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: 18

Answer

Trent's speed is 75 mph relative to Susan's reference frame.

Work Step by Step

Let $v_{Tg}$ be Trent's velocity with respect to the ground. Let $v_{Sg}$ be Susan's velocity with respect to the ground. Let $v_{TS}$ be Trent's velocity with respect to Susan. $v_{Tg} = v_{TS}+v_{Sg}$ $v_{TS} = v_{Tg} - v_{Sg}$ We can find the east component of $v_{TS}$: $v_{TS,east} = 45~mph - 0 = 45~mph$ (east) We can find the south component of $v_{TS}$: $v_{TS,south} = 0 - (-60~mph) = 60~mph$ (south) We can find Trent's speed relative to Susan's reference frame: $v_{TS} = \sqrt{(45~mph)^2+(60~mph)^2}$ $v_{TS} = 75~mph$ Trent's speed is 75 mph relative to Susan's reference frame.
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