College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 26 - Problems - Page 1010: 16


$v = 0.126~c$ $v = 3.78\times 10^7~m/s$

Work Step by Step

Let $L_0 = 1.0~m$ and let $L = 0.992~m$. We can find the speed $v$ with respect to the lab: $L = L_0~\sqrt{1-\frac{v^2}{c^2}}$ $\sqrt{1-\frac{v^2}{c^2}} = \frac{L}{L_0}$ $1-\frac{v^2}{c^2} = (\frac{L}{L_0})^2$ $\frac{v^2}{c^2} = 1-(\frac{L}{L_0})^2$ $v^2 = [1-(\frac{L}{L_0})^2]~(c^2)$ $v = \sqrt{1-(\frac{L}{L_0})^2}~\times c$ $v = \sqrt{1-(\frac{0.992~m}{1.0~m})^2}~\times c$ $v = 0.126~c$ $v = (0.126)~(3.0\times 10^8~m/s)$ $v = 3.78\times 10^7~m/s$
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