University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 2 - Motion Along a Straight Line - Problems - Exercises - Page 60: 2.26

Answer

(a) The magnitude of deceleration is $58~m/s^2$ which is 5.9$g$. (b) t = 12 ms

Work Step by Step

(a) We can calculate the rate of deceleration for the hip. $a = \frac{v^2-v_0^2}{2x} = \frac{(1.3~m/s)^2-(2.0~m/s)^2}{(2)(0.020~m)} = -58~m/s^2$ The magnitude of deceleration is $58~m/s^2$ $a = \frac{58~m/s^2}{9.80~m/s^2} = 5.9 g$ (b) $t = \frac{v-v_0}{a} = \frac{1.3~m/s-2.0~m/s}{-58~m/s^2} = 0.012~s = 12~ms$
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