## College Physics (4th Edition)

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