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: 24

Answer

The climber's speed is faster than the surfer's speed by 20 cm/s

Work Step by Step

We can find the angular speed of the earth as it rotates once each day: $\omega = \frac{2\pi~rad}{(24)(3600~s)}$ $\omega = 7.27\times 10^{-5}~rad/s$ For the surfer, the radius of rotation is $6.4\times 10^6~m$. Therefore: $v = \omega ~r$ $v = (7.27\times 10^{-5}~rad/s)(6.4\times 10^6~m)$ $v = 465.3~m/s$ For the climber, the radius of rotation is $6.403\times 10^6~m$. Therefore: $v = \omega ~r$ $v = (7.27\times 10^{-5}~rad/s)(6.403\times 10^6~m)$ $v = 465.5~m/s$ The climber's speed is faster than the surfer's speed by 20 cm/s.
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