## College Physics (4th Edition)

(a) $\Delta t = 31.25~ns$ (b) $d = 5.6~m$
(a) Let $\Delta t_0$ be the lifetime in the pion's frame. Let $\Delta t$ be the lifetime measured in the laboratory frame. We can find the lifetime measured in the laboratory frame: $\Delta t = \frac{\Delta t_0}{\sqrt{1-\frac{v^2}{c^2}}}$ $\Delta t = \frac{25~ns}{\sqrt{1-\frac{(0.60~c)^2}{c^2}}}$ $\Delta t = \frac{25~ns}{\sqrt{0.64}}$ $\Delta t = 31.25~ns$ (b) We can find the distance traveled: $d = v~t$ $d = (0.60)(3.0\times 10^8~m/s)(31.25\times 10^{-9}~s)$ $d = 5.6~m$