Answer
(a) $\Delta t = 31.25~ns$
(b) $d = 5.6~m$
Work Step by Step
(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:
$d = v~t$
$d = (0.60)(3.0\times 10^8~m/s)(31.25\times 10^{-9}~s)$
$d = 5.6~m$
