Task B-1.3. Two trains start simultaneously from their initial stations towards each other. The trains travel at constant speeds and pass each other at point $M$. One train travels twice as fast as the other. If the slower train starts 5 minutes later, it will pass the other train at a point which is $4 \mathrm{~km}$ further from point $M$. What is the speed of the slower train?
## Solution.

Let the slower train be $A$, and the faster one $B$. Let $v$ be the speed of train $A$.

Then $2v$ is the speed of train $B$. The time required for both trains to reach point $M$ is $t$ hours. The distance train $A$ travels to point $M$ is $v t \text{ km}$, and the distance train $B$ travels to point $M$ is $2 v t$.

If the slower train $A$ departs 5 minutes later ($1/12 \text{ h}$), and if $t_{1}$ is the time required for train $A$ to reach the meeting point, train $B$ has been traveling for $t_{1} + \frac{1}{12}$ hours to that point.

The distance train $A$ travels to the meeting point is $v \cdot t_{1}$, and the distance train $B$ travels is $2 v \cdot \left(t_{1} + \frac{1}{12}\right)$. Since the difference in distances traveled by the slower train in both cases is $4 \text{ km}$, we write

$$
v t - v t_{1} = 4
$$

Similarly, the difference in distances traveled by the faster train is

$$
2 v \cdot \left(t_{1} + \frac{1}{12}\right) - 2 v \cdot t = 4
$$

From the first equation, we get

$$
t - t_{1} = \frac{4}{v}
$$

and from the second equation,

$$
-2 v \cdot \left(t - t_{1}\right) + \frac{v}{6} = 4
$$

Then the speed of the slower train is $v = 72 \text{ km/h}$.