## University Physics with Modern Physics (14th Edition)

The value of $\gamma$ is $\frac{1}{\sqrt{1-v^{2}/c^{2}}}$. The relativistic kinetic energy is $(\gamma-1)mc^2$. Here, we are told that $(\gamma-1)mc^2=mc^2$, so we know that $\gamma= 2$. Solve for v when $\gamma=\frac{1}{\sqrt{1-v^{2}/c^{2}}}=2.00$ $$v=c\sqrt{1-1/\gamma^2}=0.866c$$ $$v=0.866(300m/s) = 260 m/s$$