If there are $N_0$ muons at time $t = 0$ , then at some later time $t$ , there would be
$\begin{align*}N(t) = N_0 e^{-t/\tau}\end{align*}$
So, at time $t$ , the muons decay at a rate
$\begin{align*}-\frac{d N(t)}{dt} = \frac{N_0}{\tau}e^{-t/\tau}\end{align*}$
So, in a time interval $(t,t+dt)$ ,
$\begin{align*}\frac{N_0}{\tau}e^{-t/\tau} dt\end{align*}$
muons decay. So, the probability that a muon decays in this time interval is
$\begin{align*}\frac{e^{-t/\tau}}{\tau} dt\end{align*}$
A muon that travels a distance $l$ in the laboratory will decay at time $t = l/\gamma v$ in its rest frame.
$\begin{align*}\gamma = \frac{1}{\sqrt{\displaystyle 1 - \frac{\left(\frac{\displaystyle 4c}{\displaystyle 5}\right)^2}{\displ...$
So, $t = l/\gamma v = 5l/4 \gamma c = 3 l /4c$ . So, the probability that a muon decays after traveling a distance $l$ is
$\begin{align*}\frac{e^{-3l/4c\tau}}{\tau}dt = \frac{e^{-3l/4c\tau}}{\tau}\frac{3dl}{4c}\end{align*}$
The mean distance traveled by the muons is therefore,
$\begin{align*}\int_0^{\infty}l\frac{ e^{ -3l/4c\tau}}{\tau}\frac{3dl}{4c}\\&= \left[-l e^{ -3l/4c\tau}\right]_0^{\infty} ...$
Plugging in $\tau = 2.2 \cdot 10^{-6} \mbox{ s}$ and $c = 3\cdot 10^8 \mbox{ m/s}$ gives
$\begin{align*}\frac{4 c \tau}{3} = \frac{4 \cdot 3 \cdot 10^8 \mbox{ m/s} \cdot (11/5) \cdot 10^{-6} \mbox{ s}}{3} = \boxed{8...$
Therefore, answer (C) is correct.

Solution to 2008 Problem 39