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 111: 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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.