Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 36 - Relativity - Exercises and Problems - Page 1059: 17


$v = 0.28~c$

Work Step by Step

Let $t'$ be the elapsed time on the watch and let $t_0 = \frac{24}{25}~t'$. We can find the required speed in the moving reference frame: $t' = \gamma~t_0$ $t' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}~t_0$ $\sqrt{1-\frac{v^2}{c^2}} = \frac{t_0}{t'}$ $1-\frac{v^2}{c^2} = (\frac{t_0}{t'})^2$ $\frac{v^2}{c^2} = 1-(\frac{t_0}{t'})^2$ $v^2 = [1-(\frac{t_0}{t'})^2] \times c^2$ $v = \sqrt{1-(\frac{t_0}{t'})^2}~\times c$ $v = \sqrt{1-(\frac{\frac{24}{25}~t'}{t'})^2}~\times c$ $v = \sqrt{1-(\frac{24}{25})^2}~\times c$ $v = 0.28~c$
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.