answer)
or
x = 555.5 m.
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4-4 PROJECTILE MOTION
75
Sample Problem 4.05 Launched into the air from a water slide
One of the most dramatic videos on the web (but entirely fictitious) supposedly shows a man sliding along a long wa¬ ter slide and then being launched into the air to land in a water pool. Let’s attach some reasonable numbers to such a flight to calculate the velocity with which the man would have hit the water. Figure 4-15a indicates the launch and landing sites and includes a superimposed coordinate sys¬ tem with its origin conveniently located at the launch site. From the video we take the horizontal flight distance as D = 20.0 m, the flight time as t = 2.50 s, and the launch an¬ gle as 60 = 40.0°. Find the magnitude of the velocity at launch and at landing.
KEY IDEAS
(1) For projectile motion, we can apply the equations for con¬ stant acceleration along the horizontal and vertical axes sepa¬ rately. (2) Throughout the flight, the vertical acceleration is ay = —g = —9.8 m/s and the horizontal acceleration is ax = 0.
Calculations: In most projectile problems, the initial chal¬ lenge is to figure out where to start. There is nothing wrong with trying out various equations, to see if we can somehow get to the velocities. But here is a clue. Because we are going to apply the constant-acceleration equations separately to the x and y motions, we should find the horizontal and verti¬ cal components of the velocities at launch and at landing. For each site, we can then combine the velocity components to get the velocity.
Because we know the horizontal displacement D = 20.0 m, let’s start with the horizontal motion. Since ax = 0,
( b ) (c)
Figure 4-15 (a) Launch from a water slide, to land in a water pool. The velocity at ( b ) launch and (c) landing.
we know that the horizontal velocity component vx is con¬ stant during the flight and thus is always equal to the hori¬ zontal component v0x at launch. We can relate that compo¬ nent, the displacement x — x0, and the flight time t = 2.50 s with Eq. 2-15:
x — x0 = v0xt + \axt2. (4-32)
Substituting ax = 0, this becomes Eq. 4-21. With x — xQ = D, we then write
20 m = vto(2.50 s) + ’ (0)(2.50 s)2 v0x = 8.00 m/s.
That is a component of the launch velocity, but we need the magnitude of the full vector, as shown in Fig. 4-156, where the components form the legs of a right triangle and the full vector forms the hypotenuse. We can then apply a trig definition to find the magnitude of the full velocity at launch:
„ v0x
COS0Q = - ,
n>
and so
v0x 8.00 m/s
Vn = - - - = -
cos 0O cos 40°
= 10.44 m/s ~ 10.4 m/s. (Answer)
Now let’s go after the magnitude v of the landing veloc¬ ity. We already know the horizontal component, which does not change from its initial value of 8.00 m/s. To find the verti¬ cal component vy and because we know the elapsed time t = 2.50 s and the vertical acceleration ay = —9.8 m/s2, let’s rewrite Eq. 2-11 as
Vy = V0y + Uyt
and then (from Fig. 4-156) as
vy = v0 sin 60 + ayt. (4-33)
Substituting ay = —g, this becomes Eq. 4-23. We can then write vy = (10.44 m/s) sin (40.0°) — (9.8 m/s2)(2.50 s)
= -17.78 m/s.
Now that we know both components of the landing velocity, we use Eq. 3-6 to find the velocity magnitude:
v = Vv2 + v2
= V(8.00m/s)2 + (-17.78 m/s)2 = 19.49 m/s2 ~ 19.5 m/s. (Answer)
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76
CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
4-5 UNIFORM CIRCULAR MOTION
Learning Objectives _
After reading this module, you should be able to .. .
4.16 Sketch the path taken in uniform circular motion and ex¬ plain the velocity and acceleration vectors (magnitude and direction) during the motion.
Key Ideas _
• If a particle travels along a circle or circular arc of radius r at constant speed v, it is said to be in uniform circular motion and has an acceleration ~a of constant magnitude
v2
a = — . r
The direction of ~a is toward the center of the circle or circular
4.17 Apply the relationships between the radius of the circu¬ lar path, the period, the particle's speed, and the particle's acceleration magnitude.
arc, and ~a is said to be centripetal. The time for the particle to complete a circle is
v
T is called the period of revolution, or simply the period, of the motion.
The acceleration vector always points toward the center.
A
V
The velocity vector is always tangent to the path.
Uniform Circular Motion
A particle is in uniform circular motion if it travels around a circle or a circular arc at constant ( uniform ) speed. Although the speed does not vary, the particle is accelerating because the velocity changes in direction.
Figure 4-16 shows the relationship between the velocity and acceleration vectors at various stages during uniform circular motion. Both vectors have constant magnitude, but their directions change continuously. The velocity is always directed tangent to the circle in the direction of motion. The accelera¬ tion is always directed radially inward. Because of this, the acceleration associ¬ ated with uniform circular motion is called a centripetal (meaning “center seek¬ ing”) acceleration. As we prove next, the magnitude of this acceleration ~a is
v2
a = - (centripetal acceleration), (4-34)
Figure 4-16 Velocity and acceleration vectors for uniform circular motion.
where r is the radius of the circle and v is the speed of the particle.
In addition, during this acceleration at constant speed, the particle travels the circumference of the circle (a distance of 27?r) in time
T =
2 m- v
(period).
(4-35)
T is called the period of revolution, or simply the period, of the motion. It is, in general, the time for a particle to go around a closed path exactly once.
Proof of Eq. 4-34
To find the magnitude and direction of the acceleration for uniform circular motion, we consider Fig. 4-17. In Fig. 4-17«, particle p moves at constant speed v around a circle of radius r. At the instant shown, p has coordinates xp and yp.
Recall from Module 4-2 that the velocity 7 of a moving particle is always tangent to the particle’s path at the particle’s position. In Fig. 4-17a, that means 7 is perpendicular to a radius r drawn to the particle’s position. Then the angle 9 that 7 makes with a vertical at p equals the angle 9 that radius r makes with the x axis.
4-5 UNIFORM CIRCULAR MOTION
77
The scalar components of 7 are shown in Fig. 4-176. With them, we can write the velocity 7 as
7 = vxi + vy] = ( — v sin 9)i + (v cos 9) j.
(4-36)
Now, using the right triangle in Fig. 4-1 7«, we can replace sin 0 with yplr and cos 0 with xp!r to write
vyP
i +
vxn
U-
(4-37)
To find the acceleration ~a of particle p, we must take the time derivative of this equation. Noting that speed v and radius r do not change with time, we obtain
a =
dr
v dyp
r dt
i +
dx„
dt
J-
(4-38)
Now note that the rate dyp/dt at which yp changes is equal to the velocity component vy. Similarly, dxp/dt = vx, and, again from Fig. 4-17 b, we see that vx = — v sin dand vy = v cos 6. Making these substitutions in Eq. 4-38, we find
~a = ^ — — cos f/ji + ^ - sin 0^j. (4-39)
Tliis vector and its components are shown in Fig. 4-17c. Following Eq. 3-6, we find
2 _ ,2 ,2
a = Va? + a\ = — V (cos 9)2 + (sin 6)2 = — VI = — ,
y y Y Y
as we wanted to prove. To orient 7, we find the angle cf> shown in Fig. 4-17c:
tan <f) = — =
Cly-
~(v2lr) sin 9 -(v2/r) cos 9
= tan 9.
Thus, (f>= 9, which means that ~a is directed along the radius r of Fig. 4- 1 7«, toward the circle’s center, as we wanted to prove.
Checkpoint 5
An object moves at constant speed along a circular path in a horizontal xy plane, with the center at the origin. When the object is at x = —2 m, its velocity is —(4 m/s)j. Give the object’s (a) velocity and (b) acceleration aty = 2 m.
0
y
Figure 4-17 Particle p moves in counter¬ clockwise uniform circular motion, (a) Its position and velocity 7 at a certain instant. ( b ) Velocity 7. (c) Acceleration 7.
Sample Problem 4.06 Top gun pilots in turns
“Top gun” pilots have long worried about taking a turn too tightly. As a pilot’s body undergoes centripetal acceleration, with the head toward the center of curvature, the blood pres¬ sure in the brain decreases, leading to loss of brain function.
There are several warning signs. When the centripetal acceleration is 2 g or 3 g, the pilot feels heavy. At about 4g, the pilot’s vision switches to black and white and narrows to “tunnel vision.” If that acceleration is sustained or in¬ creased, vision ceases and, soon after, the pilot is uncon¬ scious — a condition known as g-LOC for “g-induced loss of consciousness.”
What is the magnitude of the acceleration, in g units, of a pilot whose aircraft enters a horizontal circular turn with a velocity of 7, = (400i + 500j ) m/s and 24.0 s later leaves the turn with a velocity of 7^ = (— 400i — 500 j ) m/s?
KEY IDEAS
We assume the turn is made with uniform circular motion. Then the pilot’s acceleration is centripetal and has magni¬ tude a given by Eq. 4-34 ( a = v2IR), where R is the circle’s radius. Also, the time required to complete a full circle is the period given by Eq. 4-35 (T = 2i tRIv).
Calculations: Because we do not know radius R , let’s solve Eq. 4-35 for R and substitute into Eq. 4-34. We find
2ttv
To get the constant speed v, let’s substitute the components of the initial velocity into Eq. 3-6:
v = V(400 m/s)2 + (500 m/s)2 = 640.31 m/s.
78
CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
To find the period T of the motion, first note that the final 24.0 s. Thus a full circle would have taken T = 48.0 s.
velocity is the reverse of the initial velocity. This means the aircraft leaves on the opposite side of the circle from the ini¬ tial point and must have completed half a circle in the given
^WILEv"hs
PLUS
Substituting these values into our equation for a, we find 277(640.31 m/s)
48.0 s
= 83.81 m/s2 ~ 8.6g. (Answer)
Additional examples, video, and practice available at WileyPLUS
4-6 RELATIVE MOTION IN ONE DIMENSION
Learning Objective _
After reading this module, you should be able to . . .
4.1 Apply the relationship between a particle's position, ve- frames that move relative to each other at constant velocity
locity, and acceleration as measured from two reference and along a single axis.
Key Idea _
• When two frames of reference A and B are moving relative vPA = vpb + vba>
to each other at constant velocity, the velocity of a particle P , — ■ ,, . .. , D ... . . a d u
,, , ■ r . „ r wh ere ,, is t h e ve loci ty of # wi t h res pect to A . B ot h ob-
as measured by an observer in frame A usually differs from ., . .. , ., .. .
, ’ , servers measure the same acceler