Answer
See explanation.
Work Step by Step
a. The lifetime, measured in the muon frame, is the proper time $\Delta t_o$.
The value of $\gamma$ is $\frac{1}{\sqrt{1-v^{2}/c^{2}}}=2.29$.
The lab personnel measure a dilated time interval of $\gamma \Delta t_o =5.05\times10^{-6}s$.
b. The lab personnel measure a distance of the muon’s speed multiplied by its lifetime in that frame.
$d=v\Delta t = (0.900)(3.00\times10^8m/s)( 5.05\times10^{-6}s)=1.36 km$
