Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 2 - Kinematics in One Dimension - Exercises and Problems - Page 66: 58


The train's acceleration is $-1.0~m/s^2$

Work Step by Step

In the 2.0 seconds that the train moved at a constant speed, the train traveled 4.0 meters. Since the train comes to a stop a distance of 6.0 meters from the point of release, the train decelerates over a distance of 2.0 meters. We can find the rate of deceleration as: $a = \frac{v^2-v_0^2}{2d}$ $a = \frac{0-(2.0~m/s)^2}{(2)(2.0~m)}$ $a = -1.0~m/s^2$ The train's acceleration is $-1.0~m/s^2$.
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