Randall d knight instructors solutions manual to physics for scientists engineers (2012)

The particle’s speed is increasing as it moves to the right, so its acceleration vector points in the same direction as its velocity vector i.e., to the right.. The particle’s speed is i

PhD-Physics, University of Santa Barbara

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Contents

Preface v

Chapter 1 Concepts of Motion 1-1

Chapter 2 Kinematics in One Dimension 2-1

Chapter 3 Vectors and Coordinate Systems 3-1

Chapter 4 Kinematics in Two Dimensions 4-1

Chapter 5 Force and Motion 5-1

Chapter 6 Dynamics I: Motion Along a Line 6-1

Chapter 7 Newton’s Third Law 7-1

Chapter 8 Dynamics II: Motion in a Plane 8-1

Chapter 9 Impulse and Momentum 9-1

Chapter 10 Energy 10-1

Chapter 11 Work 11-1

Chapter 12 Rotation of a Rigid Body 12-1

Chapter 13 Newton’s Theory of Gravity 13-1

Chapter 14 Oscillations 14-1

Chapter 15 Fluids and Elasticity 15-1

Chapter 16 A Macroscopic Description of Matter 16-1

Chapter 17 Work, Heat, and the First Law of Thermodynamics 17-1

Chapter 18 The Micro/Macro Connection 18-1

Chapter 19 Heat Engines and Refrigerators 19-1

Chapter 20 Traveling Waves 20-1

Chapter 21 Superposition 21-1

Chapter 22 Wave Optics 22-1

Chapter 23 Ray Optics 23-1

Chapter 24 Optical Instruments 24-1

Chapter 25 Electric Charges and Forces 25-1

Chapter 26 The Electric Field 26-1

Chapter 27 Gauss’s Law 27-1

Chapter 28 The Electric Potential 28-1

Chapter 29 Potential and Field 29-1

Chapter 30 Current and Resistance 30-1

Chapter 31 Fundamentals of Circuits 31-1

Chapter 32 The Magnetic Field 32-1

Chapter 33 Electromagnetic Induction 33-1

Chapter 34 Electromagnetic Fields and Waves 34-1

Chapter 39 Wave Functions and Uncertainty 39-1

Chapter 40 One-Dimensional Quantum Mechanics 40-1

Chapter 41 Atomic Physics 41-1

Chapter 42 Nuclear Physics 42-1

Preface

This Instructor Solutions Manual has a twofold purpose First, and most obvious, is to

pro-vide worked solutions for the use of instructors Second, but equally important, is to propro-vide examples of good problem-solving techniques and strategies that will benefit your students if you post these solutions

Far too many solutions manuals simply plug numbers into equations, thereby reinforcing one

of the worst student habits The solutions provided here, by contrast, attempt to:

• Follow, in detail, the problem-solving strategies presented in the text

• Articulate the reasoning that must be done before computation

• Illustrate how to use drawings effectively

• Demonstrate how to utilize graphs, ratios, units, and the many other “tactics” that must be successfully mastered and marshaled if a problem-solving strategy is to be effective

• Show examples of assessing the reasonableness of a solution

• Comment on the significance of a solution or on its relationship to other problems

Most education researchers believe that it is more beneficial for students to study a smaller number of carefully chosen problems in detail, including variations, than to race through a larger number of poorly understood calculations The solutions presented here are intended

to provide a basis for this practice

So that you may readily edit and/or post these solutions, they are available for download as editable Word documents and as pdf files via the “Resources” tab in the textbook’s Instruc- tor Resource Center (www.pearsonhighered.com/educator/catalog/index.page) or from the textbook’s Instructor Resource Area in MasteringPhysics® (www.masteringphysics.com)

We have made every effort to be accurate and correct in these solutions However, if you do find errors or ambiguities, we would be very grateful to hear from you Please contact your Pearson Education sales representative

Acknowledgments for the First Edition

We are grateful for many helpful comments from Susan Cable, Randall Knight, and Steve Stonebraker We express appreciation to Susan Emerson, who typed the word-processing manuscript, for her diligence in interpreting our handwritten copy Finally, we would like to acknowledge the support from the Addison Wesley staff in getting the work into a publish- able state Our special thanks to Liana Allday, Alice Houston, and Sue Kimber for their will- ingness and preparedness in providing needed help at all times

Pawan Kahol

Missouri State University

Donald Foster

Wichita State University

Acknowledgments for the Second Edition

I would like to acknowledge the patient support of my wife, Holly, who knows what is portant

Northern Kentucky University

Acknowledgments for the Third Edition

To Holly, Ryan, Timothy, Nathan, Tessa, and Tyler, who make it all worthwhile

Conceptual Questions

(b) 2 significant figures This is more clearly revealed by using scientific notation:

P

2 sig figs.

1

0.53= 5.3 ×10−

(c) 4 significant figures The trailing zero is significant because it indicates increased precision

(d) 3 significant figures The leading zeros are not significant but just locate the decimal point

if it is speeding up or slowing down If the particle is moving to the right it is slowing down If it is moving to the left

it is speeding up

left The acceleration vector must therefore point in the same direction as the velocity, so the acceleration vector also points to the left Thus,a xis negative as per our convention (see Tactics Box 1.4)

the⫺ direction (down) The acceleration vector must therefore point in the direction opposite to the velocity; y

namely, in the +y direction (up) Thus, a yis positive as per our convention (see Tactics Box 1.4)

right, so its velocity is positive The particle’s speed is increasing as it moves to the right, so its acceleration vector points in the same direction as its velocity vector (i.e., to the right) Thus, the acceleration is also positive

its velocity is negative The particle’s speed is increasing as it moves in the negative direction, so its acceleration vector points in the same direction as its velocity vector (i.e., down) Thus, the acceleration is also negative

1.8 The particle position is above zero on the y-axis, so its position is positive The particle is moving down, so its

velocity is negative The particle’s speed is increasing as it moves in the negative direction, so its acceleration vector points in the same direction as its velocity vector (i.e., down) Thus, the acceleration is also negative

Exercises and Problems

Section 1.1 Motion Diagrams

between each movie frame (or between each snapshot)

Solve:

Assess: As we go from left to right, the distance between successive images of the car decreases Because the time interval between each successive image is the same, the car must be slowing down

upward with a constant acceleration

Solve:

Assess: Notice that the length of the velocity vectors increases each step by approximately the length of the acceleration vector

1.3 Model: We will assume that the term “quickly” used in the problem statement means a time that is short compared to 30 s

Solve:

Assess: Notice that the acceleration vector points in the direction opposite to the velocity vector because the car is decelerating

Section 1.2 The Particle Model

into a single point The size and shape of the object will not be considered This is a reasonable approximation of reality if (i) the distance traveled by the object is large in comparison to the size of the object and (ii) rotations and internal motions are not significant features of the object’s motion The particle model is important in that it allows

us to simplify a problem Complete reality—which would have to include the motion of every single atom in the

object—is too complicated to analyze By treating an object as a particle, we can focus on the most important aspects

of its motion while neglecting minor and unobservable details

(b) The particle model is valid for understanding the motion of a satellite or a car traveling a large distance

(c) The particle model is not valid for understanding how a car engine operates, how a person walks, how a bird flies,

or how water flows through a pipe

Section 1.3 Position and Time

Section 1.4 Velocity

instant before it hits the ground, when it will have its maximum velocity

Solve:

Assess: Notice that the “particle” we have drawn has a finite dimensions, so it appears as if the bottom half of this

“particle” has penetrated into the ground in the bottom frame This is not really the case; our mental particle has no size and is located at the tip of the velocity vector arrow

Section 1.5 Linear Acceleration

1 and 2 Speedv1is greater than speedv0because more distance is covered in the same interval of time

(b) To find the acceleration, use the method of Tactics Box 1.3:

Assess: The acceleration vector points in the same direction as the velocity vectors, which makes sense because the speed is increasing

and 2 Speedv1is greater than speedv0because more distance is covered in the same interval of time

(b) Acceleration is found by the method of Tactics Box 1.3

Assess: The acceleration vector points in the same direction as the velocity vectors, which makes sense because the speed is increasing

1.10 Solve:

Visualize: The dots are equally spaced until brakes are applied to the car Equidistant dots on a single line indicate constant average velocity Upon braking, the dots get closer as the average velocity decreases, and the distance between dots changes by a constant amount because the acceleration is constant

Visualize: The dots in the figure are equally spaced until the sled encounters a rocky patch Equidistant dots on a single line indicate constant average velocity On encountering a rocky patch, the average velocity decreases and the sled comes to a stop This part of the motion is indicated by a decreasing separation between the dots

Visualize: The dots become more closely spaced because the particle experiences a downward acceleration The distance between dots changes by a constant amount because the acceleration is constant

Visualize: Starting from rest, the tile’s velocity increases until it hits the water surface This part of the motion is represented by dots with increasing separation, indicating increasing average velocity After the tile enters the water,

it settles to the bottom at roughly constant speed, so this part of the motion is represented by equally spaced dots

Visualize: The ball falls freely for three stories Upon impact, it quickly decelerates to zero velocity while ressing, then accelerates rapidly while re-expanding As vectors, both the deceleration and acceleration are an upward vector The downward and upward motions of the ball are shown separately in the figure The increasing length between the dots during downward motion indicates an increasing average velocity or downward acceleration On the other hand, the decreasing length between the dots during upward motion indicates acceleration in a direction opposite to the motion, so the average velocity decreases

comp-Assess: For free-fall motion, acceleration due to gravity is always vertically downward Notice that the acceleration due to the ground is quite large (although not to scale—that would take too much space) because in a time interval much shorter than the time interval between the points, the velocity of the ball is essentially completely reversed

1.17 Model: Represent the toy car as a particle

Visualize: As the toy car rolls down the ramp, its speed increases This is indicated by the increasing length of the velocity arrows That is, motion down the ramp is under a constant acceleration aG At the bottom of the ramp, the toy car continues with a constant velocity and no acceleration

Section 1.6 Motion in One Dimension

covers a distance of 100 m in 200 s He then stops and chats with a student for 200 s Suddenly, he realizes he is going to be late for his next class, so the hurries on and covers the remaining 200 m in 200 s to get to class on time

stopped at a rest area one Saturday After staying there for one hour, he headed back home thinking that he was supposed to go on this trip on Sunday Absent-mindedly he missed his exit and stopped after one hour of driving at another rest area 20 miles south of El Dorado After waiting there for one hour, he drove back very slowly, confused and tired as he was, and reached El Dorado two hours later

Section 1.7 Solving Problems in Physics

1.5 m/s each second of motion

1.22 Visualize: The rocket moves upward with a constant acceleration aG The final velocity is 200 m/s and is reached at a height of 1.0 km

Section 1.8 Units and Significant Figures

1.25 Solve: (a) 3 hour (3 hour) 3600 s 10,800 s 1 10 s4

1.31 Solve: The height of a telephone pole is estimated to be around 50 ft or (using 1 m ~ 3 ft) about 15 m This height is approximately 8 times my height

campus My average speed is

1inch 2.54 cm 10 m 1 month 1 day 1 h

speed when it coasts

Visualize:

Visualize:

1.36 Model: Represent (Sam + car) as a particle for the motion diagram

Visualize:

Visualize:

1.38 Model: Represent the speed skater as a particle for the motion diagram

Visualize:

Visualize:

1.40 Model: Represent the motorist as a particle for the motion diagram

1.43 Model: Represent the cars of David and Tina and as particles for the motion diagram

Visualize:

next stop light 100 m away before it turns red When she’s about 30 m away, the light turns yellow, so she starts to brake, knowing that she cannot make the light

top, the road levels and the car continues coasting along the road at a reduced speed

the terrain becomes flat and the skier continues at constant velocity

height

incline, over the peak, and down the other side

(a)

(b) A train moving at 100 km/hour slows down in 10 s to a speed of 60 km/hour as it enters a tunnel The driver maintains this constant speed for the entire length of the tunnel that takes the train a time of 20 s to traverse Find the length of the tunnel

(c)

smaller width Similarly, the largest area will correspond to the larger length and the larger width Therefore, the smallest area is (64 m)(100 m) = 6.4⫻10 m3 2and the largest area is (75 m)(110 m) = 8.3⫻10 m 3 2

3 3

(b) Likewise, the mass density of alcohol is

leisurely pace, covering the first 10 m in 10 seconds But then Susan notices that Ella is heading toward the same seat, so Susan walks more quickly to cover the remaining 30 m in another 10 seconds, beating Ella to the seat Susan stands next to the seat for 10 seconds to remove her backpack

15 m, then he takes 4 s to make a safety check He then continues raising the bricks the remaining 15 m, which takes 4 s

Conceptual Questions

to town My stop for gas took 10 min Then I headed back east at 60 mph before I encountered a construction zone Traffic was at a standstill for 10 min and then I was able to move forward (east) at 30 mph until I got to my destination 30 mi east of town

was still able to finish his 100 m in only just over 9 s by running a world record pace for the rest of the race

second baseman who is 100 ft from home plate The second baseman then fires the ball at 200 ft/s to the catcher at home plate

greater (Both are positive slopes.)

(b) No, the speeds are never the same Each has a constant speed (constant slope) and A’s speed is always greater

A’s line

(b) A and B have the same speed at just about t=3 s At that time, the slope of the tangent to the curve representing B’s motion is equal to the slope of the line representing A

(b) D The slope is greatest at D

(c) At points A, C, and E the slope of the curve is zero, so the object is not moving

(d) At point D the slope is negative, so the object is moving to the left

(b) The slope is negative at points C, D, and E, meaning the object is moving to the left at these points

(c) At point C the slope is increasing in magnitude (getting more negative), meaning that the object is speeding up to the left

(d) At point B the object is not moving since the slope is zero Before point B, the slope is positive, while after B it is negative, so the object is turning around at B

and B are at the same locations at the time corresponding to dot 3 and again at that of dot 6

(b) The spacing of dots 4 and 5 in both diagrams is the same, so the cars are traveling at the same speeds between times corresponding to dots 4 and 5

2.9 No, though you have the same position along the road, his velocity is greater because he is passing you If his velocity were not greater, then he would remain even with the front of your car

2.11 (a) As a ball tossed upward moves upward, its vertical velocity is positive, while its vertical acceleration is negative, opposite the velocity, causing the ball to slow down

(b) The same ball on its way down has downward (negative) velocity The downward negative acceleration is pointing in the same direction as the velocity, causing the speed to increase

direction of the velocity of the ball do not matter Gravity pulls down at constant acceleration (Air friction is ignored.)

velocity The acceleration only tells how the velocity is changing

(b) The magnitude of the acceleration is still g because the rock is still in free fall The speed is increasing at the same rate each instant, that is, by the same vΔ each second

in contact with the floor When it hits the floor, it is accelerated very rapidly in the upward direction as it bounces

Exercises and Problems

Section 2.1 Uniform Motion

Beth 1 Beth 0 Beth 1 Beth 0

400 mile8:00 AM 8:00 AM 8 hr 4:00 PM

50 miles/hour

400 mile9:00 AM 9:00 AM 6.67 hr 3:40 PM

(a) Beth arrives first

(b) Beth has to wait tAlan 1−tBeth 1=20 minutes for Alan

Assess: Times of the order of 7 or 8 hours are reasonable in the present problem

Visualize:

Solve: Since Larry’s speed is constant, we can use the following equation to calculate the velocities:

f i s

f i

s s v

average speed to the house is

100 mi

48 mph5/6 h 5/4 h=+

(b) Julie drives the distance Δx1 in time Δt1 at 40 mph She then drives the distance Δx2 in time Δt2 at 60 mph She spends the same amount of time at each speed, thus

2.4 Model: The jogger is a particle

Solve: The slope of the position-versus-time graph at every point gives the velocity at that point The slope at t=10 s is

50 m 25 m

1.25 m/s

20 s

s v t

Section 2.2 Instantaneous Velocity

Section 2.3 Finding Position from Velocity

(b) There is only one turning point At t=1 s the velocity changes from 20 m/s+ to−5 m/s, thus reversing the direction

of motion At t=3 s, there is an abrupt change in motion from 5 m/s− to rest, but there is no reversal in motion

the –x direction The speed decreases as time increases during the first second, is zero at t=1 s, and then increases after the particle reverses direction

Solve: (a) The particle reverses direction at t=1 s, when v changes sign x

(b) Using the equation xf=x0+ area of the velocity graph between t and 1 t f,

1

16 m (8 m/s 12 m/s)(1 s) 26 m2

Assess: This distance seems reasonable for one beat

f i

x = +x area under the velocity-versus-time graph between t and i t f

The particle starts from the origin at t=0 s, so x i=0 m Notice that the each square of the grid in Figure EX2.8 has

“area” (5 m/s) (2 s) 10 m.× = We can find the area under the curve, and thus determine x, by counting squares You can see that x = 35 m at t=4 s because there are 3.5 squares under the curve In addition, x=35 m at t = 8 s

because the 5 m represented by the half square between 4 and 6 s is cancelled by the –5 m represented by the half square between 6 and 8 s Areas beneath the axis are negative areas The particle passes through x=35 m at t=4 sand again at t=8 s

(b) The particle moves to the right for 0 s≥ ≥t 6 s,where the velocity is positive It reaches a turning point at

40 m

x= at t=6 s The motion is to the left for t>6 s This is shown in the motion diagram below

Section 2.4 Motion with Constant Acceleration

Solve: A constant velocity from t=0 s to t=2 s means zero acceleration On the other hand, a linear increase in velocity between t=2 s and t=4 s implies a constant positive acceleration which is the slope of the velocity line

Solve: When the blood is speeding up the acceleration is

2 m/s m/s0.05s

y y

a t

y y a t

υ

Δ

Assess: 16 m/s is an impressive but reasonable acceleration 2

2 s 2.0 m area under velocity graph from 0 s to 2.0 s

12.0 m (4.0 m/s)(2.0 s) 6.0 m2

Reading from the velocity-versus-time graph, (at v x t=1 s) 4 m/s.= Also, a x=slope= Δ Δ =v t/ 0 m/s 2

(b) x(at t=3.0 s)=x(at t=0 s)+ area between t=0 s and t=3 s

2.13 Model: Represent the jet plane as a particle

The acceleration of the jet is not quite equal to g, the acceleration due to gravity; this seems reasonable for a jet

Visualize: Use the definition of acceleration and then convert units

Solve:

2

150 km/h 1000 m 1 h 1 min

83 m/s0.50 s 1 km 60 min 60 s

x

a t

Δ

Assess: 83 m/s is a remarkable acceleration 2

Assess: A deceleration of 2.8 m/s is reasonable 2

Visualize:

Solve: The Porsche’s time to finish the race is determined from the position equation

2 P1 P0 P0 P1 P0 P P1 P0

So, the Porsche wins

Assess: The numbers are contrived for the Porsche to win, but the time to go 400 m seems reasonable

Section 2.5 Free Fall

Visualize:

Solve: (a) The shot is in free fall, so we can use free fall kinematics with a= −g The height must be such that the shot takes 4 s to fall, so we choose t1=4 s Then,

(b) The impact velocity is v1= −v0 g t(1−t0)= −gt1= −39.2 m/s

Assess: Note the minus sign The question asked for velocity, not speed, and the y-component of vG is negative because the vector points downward

(at 3.0 s) 0 m (19.6 m/s)(3.0 s 0 s) 1/2( 9.8 m/s )(3.0 s 0 s) 14.7 m(at 4.0 s) 19.6 m/s (

y t

(b)

Assess: (a) A downward acceleration of 9 8 m/s 2 on a particle that has been given an initial upward velocity of 19.6 m/s

+ will reduce its speed to 9 8 m/s after 1 s and then to zero after 2 s The answers obtained in this solution are consistent with the above logic

(b) Velocity changes linearly with a negative uniform acceleration of 9 8 m/s 2 The position is symmetrical in time around the highest point which occurs at t=2 s

Visualize:

Solve: Once the ball leaves the student’s hand, the ball is in free fall and its acceleration is equal to the free-fall

acceleration g that always acts vertically downward toward the center of the earth According to the

constant-acceleration kinematic equations of motion

Assess: A time of 3.2 s is reasonable

Visualize:

Solve: (a) Substituting the known values into 1 2

Section 2.6 Motion on an Inclined Plane

Visualize:

Note that the skier’s motion on the horizontal, frictionless snow is not of any interest to us Also note that the

acceleration parallel to the incline is equal to g sin10°

Solve: Using the following constant-acceleration kinematic equations,

Assess: A time of 7.1 s to cover 64 m is a reasonable value

Visualize:

Solve: Note that the problem “ends” at a turning point, where the car has an instantaneous speed of 0 m/s before

rolling back down The rolling back motion is not part of this problem If we assume the car rolls without friction,

then we have motion on a frictionless inclined plane with an accleration a= −gsinθ = −gsin10° = −1.7 m/s 2Constant acceleration kinematics gives

Notice how the two negatives canceled to give a positive value for x 1

Assess: We must include the minus sign because the aG vector points down the slope, which is in the negative

x-direction

Section 2.7 Instantaneous Acceleration

(a) The position t=2 s is x2s=[2(2)2− +2 1] m 7 m.=

(b) The velocity is the derivative v dx dt= / and the velocity at t=2 s is calculated as follows:

f i

t x t

(b) The particle’s speed at t=1 s is v1 s=2 m/s

(c) The particle’s acceleration at t=1 s is a1 s=4 m/s 2

Assess: The acceleration is positive but decreases as a function of time The initial velocity of 8.0 m/s will therefore increase A value of 16 m/s is reasonable

2.26 Solve: (a)

(b) To be completed by student

(c) dx v x 2t 4 v x(at t 1 s) [2 m/s (1 s) 4 m/s]2 2 m/s

(d) There is a turning point at t=2 s At that time x= −2 m

(e) Using the equation in part (c),

which is the velocity at t= 7 0 s The negative sign indicates motion toward lower values on the x-axis The velocity

of particle B at t= 7 0 s can be read directly from its graph It is 20 m/s.− The velocity of particle C can be obtained from the equation

f i

v = +v area under the acceleration curve between t and i t f

This area can be calculated by adding up three sections The area between t=0 s and t=2 s is 40 m/s, the area between 2 t= s and t=5 s is 45 m/s, and the area between t=5 s and t=7 s is 20 m/s.− We get (10 m/s)+

(40 m/s) (45 m/s) (20 m/s) 75 m/s.+ − =

Solve: We will determine the object’s velocity using graphical methods first and then using calculus Graphically, v t( )= +v0 area under the acceleration curve from 0 to t In this case, v0=0 m/s The area at each time

4 s ( 4 s) (4 s)(10 m/s) 20 m/s

21

The last result arises because there is no additional area after t=6 s Let us now use calculus The acceleration

function a(t) consists of three pieces and can be written:

2.5 0 s 4 s( ) 30 4 s 6 s

These were determined by the slope and the y-intercept of each of the segments of the graph The velocity function is

found by integration as follows: For 0≤ ≤t 4 s,

2

2 0

4

5( ) ( 4 s) ( ) 20 m/s 30 2.5 30 60

graph The velocity is zero when the slope of the position-versus-time graph is zero, the velocity is most positive when the slope is most positive, and the velocity is most negative when the slope is most negative The slope is zero

at t = 0, 1 s, 2 s, 3 s, ; the slope is most positive at t = 0.5 s, 2.5 s, ; and the slope is most negative at t = 1.5 s,

3.5 s,

(a) The turning points are when the velocity changes sign Set v x=0 and check that it actually changes sign at those times The function factors into the product of two binomials:

( 2)( 5) 0 when 2 s and 5 s

v = −t t− ⇒v = t= t=

Indeed, the function changes sign at those two times

(b) The acceleration is given by the derivative of the velocity

Plug in the times from part (a): a x(2 s) 2(2) 7= − = −3 m/s2 and a x(5 s) 2(5) 7 3 m/s= − = 2

Assess: This problem does not have constant acceleration so the kinematic equations do not apply, but a dv dt= /always applies

For 0 v x= m/s, there are two solutions to the quadratic equation: t = 0 s and t = 3 s

(b) At the first of these solutions,

Solve: (a) Known information: x0=0 m, v0=0 m/s, x1=40 m, v1=11 m/s, t1=5 s If the acceleration is uniform

(constant a), then the motion must satisfy the three equations

But each equation gives a different value of a Thus the motion is not uniform acceleration

(b) We know two points on the velocity-versus-time graph, namely at t0=0 and t1=5 s What shape does the function have between these two points? If the acceleration was uniform, which it’s not, then the graph would be a

straight line The area under the graph is the displacement Δx From the figure you can see that Δ =x 27.5 m for a straight-line graph But we know that, in reality, Δ =x 40 m To get a larger Δx, the graph must bulge upward above

the straight line Thus the graph is curved, and it is concave downward

2.33 Solve: The position is the integral of the velocity

x =x +Ñ v dt x= +Ñ kt dt x= + kt =x + kt

We’re given that x0= −9.0 m and that the particle is at x1= 9 0 m at t1= 3 0 s Thus

1 3

9.0 m ( 9.0 m)= − + k(3.0 s) = −( 9.0 m)+k(9.0 s )Solving for k gives k= 2 0 m/s 3

The first solution is the initial condition Thus the particle’s velocity is again 0 m/s at t1=20 s

(b) Position is the integral of the velocity At t1=20 s, and using x0=0 m at t0=0 s, the position is

x =x +Ñ v dt= +Ñ t− t dt= t − t =

Visualize: Please refer to Figure P2.35

Solve: In the first and third segments the acceleration a is zero In the second segment the acceleration is negative and s

constant This means the velocity v will be constant in the first two segments and will decrease linearly in the third s

segment Because the velocity is constant in the first and third segments, the position s will increase linearly In the second

segment, the position will increase parabolically rather than linearly because the velocity decreases linearly with time

2.36 Model: Represent the ball as a particle

Visualize: Please refer to Figure P2.36 The ball rolls down the first short track, then up the second short track, and then down the long track s is the distance along the track measured from the left end (where s = 0) Label t = 0 at the

beginning, that is, when the ball starts to roll down the first short track

Solve: Because the incline angle is the same, the magnitude of the acceleration is the same on all of the tracks

Assess: Note that the derivative of the s versus t graph yields the v versus t graph And the derivative of the s v s

versus t graph gives rise to the a versus t graph s

Visualize: The ball moves to the right along the first track until it strikes the wall, which causes it to move to the left on a second track The ball then descends on a third track until it reaches the fourth track, which is horizontal

Solve: