Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 2 - Describing Motion: Kinematics in One Dimension - Problems - Page 43: 19


6 $m/s^2$ = 0.61 g

Work Step by Step

First, we find the initial velocity of the sports car : $$v_{0} = \frac{\Delta x}{\Delta t}$$ $$v_{0} = \frac{120 m}{5.0s} = 24 m/s$$ Then, we use equation below to find the acceleration : $$v = v_{0} + at$$ $$v-v_{0}=at$$ $$a = \frac{v-v_{0}}{t}=\frac{0-24m/s}{4.0s}=-6.0m/s^2$$ So, the magnitude of the acceleration is $6.0 m/s^2$ To convert the acceleration into g's, $(6.0m/s^2)(\frac{1g}{9.8m/s^2})=0.61g$
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