Answer
(a) $L = 79.2~m$
(b) $\Delta t = 610~ns$
(c) $\Delta t_0 = 528~ns$
Work Step by Step
(a) Let $L_0 = 91.5~m$. We can find the length measured according to the rest frame of the particle:
$L = L_0~\sqrt{1-\frac{v^2}{c^2}}$
$L = (91.5~m)~\sqrt{1-\frac{(0.50~c)^2}{c^2}}$
$L = 79.2~m$
(b) We can find the time $\Delta t$ according to Earth observers:
$\Delta t = \frac{91.5~m}{(0.50)(3.0\times 10^8~m/s)}$
$\Delta t = 610~ns$
(c) We can find the time $\Delta t_0$ in the rest frame of the particle:
$\Delta t_0 = \frac{79.2~m}{(0.50)(3.0\times 10^8~m/s)}$
$\Delta t_0 = 528~ns$
