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: 43

Answer

(a) The car is 100 meters from the intersection when the brakes are applied. (b) $a = -2.0~m/s^2$ (c) The stopping time is 10.5 seconds.

Work Step by Step

(a) It takes 0.50 seconds to apply the brakes. We can find the distance the car moves in that time. $d = v~t = (20~m/s)(0.50~s)$ $d = 10~m$ The car is 100 meters from the intersection when the brakes are applied. (b) $a = \frac{v^2-v_0^2}{2x}$ $a = \frac{0-(20~m/s)^2}{(2)(100~m)}$ $a = -2.0~m/s^2$ (c) We can find the time to stop after the brakes are applied. $t = \frac{v-v_0}{a} = \frac{0-20~m/s}{-2.0~m/s^2}$ $t = 10~s$ The total time to stop is 10 seconds plus 0.50 seconds for the reaction time. The stopping time is 10.5 seconds.
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