The quantity
$\begin{align*}(\Delta s)^2 := (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 - c^2(\Delta t)^2\end{align*}$
is a Lorentz invariant, meaning that it is the same in any reference frame. We choose as our two events the two flashes of light. We use unprimed variables for the frame of observer $O$ and primed variables for the frame of observer $O'$ . $\Delta y = \Delta z = \Delta y' = \Delta z' = 0$ , so $\left(\Delta s\right)^2 = \left(\Delta s'\right)^2$ implies that
$\begin{align*}(\Delta x)^2 - c^2(\Delta t)^2 = (\Delta x')^2 - c^2(\Delta t')^2\end{align*}$
Now,
$\begin{align*}\left|\Delta x\right| &= 10 \mbox{ m} \\\Delta t &= 0 \\\left|\Delta t'\right| &= 13 \mbox{ ns} \\\...$
so,
$\begin{align*}\left(10 \mbox{ m} \right)^2 = (\Delta x')^2 - c^2 \left(13 \mbox{ ns} \right)^2\end{align*}$
Solving for $\Delta x'$ gives
$\begin{align*}\left|\Delta x'\right| = \sqrt{115.21} \mbox{ m}\end{align*}$
From the Lorentz transformation equations,
$\begin{align*}x' &= \gamma (x - vt) \\y' &= y \\z' &= z \\t' &= \gamma \left(t - \frac{v}{c^2}x\right)\end{al...$
we also have that
$\begin{align*}\Delta x' &= \gamma \Delta x \\\Delta t' & = - \gamma \frac{v \Delta x}{c^2 }\end{align*}$
Diving the second equation by the first equation gives
$\begin{align*}\frac{\Delta t'}{\Delta x'} = - \frac{v}{c^2}\end{align*}$
or
$\begin{align*}\left|v\right| = \left|\frac{c\Delta t'}{\Delta x'} \right| c\end{align*}$
Plugging in $\left|\Delta t'\right| = 13 \mbox{ ns}$ and $\left|\Delta x'\right| = \sqrt{115.21} \mbox{ m}$ gives
$\begin{align*}\left|v\right| &= \frac{13 \cdot 10^{-9} \cdot 3 \cdot 10^8}{\sqrt{115.21}} c \\&= \frac{3.9 c}{\sqrt{1...$
Therefore, answer (C) is correct.

Solution to 2008 Problem 94