Chapter 2: One-Dimensional KinematicsAnswers to Even-Numbered Conceptual Questions2. An odometer measures the distance traveled by a car. You can tell this by the fact that an odometer has anonzero reading after a round trip.4. No. After one complete orbit the astronaut’s displacement is zero. The distance traveled, however, is roughly25,000 miles.6. A speedometer measures speed, not velocity. For example, if you drive with constant speed in a circularpath, your speedometer maintains the same reading, even though your velocity is constantly changing.8. Yes. For example, your friends might have backed out of a parking place at some point in the trip, giving anegative velocity for a short time.10. No. If you throw a ball upward, for example, you might choose the release point to be y = 0. This doesn’tchange the fact that the initial upward speed is nonzero.12. (a) Yes. The object might simply be at rest. (b) Yes. An example would be a ball thrown straight upward; atthe top of its trajectory its velocity is zero, but it has a nonzero acceleration downward.14. Yes. A ball thrown straight upward and caught when it returns to its release point has zero average velocity,but it has been accelerating the entire time.16. When she returns to her original position, her speed is the same as it was initially; that is, 4.5 m/s.18. (a) No. Displacement is the change in position, and therefore it is independent of the location chosen for theorigin. (b) Yes. In order to know whether an object’s displacement is positive or negative, we need to knowwhich direction has been chosen to be positive.Solutions to Problems and Conceptual Exercises1. 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: 0.75 +0.60 mi( )+ 0.60 mi( )=1.95 mi(b) Subtract xi from xf to find the displacement. Δx =xf−xi=0.75−0.00 mi = 0.75 miInsight: The distance traveled is always positive, but the displacement can be negative.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2 – 1Chapter 2: One-Dimensional Kinematics James S. Walker, Physics, 4th Edition2. Picture the Problem: Player A walks in the positive direction and player B walks in the negative direction.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: 5 mThe displacement of player A is positive: Δx =xf−xi=5 m−0 m= 5 m(b) Note the distance traveled by player B: 2 mThe displacement of player B is negative. Let the origin be at the initial position of player A. Δx =xf−xi=5 m−7 m= −2 mInsight: The distance traveled is always positive, but the displacement can be negative.3. 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: 10 +2.5 m( )+2.5 m=15 m(b) Subtract xi from xf to find the displacement. Δx =xf−xi=10−0 m=10 mInsight: The distance traveled is always positive, but the displacement can be negative.4. 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: 0.60 +0.35 mi( )+ 0.75+0.60+0.35 mi( )= 2.65 mi(b) Subtract xi from xf to find the displacement. Δx =xf−xi=0.75−0.00 mi = 0.75 miInsight: The distance traveled is always positive, but the displacement can be negative.5. 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.Solution: 1. (a) Add the lengths: 15 m( )+ 100 m( )+ 15 m( )=130 m2. Subtract xi from xf to find the displacement. Δx =xf−xi=100−0 m=100 m3. (b) Add the lengths: 15 +100+30+100+15 m= 260 m4. Subtract xi from xf to find the displacement. Δx =xf−xi=0−0 m= 0 mCopyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2 – 2Chapter 2: One-Dimensional Kinematics James S. Walker, Physics, 4th EditionInsight: 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.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2 – 3Chapter 2: One-Dimensional Kinematics James S. Walker, Physics, 4th Edition6. Picture the Problem: The pony walks around the circular track.Strategy: The distance is the total length of travel, and the displacement is the net change in position.Solution: (a) 1. The distance traveled is half the circumference: d =122πr( )=πr =π 4.5 m( )=14 m2. The displacement is the distance from A to B: Δx =xf−xi=2r =2 4.5 m( )= 9.0 m3. (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 whenthe child has gone halfway around, and is zero when the child returns to the starting position.5. (d) The distance traveled equals the circumference: d =2πr =2π 4.5 m( )= 28 m6. 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.7. 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