College Physics (4th Edition)

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

Chapter 26 - Problems - Page 1010: 7


$v = 0.0014~c$ $v = 4.2\times 10^5~m/s$

Work Step by Step

We can find $v$ when $\gamma = 1.000001$ $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = 1.000001$ $1-\frac{v^2}{c^2} = \frac{1}{(1.000001)^2}$ $\frac{v^2}{c^2} = 1-\frac{1}{(1.000001)^2}$ $v^2 = [1-\frac{1}{(1.000001)^2}~]~c^2$ $v = \sqrt{1-\frac{1}{(1.000001)^2}}~\times c$ $v = 0.0014~c$ $v = (0.0014)~(3.0\times 10^8~m/s)$ $v = 4.2\times 10^5~m/s$
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