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 4 - Kinematics in Two Dimensions - Exercises and Problems - Page 114: 60


The radius of the turn is 42.5 meters.

Work Step by Step

Let the centripetal acceleration $a_c$ be $1.5~g$. Then, we find $r$ as: $a_c = \frac{v^2}{r}$ $1.5~g = \frac{v^2}{r}$ $r = \frac{v^2}{1.5~g}$ $r = \frac{(25~m/s)^2}{(1.5)(9.80~m/s^2)}$ $r = 42.5~m$ The radius of the turn is 42.5 meters.
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