College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 2 - Kinematics: Description of Motion - Learning Path Questions and Exercises - Exercises - Page 63: 50



Work Step by Step

We use the following equation of kinetic motion: $v^{2}=u^{2}+2aS$ $350^{2}=210^{2}+0.08a$ $a=\frac{560\times140}{0.08}=9.8\times10^{5} m/s^{2}$ Now, $v=u+at$ $t=\frac{v-u}{a}=\frac{140}{9.8\times10^{5}}s=1.43\times10^{-4}s$
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