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Chapter 2: Introduction to Motion Answers and Solutions 1.

Picture the Problem: You walk in both the positive and negative directions along a straight line. Strategy: The distance of a trip from your house to the grocery store, then to your friend’s house, and finally back to your house is the sum of the distances traveled. Solution: Add the distances:

distance = 4.3 km + 6.4 km + 2.1 km = 12.8 km

Insight: The distance traveled is always a positive number, but the displacement can be negative. 2.

Picture the Problem: Students walk in both the positive and negative directions along a school hallway. Strategy: The displacement of a student who travels from the math classroom to the music classroom and then to the library depends only upon the initial and final positions. The initial position at the math classroom is xi = 8.0 m and the final position at the library is xf = 0 m. The distance traveled is the sum of the distances traveled on the individual legs of the journey. Δ x = xf − xi = 0.0 − 8.0 m = −8.0 m Solution: 1. Find the displacement: 2. Find the distance traveled:

distance = 6.0 m + 14 m = 20 m

Insight: The negative sign means that the displacement is toward the left of the initial position. While the displacement is −8.0 m, the total distance traveled is 6.0 m + 14 m = 20 m. 3.

(a) The distance traveled will always be positive for any journey, either round trip or one way. (b) The displacement for a round trip journey will always be zero because the initial and final positions are the same.

4.

The odometer of a car measures only the distance traveled and not the initial and final positions, so it cannot measure displacement. A Global Positioning System (GPS) navigation device can report the displacement of your car because it can track the exact initial and final positions of your car on Earth’s surface.

5.

Yes, the distance traveled and the magnitude of the displacement are equal if you hike in a straight line in only one direction. If you change direction, or even if you change altitude, the two values are not equal.

6.

(a) Yes, the displacements for you and your dog are identical when you arrive at the park because you each have the same initial and final positions. (b) No, you and your dog have not traveled the same distance when you arrive at the park. The distance your dog has traveled is greater than the distance you have traveled, because the dog zigzags around as it takes many short side trips to chase squirrels, examine fire hydrants, and so on. If you used a very short leash, the difference between the distances traveled by you and your dog would not be very different, but they would not be the same.

Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2–1

Chapter 2: Introduction to Motion

7.

Pearson Physics by James S. Walker

Picture the Problem: The ball is putted in the positive direction and then the negative direction. Strategy: The distance is the total length of travel, and the displacement is the net change in position. Solution: 1. (a) Add the lengths:

(5.0 + 1.2 m ) + 1.2 m =

2. (b) Subtract xi from xf to find the displacement:

Δ x = xf − xi = 5.0 − 0 m = 5.0 m

7.4 m

Insight: The distance traveled is always positive, but the displacement can be negative. For instance, the displacement of the second putt is −1.2 m. 8.

Picture the Problem: A billiard ball rolls in the positive direction, strikes a bumper, and then rolls in the negative direction. Strategy: The distance is the total length of travel, and the displacement is the net change in position. Let displacements to the right of the initial position be in the positive direction. Solution: 1. (a) Add the lengths:

22 + 22 + 7.5 cm = 51.5 cm

2. (b) Subtract xi from xf to find the displacement:

Δ x = xf − xi = −7.5 − 0 cm = −7.5 cm

7.5 cm

22 cm

Insight: The distance traveled is always positive, but the displacement can be negative, as it is in this example. 9.

Picture the Problem: A train travels in the positive direction and then backs up in the negative direction. Strategy: The distance is the total length of travel, and the displacement is the net change in position. The initial motion of the train is in the positive direction. If we claim the initial position of the train is xi = 0 km, its position after the first leg of the journey is +5.8 km and its final position is xf = 5.9 − 3.8 km = 2.1 km. Solution: 1. (a) The portion of the journey that is in the negative direction will decrease the magnitude of the displacement but increase the distance traveled. Therefore, the distance covered by the train is greater than the magnitude of displacement. 2. (b) Add the lengths:

5.9 + 3.8 km = 9.7 km

3. (c) Subtract xi from xf to find the displacement:

Δ x = xf − xi = 2.1 − 0 km = 2.1 km

Insight: The distance traveled is always positive, but the displacement can be negative. In this example they are both positive because the final position is greater than the initial position.

Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2–2

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

10. Picture the Problem: This is a follow up question to Guided Example 2.3. A kingfisher bird dives straight down into water to hunt for a fish. distance Strategy: Solve the average speed equation, average speed = , for the distance. elapsed time Solution: 1. Rearrange the equation:

average speed =

distance elapsed time

distance = ( average speed ) × ( elapsed time )

2. Insert numerical values:

æ mö distance = çç4.6 ÷÷÷(1.4 s ) = 6.4 m çè s ÷ø

Insight: The bird is also accelerating, but the details of its speed as a function of time are not needed for calculating distance because we know the bird’s average speed and the time elapsed.

11. Picture the Problem: On a dark and stormy night you see a flash of lightning and then hear the thunder. Strategy: The light travels to your eye nearly instantaneously, but the sound travels at a much slower speed. Solve the distance average speed equation, average speed = , for the distance. elapsed time distance Solution: 1. Rearrange the equation: average speed = elapsed time distance = ( average speed ) × ( elapsed time )

2. Insert numerical values:

distance = 340

m s

1 km = 1.2 km (3.5 s ) = 1200 m × 1000 m

Insight: The distance traveled by the sound wave is approximately ¾ mile. The speed of sound, 340 m/s, works out to approximately one mile every five seconds, a useful rule of thumb for estimating the distance to an approaching thunderstorm!

12. Picture the Problem: A red kangaroo hops at a constant speed. Strategy: Solve the average speed equation, average speed = Solution: 1. (a) Rearrange the equation:

distance , for the distance and then for the time. elapsed time

average speed =

distance elapsed time

distance = ( average speed ) × ( elapsed time )

2. Insert numerical values:

æ km ö÷æç 1 h ö÷ ÷÷ = 3.5 km ÷÷çç3.2 min ´ distance = çç65 çè h ÷øçè 60 min ÷ø

3. (b) Rearrange the equation:

average speed =

4. Insert numerical values:

elapsed time =

distance elapsed time distance elapsed time = average speed distance 0.25 km = average speed km 65 h min s = 0.00385 h × 60 × 60 = 14 s min h

Insight: The distance of 0.25 km is the same as 250 m or about 273 yards. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2–3

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

13. Picture the Problem: A finch travels a short distance on the back of the tortoise and a longer distance through the air. Both displacements are along the same direction. Strategy: First find the total distance traveled by the finch and then determine the average speed by dividing by the total time elapsed. d = s1Dt1 + s2 Dt2 Solution: 1. Determine the total distance traveled: éæ ù æ mö mö s d = êêçç0.060 ÷÷÷(1.2 min ) + çç13 ÷÷÷(1.2 min )úú ´ 60 ÷ ÷ çè ç min sø sø è ë û d = 940 m = 0.94 km 2. Divide the distance by the elapsed time:

average speed =

distance 940 m = = 6.5 m/s elapsed time 2.4 min × 60 s/ min

Insight: Most of the distance traveled by the finch occurred by air. In fact, if we neglect the 4.3 m the finch traveled while on the tortoise’s back, we still get an average speed of 6.5 m/s over the 2.4 min time interval! The bird might as well have been at rest instead of perched upon a tortoise’s back.

14. Picture the Problem: This is a follow up question to Guided Example 2.5. An athlete sprints, stops, and then walks slowly back to the starting line. During the walking portion of the trip, the initial position xi = 50.0 m and xf = 0. Strategy: Solve the average velocity equation, average velocity =

displacement , for the elapsed time. elapsed time

Solution: 1. Rearrange the equation:

average velocity =

2. Substitute the numerical values:

Δt =

displacement Δ x = elapsed time Δ t

t

Δx vav

Δ x xf − xi 0 − 50.0 m = 33.3 s = = vav vav −1.50 m /s

Insight: If the athlete walks faster, the time required to return to the starting line decreases.

15. If two tennis players run with the same speed but in opposite directions, no, their velocities are not equal because velocity is a vector that includes both speed and direction. Two velocities can only be equal if their speeds and directions are the same. 16. Picture the Problem: A person travels around the world in 80 days. Strategy: Use the definitions of average speed and average velocity to answer the questions. Solution: 1. (a) Find the average speed, noting the distance traveled is the 40,075-km circumference of the Earth:

average speed =

2. (b) Because the initial and final positions are the same, Δ x = 0 :

vav =

distance 40, 075 km 1 d = × 24 h elapsed time 80 d

= 20.9 km/h Δx = 0 Δt

Insight: The average velocity over any round trip journey is always zero because the displacement is zero.

Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2–4

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

17. Picture the Problem: Information is given regarding motion of four trains along a north-south line. Strategy: In each case, calculate the average velocity of the train, letting north be the positive direction. Evaluate the velocities to rank them from the most negative to the most positive. Solution: 1. Find the average velocity of train A:

vav, A =

Δ xA +10 m = = +10 m/s Δt 1.0 s

2. Find vav, B :

vav, B =

Δ xB −900 m 1 min = × = −15 m/s Δt 1.0 min 60 s

3. Find vav, C :

vav, C =

Δ xC 10 m = −2 × = −20 m/s 1.0 s Δt

4. Find vav, D :

vav, D =

Δ xD +24 m = = +12 m/s Δt 2.0 s

5. From the above results we can determine the ranking of the velocities from most negative to most positive:

vav, C < vav, B < vav, A < vav, D

Insight: In this scenario object C has the greatest speed but the smallest (that is, the most negative) velocity.

18. The main difference between velocity and speed is that velocity includes information about the direction of travel as well as the rate of travel. Speed only represents the rate of travel, independent of direction. 19. The SI units of speed are meters per second. 20. Average velocity only includes information about the net displacement and the time elapsed. Therefore, yes, it is possible for your friends to have a temporary velocity of −20 m/s at some point even though their average velocity is +20 m/s. For instance, suppose they begin at x = 0 and drive with a velocity of −20 m/s for 5 minutes. Their position is æ 60 s ö÷ æ mö ÷ = - 6000 m. Then they turn around and drive at +30 m/s for 20 minutes. Their now x = çç-20 ÷÷÷(5.0 min )ççç çè çè min ÷÷ø s ÷ø æ 60 s ö÷ æ mö ÷ = +30, 000 m. Their net displacement is +30,000 m and position is now xf = - 6000 m + çç30 ÷÷÷(20.0 min )ççç çè çè min ÷÷ø s ø÷

the time elapsed is 25 min × 60 s/min = 1500 s. Thus their average velocity is vav =

Δ x 30, 000 m = = +20 m/s . 1,500 s Δt

21. If you ride a bicycle around the block, returning to your starting point, your displacement is zero even though the distance you traveled would be hundreds of meters. Therefore, at the end of your trip your average speed is greater than your average velocity. This is because your average velocity is zero because your displacement is zero. 22. Picture the Problem: Usain Bolt sprints in the forward direction. Strategy: The average speed is the distance divided by elapsed time. distance 100.0 m = = 10.44 m/s = 10.4 m/s time 9.58 s

Solution: 1. Divide the distance by the time:

s=

2. Convert the units to km/h:

10.44

m 1 km 3600 s × × = 37.6 km/h 1h s 1000 m

Insight: The displacement would be also be 100 m because the runners typically run in a straight line. Therefore, the average speed and average velocity each have a magnitude of 10.4 m/s, but average velocity includes the direction of travel (south, for example).

Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2–5

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

23. Picture the Problem: The radio waves propagate in a straight line. Strategy: The time elapsed is the distance divided by the average speed. The distance to the Moon is 3.84×108 m. We must double this distance because the signal travels there and back again. Solution: Divide the distance by the average speed:

t=

(

)

8 2d 2 3.84 × 10 m = 2.56 s = s 3.00 × 108 m /s

Insight: The time is slightly shorter than this because the given distance is from the center of the Earth to the center of the Moon, but presumably any radio communications would occur between the surfaces of the Earth and Moon. When the radii of the two spheres is taken into account, the time decreases to 2.51 s.

24. Picture the Problem: A train travels in a straight line at two different speeds. Strategy: We can calculate the average speed with the given information by determining the total distance traveled and dividing by the elapsed time. However, we can arrive at a conceptual understanding of the first answer by remembering that average speed is an average over time, not an average over the distance traveled. Solution: 1. (a) The average speed will be less than 25 m/s because the train spends a longer time driving at the lower speed. The train covers the second 2 km distance in less time at the higher speed than it spent traveling the first 2 km. 2. (b) Find the distance traveled:

Δ x = 2 km + 2 km = 4 km × 1000 m/km = 4000 m

3. Find the time elapsed for the first segment:

Δ t1 =

Δ x1 2 km 1000 m = × = 100 s v1 20 m/s km

4. Find the time elapsed for the second segment:

Δ t2 =

Δ x2 2 km 1000 m = × = 66.67 s v2 30 m/s km

5. Find the average speed:

vav =

Δx 4000 m = = 24 m/s Δ t1 + Δ t2 166.67 s

Insight: Swimming up and down a river always results in a longer time of travel when compared to swimming in a pool because you spend more time swimming against the current than you do swimming with the current.

25. Picture the Problem: You ride your bicycle 1 km at 10 km/h and then another 1 km at 30 km/h. Strategy: We can calculate the average speed with the given information by determining the total distance traveled and dividing by the elapsed time. Solution: 1. Find the distance traveled:

Δ x = 1 km + 1 km = 2 km

3. Find the time elapsed for the first segment:

Δ t1 =

Δ x1 1 km = = 0.100 h v1 10 km /h

4. Find the time elapsed for the second segment:

Δ t2 =

Δ x2 1 km = = 0.033 h v2 30 km /h

5. Find the average speed:

vav =

Δx 2 km = = 15 km/h Δ t1 + Δ t2 0.133 h

Insight: The average speed is less than 20 km/h, as expected, because you spent 75% of your time at the lower speed.

Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2–6

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

26. Picture the Problem: The motion of a parade float is graphed while it moves at various speeds. Strategy: In the first segment of the float’s motion its position increases 2.0 meters every second. In the second segment its position does not change at all for five seconds. In the third segment its position increases 1.0 meter every second. Carefully prepare a graph of these three segments. Solution: 1. (a) The graph of the float’s motion is shown at the right. 2. (b) By examining the graph we can see that the position of the float at t = 2 s is 4.0 m. 3. By examining the graph we can see that the position of the float at t = 11 s is 11 m. Insight: The speed of the float is equal to the slope of the graph. The slopes of the three segments are thus 2.0 m/s, zero, and 1.0 m/s, respectively.

27. On a position-time graph the position is plotted on the vertical axis and the time is plotted on the horizontal axis. 28. The slope of a position-time graph equals the rise over the run, or change in position over the change in time. The slope thus equals the speed of the object. 29. The position-time graph for an object that is moving at constant speed is a straight line whose slope equals the speed of the object. 30. Picture the Problem: Following the motion specified in the position-time graph, a tennis player moves left, then right, then left again, if we take left to be in the negative direction. Strategy: The magnitude of the slope of the position-time graph is equal to the speed of the tennis player. When the position of the player is decreasing, the player has a negative average velocity, and when it is increasing the player has a positive average velocity. Solution: 1. (a) The magnitude of the slope of segment B is the largest of the three, and the slope of segment C is the smallest, so we arrive at the ranking: speed C < speed A < speed B. 2. (b) Find the slope of segment A:

slope A = vav, A =

Δ x −2.0 m = = −1.0 m/s Δt 2.0 s

3. Find the slope of segment B:

slope B = vav, B =

Δ x 2.0 m = = 2.0 m/s Δt 1.0 s

4. Find the slope of segment C:

slope C = vav, C =

Δ x −1.0 m = = − 0.50 m/s Δt 2.0 s

5. From the above calculations we arrive at the ranking of the velocities: velocity A < velocity C < velocity B. Insight: The calculated speed during segment B is larger than the speed during segments A and C, as predicted. Speeds are always positive because they do not involve direction, but velocities can be negative to indicate their direction.

Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2–7

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

31. Picture the Problem: Objects move according to the various positiontime graphs depicted at the right. Strategy: The magnitude of the slope of the position-time graph is equal to the speed of the object. When the position of the object is decreasing, the object has a negative average velocity, and when it is increasing the object has a positive average velocity. Solution: 1. (a) The magnitude of the slope of segment D is the largest of the four, and the slope of segment C is the smallest, so we arrive at the ranking: speed C < speed B < speed A < speed D. 2. (b) Slopes D and C are negative, with slope D having the largest magnitude. Slope A also has a larger magnitude than B. We arrive at the ranking for velocity, from most negative to most positive: velocity D < velocity C < velocity B < velocity A. Insight: If the squares on the plot represent 1.0 meter by 1.0 second, the slope of graph C is: Δ xC −2.0 m slope C = = = − 0.22 m/s Δ tC 9.0 s

32. Picture the Problem: Following the motion specified in the position-time graph, a tennis player moves left, then right, then left again, if we take left to be in the negative direction. Strategy: The average velocity of the tennis player over the time interval from 0 to 5 seconds is the net displacement divided by the time elapsed. Solution: Divide the displacement by the time:

vav, =

Δ x 2.0 m − 3.0 m = = − 0.20 m/s Δt 5.0 s − 0 s

Insight: The average velocity over segments A and B is zero because the displacement is zero. The average speed, however, is ( 2 s )(1 m/s ) + (1 s )( 2 m/s ) (3 s ) = 1.33 m/s.

33. Picture the Problem: A small-gauge train moves slowly back and forth along a straight segment of track. The position-time graph for the train is shown at the right. Strategy: The magnitude of the slope of the position-time graph is equal to the speed of the train. Solution: 1. (a) The slope of segment B is the smallest of the three, so we conclude the train has the smallest speed on portion B of its motion. 2. (b) The speed of the train on portion B is given by the slope: slope B =

Δ xB −3.0 m = = −1.5 m/s 2.0 s Δ tB

vB

1.5 m/s

Insight: The velocity of the train on portion A of its motion is +2.0 m/s and the velocity on portion C is +2.3 m/s.

Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2–8

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

34. Picture the Problem: A small-gauge train moves slowly back and forth along a straight segment of track. The position-time graph for the train is shown at the right. Strategy: The average velocity of the train is given by the displacement divided by the elapsed time. Solution: Divide the displacement by the elapsed time: vav =

Δ x 5.0 m − ( −3.0 m ) 8 m = = = 1.14 m/s 7.0 s 7s Δt

Insight: You could also find the time-weighted average of the velocities of the three portions of the train’s motion: ( 2.0 s)( 2.0 m/s ) + ( 2.0 s)( −1.5 m/s) + (3.0 s)( 2.33 m/s) = 8 m/s. 7.0 s 7

35. Picture the Problem: An equation is given to describe the motion of a bunny hopping across the yard. Strategy: The given linear equation is of the form, xf = xi + v t , where the number that multiplies the variable t is the velocity, and the number that is added to the right side of the equation is the initial position. Use the given equation to infer facts about the motion of the bunny. Solution: 1. (a) The initial position of the bunny is the number that is added to the right side of the equation:

xf = (8.3 m ) + ( 2.2 m/s ) t

xi = 8.3 m

2. (b) The velocity is the number that multiplies the variable t:

xf = (8.3 m ) + ( 2.2 m/s ) t

v = 2.2 m/s

Insight: If the bunny were accelerating there would be a nonlinear relationship between xf and t.

36. Picture the Problem: A bowling ball moves with constant velocity from an initial position of 1.6 m to a final position of 7.8 m in 3.1 s. Strategy: The desired linear equation is of the form, xf = xi + v t , where the number that multiplies the variable t is the velocity, and the number that is added to the right side of the equation is the initial position. Use the given facts to determine the velocity of the bowling ball, and use the velocity together with the initial position to write an equation. Solution: 1. (a) The velocity of the bowling ball is given by its displacement divided by the elapsed time:

vav =

2. Now write an equation according to the pattern xf = xi + v t :

xf = (1.6 m ) + ( 2.0 m/s ) t

Δ x 7.8 m − 1.6 m = = 2.0 m/s 3.1 s Δt

3. (b) Solve the equation xf = xi + v t for t by

xf − xi = v t

subtracting xi from both sides of the equation and then dividing by v:

xf − xi =t v

4. Substitute the numerical values:

t=

xf − xi 8.6 m − 1.6 m = = 3.5 s v 2.0 m /s

Insight: If the bowling ball were accelerating there would be a nonlinear relationship between xf and t.

Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2–9

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

37. Picture the Problem: The position-time graph in the figure at the right shows the motion of two cars, A and B. Strategy: The slope of the graph is the velocity of the car, and the y-intercept of the graph is the initial position of the car. Use these facts to answer questions about the motions of the two cars. Solution: 1. (a) The slope of the graph for car A is smaller than the slope of the graph for car B. Therefore, we conclude that the velocity of car A is less than the velocity of car B. 2. (b) The y-intercept of the graph for car A is greater than the y-intercept of the graph for car B. Therefore, we conclude that the initial position of car A is greater than the initial position of car B. 3. (c) The position of car A is larger than the position of car B during the time interval t = 0 to t = 1 s. Therefore, we conclude that car A is ahead of car B during that time interval. Insight: If you do a numerical analysis of the motions of the two cars, you’ll discover that car A is traveling at 0.714 m/s and car B is going 5.0 m/s (about 11 mph), pretty slow for normal cars, but reasonable speeds for toy cars. The two cars pass each other at 1.75 s when each car is at the position 13.75 m.

38. Picture the Problem: This is a follow up question to Guided Example 2.7. At the moment you step onto the walkway your friend is 7.8 m ahead and walking with a speed of 2.3 m/s. Your speed on the walkway is 4.2 m/s. Your friend stops walking 2.2 s after you step onto the walkway. Strategy: Draw a sketch of the motions of you and your friend like that shown at the right. Use the sketch together with the equations of motion to find the time at which you and your friend are at the same position. The equations of motion are given in Guided Example 2.7: xf, you = ( 4.2 m/s ) t

and xf, friend = ( 7.8 m ) + ( 2.3 m/s ) t .

Solution: 1. Find the position of your friend 2.2 s after you step onto the moving walkway. This is the position at which the two of you will eventually meet because your friend stops walking at this instant.

xf, friend = ( 7.8 m ) + ( 2.3 m/s ) t

= ( 7.8 m ) + ( 2.3 m/s )( 2.2 s ) = 12.9 m

xf, you = ( 4.2 m/s ) t

2. Find the time at which you will reach 12.9 m:

xf, you

4.2 m/s

=t=

12.9 m = 3.1 s 4.2 m /s

Insight: If you do a numerical analysis of the motions of you and your friend, you’ll discover that your friend stops walking at position 12.86 m, and that you catch your friend at exactly at t = 3.062 s.

Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2 – 10

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

39. Picture the Problem: At a particular instant your car is at the origin and is traveling in the positive direction at 26 m/s. A truck on the other side of the highway is at position 420 m and is moving at 31 m/s in a direction opposite to that of your car. Strategy: Write equations of motion for your car and for the truck, noting that the velocity of the truck is in the negative direction. Use the equations of motion to draw graphs of the positions of your car and that of the truck. Finally, use the equations of motion to determine the time at which your car and the truck have the same position. Solution: 1. (a) Write an equation of motion for your car, noting that it is at the origin at t = 0 and is moving in the positive direction:

xf, car = xi, car + v t = ( 0 m ) + ( 26 m/s ) t = ( 26 m/s ) t

2. Write an equation of motion for the truck, noting that it is at 420 m at t = 0 and is moving in the negative direction:

xf, truck = xi, truck + v t = ( 420 m ) + ( −31 m/s ) t

3. (b) The two equations of motion were used to produce the graph shown above. xf, car = xf, truck

4. (c) Find the time at which your car and the truck have the same position:

( 26 m/s) t = ( 420 m ) + ( −31 m/s) t ( 26 + 31 m/s) t = 420 m t=

420 m = 7.4 s 57 m /s

Insight: If the truck were traveling in the same direction as your car, you would never catch up to it because it would be ahead of you and moving at a higher speed.

40. The information that can be obtained from the position-time equation includes the initial position of the object, the rate at which the position is changing (that is, the velocity of the object), and the time at which the object has its final position. 41. If a straight line represents the motion of an object on a position-time graph, the slope of the graph represents the velocity of the object and the y-intercept of the graph represents the initial position of the object. 42. The equations of motion for two bicycles are given. The number in each equation that multiplies the variable t is the velocity of the bicycle. We can see that bicycle 1 has a velocity of 2.5 m/s and bicycle 2 has a velocity of −3.5 m/s. The speeds of each bicycle are the magnitudes of these velocities. We conclude that bicycle 2 has the greater speed. 43. The equations of motion for two bicycles are given. The number in each equation that is added to the right side is the initial position of the bicycle. We can see that bicycle 1 has an initial position of −4.0 m and bicycle 2 has an initial position of 6.7 m. Therefore, at time t = 0 the two bicycles are separated by 6.7 m − (−4.0 m) = 10.7 m. 44. Picture the Problem: An equation is given to describe the motion of a ball. Strategy: The given linear equation is of the form, xf = xi + v t , where the number that multiplies the variable t is the velocity, and the number that is added to the right side of the equation is the initial position. Use the given equation to find the final position of the ball. xf = (3.0 m ) + ( −5.0 m/s ) t

Solution: The final position of the ball can be found by using the given equation and inserting the time:

(

)(

)

= (3.0 m ) + −5.0 m/ s 1.5 s = − 4.5 m

Insight: If the ball had a velocity of +5.0 m/s it would be at the position 10.5 m at t = 1.5 s.

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Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

45. Picture the Problem: The equations of motion are given for two bumper cars that move with constant velocity. Strategy: Use the given equations of motion to draw graphs of the positions of the bumper cars. The graphs are not essential for answering the questions but can help you picture the situation. Finally, use the equations of motion to determine which car is moving faster and find the time at which the two bumper cars will collide. The two equations are:

xf, 1 = xi, 1 + v1 t = ( − 4.0 m ) + (1.5 m/s ) t

xf, 2 = xi, 2 + v2 t = (8.8 m ) + ( −2.5 m/s ) t

Solution: 1. (a) In the equations of motion the number that multiplies t is the velocity of the bumper car. By inspection you can see that the velocity of the first bumper car is 1.5 m/s and the velocity of the second bumper car is −2.5 m/s. The speed of each car is the magnitude of its velocity, so we conclude that bumper car 2 is traveling faster. xf, 1 = xf 2

2. (b) Set the final positions of the two bumper cars equal to each other:

xi, 1 + v1 t = xi, 2 + v2 t

3. Subtract v2 t and xi, 2 from both sides,

− v2 t + v1t = x2, i − x1, i

rearrange the equation, and solve for t:

(v1 − v2 ) t = x2, i − x1, i t=

x2, i − x1, i v1 − v2

=

8.8 − ( − 4.0) m

1.5 − ( −2.5) m /s

= 3.2 s

Insight: If bumper car 1 had a faster velocity its position-time graph would be steeper, and it would thus intersect the graph for bumper car 2 sooner than 3.2 s. In fact, if the velocity of bumper car 1 were +2.5 m/s instead of +1.5 m/s, the two cars would collide at t = 2.56 s.

46. Yes, the distance traveled by an object can be greater than its displacement. For example, if you walk all the way around the block and return to your starting position, your displacement is zero but the distance you have traveled is several hundred meters. 47. No, the distance traveled by an object cannot be less than its displacement. If you ride a bicycle in a straight line your displacement and distance traveled are equal. However, any deviation from straight line travel will result in a distance traveled that is greater than the displacement. 48. An astronaut that has just completed an orbit of the Earth has traveled a long distance (about 41,300 km for low Earth orbit) but has zero displacement because she ended her journey exactly where she started. However, you travel some distance on your way home from school (unless you live on the school campus) and so you have a small but nonzero displacement. Therefore you have a greater displacement when you travel home from school than an astronaut when she completes an orbit. 49. In this situation your displacement is xf − xi = 25 − 10 m = 15 m. Your friend’s displacement is xf − xi = 40 − 35 m = 5 m. Assuming each person travels along a straight line and in one direction, the distances traveled are equal to the displacements. We conclude that the distance you have traveled is greater than the distance traveled by your friend. 50. In this situation your displacement is xf − xi = 25 − 20 m = 5 m. Your friend’s displacement is xf − xi = 30 − 35 m = −5 m. We conclude that you have a positive displacement and your friend has a negative displacement. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

51. Picture the Problem: You walk in both the positive and negative directions along a straight line. Strategy: The distance is the total length of travel, and the displacement is the net change in position. Solution: (a) Add the lengths to find the distance traveled:

(0.75 + 0.60 km ) + (0.60 km ) = 1.95 km

(b) Subtract xi from xf to find the displacement.

Δ x = xf − xi = 0.75 − 0.00 km = 0.75 km

Insight: The distance traveled is always positive, but the displacement can be negative.

52. Picture the Problem: You walk in both the positive and negative directions along a straight line. Strategy: The distance is the total length of travel, and the displacement is the net change in position.

Solution: (a) Add the lengths to find the distance traveled:

(0.60 + 0.35 km ) + (0.35 + 0.60 + 0.75 km ) =

(b) Subtract xi from xf to find the displacement.

Δ x = xf − xi = 0.00 − 0.75 km = − 0.75 km

2.65 km

Insight: The distance traveled is always positive, but the displacement can be negative.

53. Picture the Problem: Your motion results in a displacement on some coordinate axis. Strategy: The displacement is the difference between your initial and final positions. Use this fact to determine your final position given your initial position and displacement. Δ x = xf − xi

Solution: Solve the displacement equation for the final position by adding xi to both sides:

Δ x + xi = xf

(6.2 m ) + ( 4.5 m ) = xf

= 10.7 m

Insight: The displacement simply represents the change in your position. Here you changed your position in the positive direction and ended up at a larger position value.

54. Picture the Problem: Your motion results in a displacement on some coordinate axis. Strategy: The displacement is the difference between your initial and final positions. Use this fact to determine your final position given your initial position and displacement. Δ x = xf − xi

Solution: Solve the displacement equation for the final position by adding xi to both sides:

Δ x + xi = xf

( −8.3 m ) + (7.5 m ) = xf

= − 0.8 m

Insight: The displacement simply represents the change in your position. Here you changed your position in the negative direction and ended up at a smaller position value.

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2 – 13

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

55. Picture the Problem: A train moves so that it has a displacement on some coordinate axis. Strategy: The displacement is the difference between the train’s initial and final positions. Use this fact to determine the initial position of the train given its final position and displacement. Δ x = xf − xi

Solution: Solve the displacement equation for the initial position by adding xi to both sides and then subtracting Δ x from both sides:

xi = xf − Δ x xi = ( 4.3 m ) − ( −26 m ) = 30.3 m = 30 m

Insight: The displacement simply represents the change in the train’s position. Here the train changed its position in the negative direction and ended up at a smaller position value than where it started.

56. Picture the Problem: A train moves so that it has a displacement on some coordinate axis. Strategy: The displacement is the difference between the train’s initial and final positions. Use this fact to determine the initial position of the train given its final position and displacement. Δ x = xf − xi

Solution: Solve the displacement equation for the initial position by adding xi to both sides and then subtracting Δ x from both sides:

xi = xf − Δ x xi = ( −2.2 m ) − (17 m ) = −19.2 m = −19 m

Insight: The displacement simply represents the change in the train’s position. Here the train changed its position in the positive direction and ended up at a larger (more positive) position value than where it started.

57. Picture the Problem: Tennis player A walks in the positive direction and player B walks in the negative direction, as shown in the figure at the right. Strategy: In each case the distance is the total length of travel, and the displacement is the net change in position. Solution: (a) Note the distance traveled by player A:

5m

The displacement of player A is positive:

Δ x = xf − xi = 5 m − 0 m = 5 m

(b) Note the distance traveled by player B:

2m

The displacement of player B is negative. Let the origin be at the initial position of player A.

Δ x = xf − xi = 5 m − 7 m = −2 m

Insight: The distance traveled is always positive, but the displacement can be negative.

58. Picture the Problem: The ball is putted in the positive direction and then the negative direction. Strategy: The distance is the total length of travel, and the displacement is the net change in position.

Solution: (a) Add the lengths to find the distance traveled:

(10 + 2.5 m) + 2.5 m = 15 m

(b) Subtract xi from xf to find the displacement.

Δ x = xf − xi = 10 − 0 m = 10 m

Insight: The distance traveled is always positive, but the displacement can be negative.

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2 – 14

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

59. Picture the Problem: The runner moves along the oval track. Strategy: The distance is the total length of travel, and the displacement is the net change in position. For the distance traveled, the runner first goes 15 m in the –y direction, then 100 m in the +x direction, and then 15 m in the +y direction. Solution: 1. (a) Add the lengths to find the distance traveled:

(15 m ) + (100 m ) + (15 m ) = 130 m

2. Subtract xi from xf to find the displacement.

Δ x = xf − xi = 100 − 0 m = 100 m

3. (b) Add the lengths to find the distance traveled:

15 + 100 + 30 + 100 + 15 m = 260 m

4. Subtract xi from xf to find the displacement.

Δ x = x f − xi = 0 − 0 m = 0 m

Insight: The distance traveled is always positive, but the displacement can be negative. The displacement is always zero for a complete circuit, as in this case.

60. The average velocity of your dog is its displacement divided by the time elapsed. If the average velocity of the dog is zero, we can conclude that its displacement is also zero. 61. An astronaut that has just completed an orbit of the Earth has traveled a long distance (about 41,300 km for low Earth orbit) at very high speed (about 8 km/s) but has zero displacement because she ended her journey exactly where she started. However, you travel some distance on your way home from school (unless you live on the school campus) and so you have a small but nonzero displacement. Because average velocity is the displacement divided by the elapsed time, we conclude that you have a greater average velocity when you travel home from school than when an astronaut completes an orbit. 62. Yes, it is possible for two different objects to have the same speed but different velocities if they are traveling in different directions. 63. No, it is not possible for two different objects to have the same velocity but different speeds, because the magnitude of the velocity vector is the speed of an object. If two objects have the same velocity, they must also have the same speed and be traveling in the same direction. 64. Picture the Problem: You drive your car in a straight line at two different speeds. Strategy: We could calculate the average speed with the given information by determining the total distance traveled and dividing by the elapsed time. However, we can arrive at a conceptual understanding of the answer by remembering that average speed is an average over time, not an average over the distance traveled. Solution: (a) The average speed will be less than 20 m/s because you will spend a longer time driving at the lower speed. You will cover the 10 km distance in less time at the higher speed than you did at the lower speed. (b) The best answer is A. More time is spent driving at 15 m/s than at 25 m/s because the distances traveled at each speed are the same, and it will take a longer time at the slower speed to cover the same distance. Statement B is true but irrelevant, and statement C is false. Insight: The time elapsed at the lower speed is (10, 000 m ) (15 m/s ) = 667 s and the time elapsed at the higher speed is

(10, 000 m ) ( 25 m/s) = 400 s, so the average speed is ( 20, 000 m) (1067 s) = 18.7 m/s.

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2 – 15

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

65. Picture the Problem: You drive your car in a straight line at two different speeds. Strategy: We could calculate the average speed with the given information by determining the total distance traveled and dividing by the elapsed time. However, we can arrive at a conceptual understanding of the answer by remembering that average speed is an average over time, not an average over the distance traveled. Solution: (a) The average speed will be equal to 20 m/s because you will spend an equal amount of time driving at the lower speed as at the higher speed. The average speed is therefore the mean value of the two speeds. (b) The best answer is C. Equal time is spent at 15 m/s and 25 m/s because that fact is stated in the question. Statements A and B are both false. Insight: The distance traveled at the lower speed would be (15 m/s )( 600 s ) = 9000 m and the distance traveled at the higher speed would be ( 25 m/s )( 600 s ) = 15, 000 m so the average speed is ( 24, 000 m ) (1200 s ) = 20.0 m/s.

66. Picture the Problem: A swimmer swims in the forward direction. Strategy: The average speed is the distance divided by elapsed time. average speed =

Solution: Divide the distance by the time:

distance 100.0 m = = 1.883 m/s time 53.12 s

1.883

m 1 mi 3600 s mi × × = 4.212 1h h s 1609 m

Insight: The displacement would be zero in this case because the swimmer swims either two lengths of a 50-m pool or four lengths of a 25-m pool, returning to the starting point each time. However, the average speed depends upon distance traveled, not displacement. 67. Picture the Problem: Rubber ducks drift along the ocean surface. Strategy: The average speed is the distance traveled divided by elapsed time. Solution: Divide the distance by the time:

s=

d 2600 km 1000 m 1 mo 1 d = 0.099 m/s = × × × t 1 km 10 mo 30.5 d 8.64 × 104 s

Insight: The instantaneous speed might vary from 0.099 m/s, but we can calculate only average speed from the total distance traveled and the total time elapsed. 68. Picture the Problem: A roller coaster moves along a track. Strategy: The average speed is the distance traveled divided by elapsed time. Use this fact to solve for the distance traveled given the speed and the time. Solution: Solve the average speed equation for the distance traveled by multiplying both sides by the time:

distance elapsed time æ mö speed ´ elapsed time = distance = çç12 ÷÷´ ÷ (5.5 s ) = 66 m çè s ÷ø average speed =

Insight: The roller coaster is moving at a good clip. In more familiar units, 12 m/s is equivalent to 27 mph.

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2 – 16

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

69. Picture the Problem: A billiard ball rolls along the surface of the table. Strategy: The average speed is the distance traveled divided by elapsed time. Use this fact to solve for the elapsed time traveled given the speed and the distance. Solution: Solve the average speed equation for the elapsed time by multiplying both sides by the time and dividing by the average speed:

average speed = elapsed time =

distance elapsed time distance 0.23 m = 0.30 s = average speed 0.76 m /s

Insight: The average speed equation relates the three quantities speed, time, and distance. Given any two of these quantities, the third can be found with this equation.

70. Picture the Problem: You travel 8.00 km on foot and then an additional 16.0 km by car, with both displacements along the same direction. Strategy: First find the total elapsed time by dividing the distance traveled by the average speed. Also find the time elapsed while jogging by dividing the jogging distance by the jogging speed. Subtract the jogging time from the total time to find the driving time, and use the driving distance and driving time to find the driving speed. distance 8.00 + 16.0 km = 1.09 h = average speed 22.0 km /h

Solution: 1. Use the definition of average speed to determine the total time elapsed.

elapsed time =

2. Find the time elapsed while jogging:

Δ t jog =

3. Find the time elapsed while in the car:

Δ tcar = Δ t − Δ t jog = 1.09 h − 0.842 h = 0.25 h

4. Find the speed of the car:

vcar =

d jog v jog

=

8.00 km = 0.842 h 9.50 km /h

d car 16.0 km = = 64 km/h Δ tcar 0.25 h

Insight: This problem illustrates the limitations that significant figures occasionally impose. If there were only two significant figures, the total elapsed time would be 1.1 h and the elapsed time for the car trip would be 0.3 h, not 0.25 h, and the speed of the car would be only 53 km/h, which would have to be rounded to 50 km/h to keep only one significant digit.

71. Picture the Problem: The dog continuously runs back and forth as the owners close the distance between each other. Strategy: First find the time that will elapse before the owners meet each other. Then determine the distance the dog will cover if it continues running at constant speed over that time interval. Solution: 1. Find the time it takes each owner to walk 5.00 m before meeting the other:

elapsed time =

2. Find the distance the dog runs:

d = vav Δ t = 3.0 m/ s 3.8 s = 11 m

(

distance 5.00 m = 3.8 s = average speed 1.3 m /s

)(

)

Insight: The dog will actually run a shorter distance than this, because it is impossible for it to maintain the same 3.0 m/s as it turns around to run to the other owner. It must first slow down to zero speed and then accelerate again.

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2 – 17

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

72. Picture the Problem: You drive in a straight line at two different speeds during the specified time interval. Strategy: Determine the average speed by first calculating the total distance traveled and then dividing it by the total time elapsed. Solution: 1. (a) Because the time intervals are the same, you spend equal times at 20 m/s and 30 m/s, and your average speed will be equal to 25.0 m/s. v Δt + v Δt 2. (b) Divide the total distance vav = 1 1 2 2 by the time elapsed: Δt1 + Δt2 =

(20.0 m/ s ) (10.0 min × 60 s / min ) + (30.0 m/ s ) (600 s ) = 25.0 m/s 600 + 600 s

Insight: The average speed is a weighted average according to how much time you spend traveling at each speed.

73. Picture the Problem: You drive in a straight line at two different speeds during the specified time interval. Strategy: Determine the average speed by first calculating the total distance traveled and then dividing it by the total time elapsed. Solution: 1. (a) The distance intervals are the same but the time intervals are different. You will spend more time at the lower speed than at the higher speed. Because the average speed is a time weighted average, it will be less than 25.0 m/s. 2. (b) Divide the total distance by the time elapsed:

vav =

d1 + d 2 d + d2 20.0 km = 1 = æ ö d1 d 2 Dt1 + Dt2 çç10.0 km + 10.0 km ÷÷ + s1 s2 ççè 20.0 m/s 30.0 m/s ø÷÷

vav = 24.0 m/s

Insight: Notice that in this case it is not necessary to convert miles to meters in both the numerator and denominator because the units cancel out and leave m/s in the numerator.

74. (a) Yes, an object’s position-time graph can be horizontal. That would correspond to zero distance traveled over a very long time, which means the object is at rest. (b) No, an object’s position-time graph cannot be vertical. That would correspond to a very large distance traveled over zero time, which means the object is traveling at infinite speed. We will later learn that the speed of light, 300,000 km/s, is the speed limit for any mass in our universe! 75. The slope of an object’s position-time graph is equal to its velocity. Therefore, we conclude that the velocity of an object whose position-time graph is a straight line with a positive slope is positive. 76. The speed of an object is always positive. If the object has a position-time graph that is a straight line with a negative slope, we can conclude the velocity is negative, but the speed is positive. 77. The slope of an object’s position-time graph is equal to its velocity. Therefore, we conclude that the instantaneous velocity of an object is negative where the slope of the tangent line to its position-time graph is negative.

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2 – 18

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

78. Picture the Problem: Two trains move on a track according to the position-time graphs shown below.

A

B

C

Strategy: Use the position-time graphs to answer the questions about the motions of the two trains. Note that the slope of the position-time graph is equal to the velocity of the train. Solution: 1. (a) The trains travel in opposite directions when one has a positive velocity and the other has a negative velocity. This is the case when one position-time graph has a positive slope and the other has a negative slope. We conclude the trains travel in opposite directions in cases B and C. 2. (b) The trains get farther apart when the vertical spacing between the two position-time graphs increases as time increases. We conclude the trains get farther apart as they travel in cases A and C. In case B the trains first get closer together and then get further apart (unless a collision occurs!) 3. (c) The trains will collide if they are running on the same track and their position-time graphs intersect. In such a case the trains would have the same position at the same instant of time. We conclude the trains collide in case B. Insight: The case B graph would not result in a collision if the two trains were on parallel tracks and their positions were measured from the same point of origin.

79. Picture the Problem: Two trains move on a track according to the position-time graphs shown below.

A

B

C

Strategy: Use the position-time graphs to answer the questions about the motions of the two trains. Note that the slope of the position-time graph is equal to the velocity of the train. Solution: 1. (a) In case A both trains 1 and 2 have positive velocities. In case B train 2 has a positive velocity. In case C train 1 has a positive velocity. 2. (b) In case A neither train has a negative velocity. In case B train 1 has a negative velocity. In case C train 2 has a negative velocity. Insight: The case B graph would not result in a collision if the two trains were on parallel tracks and their positions were measured from the same point of origin.

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2 – 19

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

80. Picture the Problem: Following the motion specified in the position-time graph shown at the right, the father walks forward, stops, walks forward again, and then walks backward. Strategy: The direction of the velocity (positive or negative) matches the sign of the slope of the graph (positive or negative) at each point. Solution: 1. (a) The slope of segment A is positive so the velocity on segment A is positive. 2. (b) The velocity on segment B is zero. 3. (c) The velocity on segment C is positive. 4. (d) The velocity on segment D is negative. Insight: With practice you can form both a qualitative and quantitative “movie” of the father’s motion in your head simply by examining the position-time graph.

81. Picture the Problem: Following the motion specified in the position-time graph shown at the right, the father walks forward, stops, walks forward again, and then walks backward. Strategy: Determine the magnitude of the velocity by calculating the slope of the graph at each specified point. Solution: 1. (a) Find the slope of the graph at A:

vav =

Δx 2.0 m = = 2.0 m/s Δt 1.0 s

2. (b) Find the slope of the graph at B:

vav =

Δx 0.0 m = = 0.0 m/s Δt 1.0 s

3. (c) Find the slope of the graph at C:

vav =

Δx 1.0 m = = 1.0 m/s Δt 1.0 s

Δx −3.0 m = = −1.5 m/s Δt 2.0 s Insight: The signs of each answer in (a) through (d) match those predicted in parts (a) through (d) in the previous question.

4. (d) Find the slope of the graph at D:

vav =

82. Picture the Problem: A particle is at x = 5.0 m at t = 0 and moves with a constant velocity of 3.5 m/s. Strategy: Construct an equation to describe the motion of the particle and use it to either generate (x, y) points and plot them by hand, or to generate a graph by means of a computer spreadsheet or plotting program. Solution: 1. Write a formula to describe the particle’s motion:

xf = xi + v t xf = (5.0 m ) + (3.5 m/s ) t

2. Use the formula to generate a position-time graph like that shown at the right. Insight: A particle that moves at constant speed has a linear position-time graph.

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2 – 20

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

83. Picture the Problem: A particle is at x = 3.1 m at t = 0 and moves with a constant velocity of −2.7 m/s. Strategy: Construct an equation to describe the motion of the particle and use it to either generate (x, y) points and plot them by hand, or to generate a graph by means of a computer spreadsheet or plotting program. Solution: 1. Write a formula to describe the particle’s motion:

xf = xi + v t xf = (3.1 m ) + ( −2.7 m/s ) t

2. Use the formula to generate a position-time graph like that shown at the right. Insight: A particle that moves at constant speed has a linear position-time graph.

84. Picture the Problem: An object starts at x = 1.5 m, moves with a velocity of 2.2 m/s from t = 0 to t = 1 s, has a velocity of 0 m/s from t = 1 s to t = 2 s, and has a velocity of –3.7 m/s from t = 2 s to t = 5 s. Strategy: Construct equations to describe the motion of the particle and use them to either generate (x, y) points and plot them by hand, or to generate a graph by means of a computer spreadsheet or plotting program. Solution: 1. Write a formula to describe the first part of the particle’s motion:

xf = xi + v t xf = (1.5 m ) + ( 2.2 m/s ) t

2. In the second part of its motion, the particle is at rest. Therefore, at t = 2.0 s it will still have the position it attained after 1.0 s had elapsed: xf = (1.5 m ) + ( 2.2 m/s )(1.0 s ) = 3.7 m. 3. The slope of the third line will be −2.7 m/s, but what about the intercept? Use the knowledge that the object has a position of 3.7 m at t = 2.0 s, together with the known slope, to solve for the intercept xi . 3. Use the known point and slope to find the intercept:

xf = xi + v t

(

)(

3.7 m = xi + −3.7 m/ s 2.0 s

)

3.7 m = xi − 7.4 m 3.7 + 7.4 m = xi = 11.1 m thus the equation of motion is: xf = (11.1 m ) + ( −3.7 m/s ) t

Insight: The only reason you would need to determine the intercept for the third segment would be to enter the appropriate formula into a computer spreadsheet. You could avoid the calculation and simply plot that segment by hand and still capture the essence of the object’s motion: It moves in the positive direction, stops, then moves in the reverse direction at a constant speed.

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2 – 21

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

85. Picture the Problem: The figure shown at the right displays the position-time graphs for two different objects, A and B. Strategy: The position-time graph for object A has a negative slope and therefore it has a negative velocity. Meanwhile, object B has a positive velocity. The initial positions for each can be determined by finding the location where the position-time graphs intersect the y-axis, where t = 0. Solution: 1. (a) By inspecting the y-intercept of the position-time graph for object A we see that it has a value of 35 m at t = 0. 2. (b) By inspecting the y-intercept of the position-time graph for object B we see that it has a value of 10 m at t = 0. Insight: Although they start at different positions, objects A and B will have the same position when the two lines cross.

86. Picture the Problem: The figure shown at the right displays the positiontime graphs for two different objects, A and B. Strategy: The velocity of an object equals the slope of its position-time graph. The position-time graph for object A has a negative slope and therefore it has a negative velocity. Meanwhile, object B has a positive velocity. To find the velocities, pick two points on the line and calculate the slope by dividing the rise by the run. Solution: 1. (a) Choose the points (0 s, 35 m) and (3.5 s, 5 m) to find the velocity:

vav, A =

rise 5 − 35 m = = −8.6 m/s run 3.5 − 0 s

2. (b) Choose the points (0 s, 10 m) and (3.5 s, 25 m) to find the velocity:

vav, B =

rise 25 − 10 m = = 4.3 m/s run 3.5 − 0 s

+ + +

+

Insight: We can now write formulas for the positions of each object: xf, A = (35 m ) − (8.6 m/s ) t and

xf, B = (10 m ) + ( 4.3 m/s ) t . By setting these two formulas equal to each other and solving for t, we can determine that

the two objects will have the same position at t = 1.94 s. 87. Picture the Problem: Following the motion specified in the position-time graph shown at the right, the father walks forward, stops, walks forward again, and then walks backward. Strategy: Determine the magnitude of the velocity by dividing the displacement by the elapsed time. At t = 1 s the position of the father is 2 m, and at t = 3 s his position is 3 m. Solution: Divide the displacement by the elapsed time:

vav =

Δ x 3−2 m = = 0.50 m/s Δt 3 −1 s

Insight: The father moved four times faster in the interval between t = 0 and t = 1 s, when his average velocity was 2.0 m/s.

88. Yes, it is possible for two different objects to have the same initial position but different velocities because an object’s velocity is completely distinct from its initial position. 89. Yes, it is possible for two different objects to have the same velocity but different initial positions, because an object’s velocity is completely distinct from its initial position. In order for the objects to have the same velocity they need only have the same speed and be traveling in the same direction.

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Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

90. When the initial position of an object changes, the intercept of its position-time graph changes as well. However, the slope of the position-time graph simply represents the velocity of the object, and it does not change when the object’s initial position changes. 91. When the velocity of an object changes, the slope of its position-time graph changes as well. However, the intercept of the position-time graph simply represents the initial position of the object, and it does not change when the object’s velocity changes. 92. The equations of motion for two fish in a river are given. The number in each equation that multiplies the variable t is the velocity of the fish. We can see that fish 1 has a velocity of −1.2 m/s and fish 2 has a velocity of −2.7 m/s. The speeds of each fish are the magnitudes of these velocities. We conclude that fish 2 has the faster speed. 93. The equations of motion for two people walking on a sidwalk are given. (a) The number in each equation that multiplies the variable t is the velocity of the person. We can see that person 1 has a velocity of −1.1 m/s and person 2 has a velocity of 1.7 m/s. The speeds of each person are the magnitudes of these velocities. We conclude that person 2 has the fastest speed. (b) Both people have a positive initial position, but the velocity of person 1 is negative, and we conclude that person 1 will eventually reach the x = 0 position. 94. Picture the Problem: The figure shown at the right displays the positiontime graphs for two different objects, A and B. Strategy: The velocity of an object equals the slope of its position-time graph. The position-time graph for object A has a negative slope so that it must have a negative velocity. To find the velocity, pick two points on the line and calculate the slope by dividing the rise by the run. Solution: 1. Choose the points (0 s, 35 m) and (3.5 s, 5 m) to find the velocity:

vav, A =

2. Use the initial position of 35 m to write the equation of motion:

xf, A = (35 m ) − (8.6 m/s ) t

rise 5 − 35 m = = −8.6 m/s run 3.5 − 0 s

+ + +

+

Insight: We can now use the equation of motion to discover things like object A will arrive at xf, A = 0 at t = 4.08 s.

95. Picture the Problem: The figure shown at the right displays the positiontime graphs for two different objects, A and B. Strategy: The velocity of an object equals the slope of its position-time graph. The position-time graph for object B has a positive slope so that it must have a positive velocity. To find the velocity, pick two points on the line and calculate the slope by dividing the rise by the run. Solution: 1. Choose the points (0 s, 10 m) and (3.5 s, 25 m) to find the velocity:

vav, B =

2. Use the initial position of 10 m to write the equation of motion:

xf, B = (10 m ) + ( 4.3 m/s ) t

rise 25 − 10 m = = 4.3 m/s run 3.5 − 0 s

+ + +

+

Insight: We can now compare the equations of motion of each object, recalling that xf, A = (35 m ) − (8.6 m/s ) t from the

previous question. By setting these two formulas equal to each other and solving for t, we can determine that the two objects will have the same position at t = 1.94 s.

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2 – 23

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

96. Picture the Problem: A rabbit moves with a known initial position and a constant velocity. Strategy: The velocity of an object is the number that multiplies t in its equation of motion. The number that is added to the equation is the initial position. Use the given initial position and velocity to write the equation of motion for the rabbit. Solution: Use the initial position of 8.1 m and the velocity of −1.6 m/s to write the equation of motion:

xf = (8.1 m ) + ( −1.6 m/s ) t

Insight: We can use the equation of motion to discover things like the rabbit will arrive at xf = 0 at t = 5.1 s.

97. Picture the Problem: The equation of motion for a bicycle is given as: xf = ( 6.0 m ) + ( 4.5 m/s ) t . Strategy: The velocity of an object is the number that multiplies t in its equation of motion. The number that is added to the equation is the initial position. We can therefore examine the equation of motion to learn that the bicycle starts at 6.0 m at t = 0 s and travels with a constant velocity of 4.5 m/s. Use these facts to answer the questions. Solution: 1. (a) Use the equation of motion to find the bicycle’s position at t = 2.0 s:

xf = ( 6.0 m ) + ( 4.5 m/s )( 2.0 s ) = 15.0 m

xf = xi + v t

2. (b) Solve the equation of motion for t to find the time when xf = 24 m:

xf − xi = v t xf − xi 24 − 6.0 m =t= = 4.0 s v 4.5 m /s

Insight: We can use the equation of motion to discover facts like the bicycle was at x = 0 m at t = −1.33 s.

98. Picture the Problem: The equation of motion for a parade float is given as: xf = ( −9.2 m ) + (1.5 m/s ) t . Strategy: The velocity of an object is the number that multiplies t in its equation of motion. The number that is added to the equation is the initial position. We can therefore examine the equation of motion to learn that the parade float starts at −9.2 m at t = 0 s and travels with a constant velocity of 1.5 m/s. Use these facts to answer the questions. Solution: 1. (a) Use the equation of motion to find the float’s position at t = 3.5 s:

xf = ( −9.2 m ) + (1.5 m/s )(3.5 s ) = − 4.0 m xf = xi + v t

2. (b) Solve the equation of motion for t to find the time when xf = 0:

xf − xi = v t

(

)

0 − −9.2 m xf − xi = 6.1 s =t= v 1.5 m /s

Insight: If the parade float maintains its constant speed, it will travel a city block (about 100 m) in 67 seconds.

99. Picture the Problem: The equation of motion for Cleo the black lab is given as: xf = ( −12.1 m ) + (5.2 m/s ) t . Strategy: The velocity of an object is the number that multiplies t in its equation of motion. The number that is added to the equation is the initial position. We can therefore examine the equation of motion to learn that the dog starts at −12.1 m at t = 0 s and travels with a constant velocity of 5.2 m/s. Use these facts to answer the questions. Solution: 1. (a) Use the equation of motion to find Cleo’s position at t = 1.6 s:

xf = ( −12.1 m ) + (5.2 m/s )(1.6 s ) = −3.8 m xf = xi + v t

2. (b) Solve the equation of motion for t to find the time when Cleo’s position is the same as the stick’s position at xf = 3.0 m:

xf − xi = v t

(

)

3.0 − −12.1 m xf − xi = 2.9 s =t= v 5.2 m /s

Insight: We can use the equation of motion to discover facts like Cleo will be at x = 0 m at t = 2.33 s. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2 – 24

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

100. Picture the Problem: The equations of motion are given for two football players that move with constant velocity. Strategy: Use the given equations of motion to draw graphs of the positions of the football players. The graphs are not essential for answering the questions but can help you picture the situation. Finally, use the equations of motion to determine which player is moving faster and find the time at which the two football players will collide. The two equations are:

xf, 1 = xi, 1 + v1 t = ( 0.1 m ) + ( −3.1 m/s ) t

xf, 2 = xi, 2 + v2 t = ( − 6.3 m ) + ( 2.8 m/s ) t Solution: 1. (a) In the equations of motion the number that multiplies t is the velocity of the football player. By inspection you can see that the velocity of the first football player is −3.1 m/s and the velocity of the second football player is 2.8 m/s. The speed of each player is the magnitude of his velocity, so we conclude that football player 1 is traveling faster than football player 2. xf, 1 = xf 2

2. (b) Set the final positions of the two football players equal to each other:

xi, 1 + v1 t = xi, 2 + v2 t

3. Subtract v2 t and xi, 2 from both sides,

− v2 t + v1t = x2, i − x1, i

rearrange the equation, and solve for t:

(v1 − v2 ) t = x2, i − x1, i t=

x2, i − x1, i v1 − v2

=

− 6.3 − ( 0.1) m

−3.1 − ( 2.8) m /s

= 1.1 s

Insight: If football player 1 had a faster velocity its position-time graph would be steeper, and it would thus intersect the graph for football player 2 sooner than 1.1 s. In fact, if the velocity of football player 1 were to double to −6.2 m/s from −3.1 m/s, the two players would collide at t = 0.71 s.

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2 – 25

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

101. Picture the Problem: The equations of motion are given for two soccer players that move with constant velocity. Strategy: Use the given equations of motion to draw graphs of the positions of the soccer players. The graphs are not essential for answering the questions but can help you picture the situation. Finally, use the equations of motion to determine which player is closer to the ball and find the time at which the two soccer players will pass each other. The two equations are: xf, 1 = xi, 1 + v1 t = ( −8.2 m ) + ( 4.2 m/s ) t

xf, 2 = xi, 2 + v2 t = ( −7.3 m ) + (3.9 m/s ) t

Solution: 1. (a) In the equations of motion the number that is added to the right side is the initial position of the player. The ball is at +5.0 m, so it is clear that soccer player 2 is closer to the ball at t = 0 because −7.3 m is closer to +5.0 m than is −8.2 m. xf, 1 = xf 2

2. (b) Set the final positions of the two soccer players equal to each other:

xi, 1 + v1 t = xi, 2 + v2 t

3. Subtract v2 t and xi, 2 from both sides,

− v2 t + v1t = x2, i − x1, i

rearrange the equation, and solve for t:

(v1 − v2 ) t = x2, i − x1, i t=

4. (c) Insert the time from part (b) into the equation of motion of either player (player 1 equation was used here) to find the position of the two players when they pass each other:

x2, i − x1, i v1 − v2

=

−7.3 − ( −8.2) m 4.2 − (3.9) m /s

= 3.0 s

æ mö xf, 1 = (-8.2 m) + ççç4.2 ÷÷÷(3.0 s ) = 4.4 m s ø÷ è

Insight: You can verify with the equations of motion that soccer player 1 will reach the ball at x = 5.0 m first, at time t1 = 3.14 s, and player 2 will arrive shortly thereafter at t2 = 3.15 s

102. The velocity of an object is its displacement per time interval. If the velocity remains the same, and the time intervals are the same, then the displacements for each time interval will be the same. We conclude that the displacement of the golf cart from t = 0 to t = 5 s is equal to the displacement of the golf cart from t = 5 s to t = 10 s. You can verify that if the velocity of the golf cart is 8 m/s, the displacement in each time interval is 40 m. 103. The average velocity of an object is the total displacement divided by the time elapsed. It contains no information about the motion of the object at a particular instant of time during the time interval. Therefore, yes, it is possible that you were at rest at some point during the 10-min interval, but at other times you must have moved fast enough to make your average velocity equal to 2.2 m/s over the 10-min time interval.

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2 – 26

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

104. Picture the Problem: The equations of motion are given for two dragonflies that move with constant velocity. Strategy: Use the given equations of motion to draw graphs of the positions of the dragonflies. The graphs are not essential for answering the questions but can help you picture the situation. Finally, use the equations of motion to determine which dragonfly is moving faster and find the time at which the two dragonflies will pass each other. The two equations are:

2.2 m −3.1 m

xf, 1 = xi, 1 + v1 t = ( 2.2 m ) + ( 0.75 m/s ) t

xf, 2 = xi, 2 + v2 t = ( −3.1 m ) + ( −1.1 m/s ) t Solution: 1. (a) In the equations of motion the number that multiplies t is the velocity of the dragonfly. By inspection you can see that the velocity of the first dragonfly is 0.75 m/s and the velocity of the second dragonfly is −1.1 m/s. The speed of each dragonfly is the magnitude of its velocity, so we conclude that dragonfly 2 is moving faster. 2. (b) In the equations of motion the number that is added to the right side is the initial position of the dragonfly. By inspection it is clear that dragonfly 1 is closer to x = 0 at t = 0 because 2.2 m is closer to zero than is −3.1 m. The two initial positions are marked on the position-time graph with arrows. Insight: You can verify with the equations of motion that the two dragonflies shared the same position at t = −2.9 s.

105. Picture the Problem: A snail crawls at a slow pace through a garden. Strategy: To make an estimate of the speed we will choose a time interval and then estimate the displacement during that interval. Solution: If you imagine watching a garden snail crawling over a period of a ten seconds, it might travel a distance Δ x 0.02 m = = 2 × 10 − 3 m/s ≅ 10− 3 m/s about equal to its 2-cm body length. Its estimated speed is thus: vav = Δt 10 s Insight: A Wikipedia article reports that 1 mm/s is a typical speed for an adult snail of species helix lucorum. This is the same order of magnitude as our estimate of 2 mm/s.

106. Picture the Problem: Nerve impulses propagate at a fixed speed through a human body. Strategy: The time elapsed is the displacement divided by the average speed. The distance from your finger to your brain is on the order of one meter.

Δt =

Solution: Divide the distance by the average speed:

Δx 1 m = = 0.010 s vav 1 × 102 m /s

Insight: This nerve impulse travel time is not the limiting factor for human reaction time, which is about 0.2 s.

107. The displacement of an object is the difference between its initial and final positions. Therefore we conclude that object 1, which moves from 5.0 m to 7.0 m, has a greater displacement than object 2, which moves from 15 m to 16 m. The time information is irrelevant to the calculation of the displacement. 108. Picture the Problem: An object moves with a constant velocity. Strategy: Use the equation of motion and the given information to find the final position of the object. Solution: Insert numerical values into the equation of motion:

(

)(

)

xf = xi + v t = ( 7.3 m ) + −1.1 m/ s 3.5 s = 3.5 m

Insight: The same equation of motion can be used to predict that the object will pass the x = 0 position at t = 6.6 s

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2 – 27

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

109. Picture the Problem: An object moves with a constant velocity. Strategy: Use the given information to find the velocity by dividing the displacement by the time elapsed. Δ x xf − xi 62 − 73 m = = = − 0.92 m/s t f − ti 12 − 0 s Δt

Solution: 1. (a) Divide the displacement by the time elapsed

vav =

2. (b) The speed is the magnitude of the velocity:

average speed = vav = 0.92 m/s

Insight: The equation of motion can be used to predict that the object will pass the x = 0 position at t = 80 s

110. Picture the Problem: A horse is at x = 4.3 m at t = 0 and moves with a constant velocity of 6.7 m/s. Strategy: Construct an equation to describe the motion of the horse and use it to either generate (x, y) points and plot them by hand, or to generate a graph by means of a computer spreadsheet or plotting program. Solution: 1. (a) Use the given information to create a positiontime graph like that shown at the right. 2. (b) Write a formula to describe the horse’s motion:

xf = xi + v t xf = ( 4.3 m ) + ( 6.7 m/s ) t

Insight: A horse that moves at constant speed has a linear position-time graph.

111. Picture the Problem: A train moves at constant speed along a track. Strategy: Use the given equation of motion to either generate (x, y) points and plot them by hand, or to generate a graph by means of a computer spreadsheet or plotting program. The given equation is: xf = xi + v t = (11 m ) + ( 6.5 m/s ) t . Solve the equation for t to find the time the train is at x = 32 m. Solution: 1. (a) Use the equation of motion to create a positiontime graph like that shown at the right. 2. (b) Solve the equation of motion for t: 3. Insert numerical values:

xf = xi + v t xf − xi =t v t=

32 − 11 m = 3.2 s 6.5 m /s

Insight: You can use the equation of motion to calculate that the train passed x = 0 at t = −1.7 s, or 1.7 s before the clock was started in this experiment to measure the position of the train.

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2 – 28

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

112. Picture the Problem: A pony walks around the circular track. 4.5 m

Strategy: The distance is the total length of travel, and the displacement is the net change in position. The distance around the track is the circumference of the circle, or 2 π r.

A

B

( 2π r ) = π r = π ( 4.5 m ) = 14 m

Solution: (a) 1. The distance traveled is half the circumference:

d=

2. The displacement is the distance from A to B:

Δx = x f − xi = 2r = 2 ( 4.5 m ) = 9.0 m

1 2

3. (b) The distance traveled will increase when the child completes one circuit, because the pony will have taken more steps. 4. (c) The displacement will decrease when the child completes one circuit, because the displacement is maximum when the child has gone halfway around, and is zero when the child returns to the starting position. d = 2π r = 2π ( 4.5 m ) = 28 m

5. (d) The distance traveled equals the circumference:

6. The displacement is zero because the child has returned to her starting position. Insight: The distance traveled is always positive, but the displacement can be negative. The displacement is always zero for a complete circuit, as in this case.

113. Picture the Problem: In heavy rush-hour traffic you travel in a straight line at two different speeds. Strategy: Determine the distance traveled during each leg of the trip in order to plot the graph. Solution: 1. (a) Calculate the distance traveled in the first leg:

Δ x1 = v1 Δ t1 = (12 m/s ) 1.5 min × 60 s/ min = 1080 m

2. Calculate the distance traveled in the second leg:

Δ x2 = v2 Δ t2 = ( 0 m/s )(3.5 min ) = 0 m

3. Calculate the distance traveled in the third leg:

Δ x3 = v3 Δ t3 = (15 m/s ) 2.5 min × 60 s/ min = 2250 m

4. Calculate the total distance traveled:

Δ x = Δ x1 + Δ x2 + Δ x3 = 1080 + 0 + 2250 m = 3330 m

(

(

)

)

5. Draw the graph:

6. (b) Divide the total distance by the time elapsed:

vav =

Δ x1 + Δ x2 + Δ x3 3330 m = 7.4 m/s = Δ t1 + Δ t2 + Δ t3 7.5 min × 60 s/ min

Insight: The average speed is a weighted average according to how much time you spend traveling at each speed. Here you spend the most amount of time at rest, so the average speed is less than either 12 m/s or 15 m/s.

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2 – 29

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

114. Picture the Problem: The initial positions and velocities are given for two objects that move directly toward one another. Strategy: Use the given data to write equations of motion and draw position-time graphs for the two objects. The graphs are not essential for answering the questions but can help you picture the situation. Finally, use the equations of motion to find the time at which the two objects will collide and their position at the time of collision.. The two equations are:

xf, 1 = xi, 1 + v1 t = (5.4 m ) + (1.3 m/s ) t

xf, 2 = xi, 2 + v2 t = (8.1 m ) + ( −2.2 m/s ) t xf, 1 = xf 2

Solution: 1. (a) Set the final positions of the two objects equal to each other:

xi, 1 + v1 t = xi, 2 + v2 t

2. Subtract v2 t and xi, 2 from both sides,

− v2 t + v1t = x2, i − x1, i

rearrange the equation, and solve for t:

(v1 − v2 ) t = x2, i − x1, i t=

3. (b) Use either equation of motion to find the position of the objects. The equation for the second object is used here:

x2, i − x1, i v1 − v2

=

8.1 − (5.4) m

1.3 − ( −2.2) m /s

= 0.77 s

xf, 2 = (8.1 m ) + ( −2.2 m/s )( 0.77 s ) = 6.4 m

Insight: If the objects bounced off each other, the graph after the collision would look different. The object 2 graph would suddenly acquire a positive slope toward larger positions, and the object 1 graph would suddenly acquire a negative slope and its position would decrease after the collision.

115. Picture the Problem: Two objects move with constant speeds and then collide with each other. Strategy: Set the positions of the two objects equal to each other at the instant of collision, and use the given information to determine the velocity of object 2. Then use an equation of motion to find the position of the two objects at the moment of collision. While it is not necessary for solving this problem, position-time graphs for the two objects are shown at the right to help you picture the problem.

Solution: 1. (a) Set the final positions of the two objects equal to each other:

xf, 1 = xf 2 xi, 1 + v1 t = xi, 2 + v2 t

2. Subtract xi, 2 from both sides and

x1, i − x2, i + v1t = v2 t

divide by t to solve for v2:

x1, i − x2, i + v1t t

= v2 =

( 25 m) − (13 m) + ( −5.6 m/ s ) (0.61 s ) 0.61 s

= 14 m/s

3. (b) Use either equation of motion to find the position of the objects. The equation for the first object is used here:

(

)(

)

xf, 1 = ( 25 m ) + −5.6 m/ s 0.61 s = 22 m

Insight: You could also solve part (b) first and then use the known position of object 2 at 0.61 s to find its velocity. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

116. Picture the Problem: You travel in the forward direction along the roads leading to the wedding ceremony, but your average speed is different during the first and second portions of the trip. Strategy: First find the distance traveled during the first 15 minutes in order to calculate the distance yet to travel. Then determine the speed you need during the second 15 minutes of travel. Solution: 1. Use the definition of average speed to determine the distance traveled:

æ km ö÷æç 1 h ö÷ ÷÷ = 3.00 km ÷÷çç15.0 min ´ Δ x1 = vav, 1 Δt1 = çç12.0 çè h ÷øçè 60 min ÷ø

2. Find the remaining distance to travel:

Δ x2 = Δ xtotal − Δ x1 = 17.0 − 3.00 km = 14.0 km

3. Find the required speed for the second part of the trip:

vav, 2 =

Δ x2 14.0 km = = 56.0 km/h Δ t2 0.250 h

Insight: The car needs an average speed of 17.0 km/0.500 h = 34.0 km/h for the entire trip. However, in order to make it on time it must go 4.7 times faster in the second half (time-wise) of the trip than it did in the first half of the trip.

117. Answers will vary. The San Andreas fault slips at an average speed of 35 mm per year. The cities of Los Angeles and San Francisco are separated by 560 km. The time for the fault to slip by 560 km is: Δx 560 km 1000 mm 1000 m t= = × × = 1.6 × 107 yr or about 16 million years. vav 35 mm /yr m km 118. Equations will vary depending on the situation. Students should measure or estimate the speeds of moving objects around the school and set up a coordinate system to determine the initial postions of the objects. Then, they should use these measurments to write an equation of motion for each object. For instance, if at t = 0 a janitor’s cart is outside the physics room door, 8.0 m from the end of the hallway, and moves at a speed of 1.5 m/s as it is pushed down the hallway in the positive direction, its equation of motion would be xf = (8.0 m ) + (1.5 m/s ) t . 119. Picture the Problem: A robot moves forward, then stops, then moves backward. Its position-time graph is shown at the right. Strategy: Find the slope of the line between t = 0 and t = 2 s, the time interval during which the robot is walking forward. The slope of the line is the velocity of the robot during that time interval. rise Δ x = Solution: Find the slope. The vav = run Δ t calculated answer 2−0 m corresponds to choice B. = = 1.0 m/s 2−0 s

Insight: The average human walking pace is about 20 minutes per mile, which is equivalent to 1609 m per 1200 s, or 1.3 m/s.

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2 – 31

Chapter 2: Introduction to Motion

Pearson Physics by James S. Walker

120. Picture the Problem: A robot moves forward, then stops, then moves backward. Its position-time graph is shown at the right. Strategy: Find the slope of the line between t = 5 s and t = 8 s, the time interval during which the robot is walking backward. The slope of the line is the velocity of the robot during that time interval. rise Δ x = Solution: Find the slope. The vav = run Δ t calculated result is answer 1− 2 m choice A. = = − 0.33 m/s 8−5 s

Insight: The average human walking pace is about 20 minutes per mile, which is equivalent to 1609 m per 1200 s, or 1.3 m/s.

121. Picture the Problem: A robot moves forward, then stops, then moves backward. Its position-time graph is shown at the right. Strategy: Find the change in position of the robot between the times t = 2 s and t = 8 s. The displacement is the change in position over that time interval. Solution: Find the change in position. The calculated result is answer choice B.

Δ x = 1 m − 2 m = −1.0 m

Insight: The distance traveled by the robot from t = 0 to t = 8 s is 3.0 m, while its displacement over that time interval is +1.0 m.

122. Picture the Problem: A robot moves forward, then stops, then moves backward. Its position-time graph is shown at the right. Strategy: In the previous question we discovered the displacement of the robot over the time interval t = 2 s and t = 8 s is −1.0 m. Use this displacement together with the elapsed time to find the average velocity of the robot. Solution: Find the average velocity. The calculated result is answer choice A.

vav =

Δ x −1.0 m = = − 0.17 m/s Δt 8−2 s

Insight: The average velocity of the robot over the entire time interval t = 0 to t = 8 s is +0.125 m/s.

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