The Two Trains

Let the speed of the two trains be $s$ and $2s$ and the distance between starting points of the two trains be $d$ . Therefore, $M$ is $\frac {1}{3} d$ from the starting point of the slow train.

Assuming the faster train starts $5$ minutes later. And if the time the trains meet is at $t$ , then, we have:

Distance traveled by slow train:

$\quad st=\dfrac {d}{3}+2$

Distance traveled by fast train:

$\quad 2s\left( t - \dfrac {5}{60} \right) =d - \left(\dfrac {d}{3}+2 \right)$

$\Rightarrow 2st - \dfrac {s}{6} = \dfrac{2d}{3}-2$

$\Rightarrow 2 \left( \dfrac {d}{3}+2 \right) - \dfrac {s}{6} = \dfrac{2d}{3}-2$

$\Rightarrow \dfrac {s}{6} = 6 \quad \Rightarrow s = \boxed{36} \space$ miles per hour

We get the same result is we assume the slow train starts later.