Problem 2. A traveler was walking from a village to a city. At 14:00, when the traveler had covered a quarter of the distance, a motorcyclist set off from the village to the city, and a truck set off from the city to the village. At 15:00, the motorcyclist caught up with the traveler, and at 15:30, the motorcyclist met the truck. When will the traveler meet the truck?
Solution. Answer: 15:48.

Let the entire distance be denoted as $S$, and the speeds of the pedestrian, motorcycle, and truck as $V_{p}, V_{m}$, and $V_{g}$, respectively (distance is measured in kilometers, and speed in kilometers per hour). According to the problem, the motorcyclist caught up with the pedestrian in one hour. Their relative speed is $V_{m}-V_{p}$, and the distance between them is $S / 4$, so we have the equation

$$
\frac{S / 4}{V_{m}-V_{p}}=1
$$

One and a half hours after the start of the movement, the motorcycle and the truck met. Their relative speed is $V_{m}+V_{g}$, and the total distance they covered is $S$, so we have the equation

$$
\frac{S}{V_{m}+V_{g}}=1.5
$$

Transforming both equations, we get

$$
\left\{\begin{array}{c}
V_{m}-V_{p}=\frac{S}{4} \\
V_{m}+V_{g}=\frac{2 S}{3}
\end{array}\right.
$$

Subtract the first equation from the second and we get

$$
V_{g}+V_{p}=\frac{5 S}{12}
$$

from which we find

$$
\frac{3 S / 4}{V_{g}+V_{p}}=\frac{9}{5}
$$

Therefore, the pedestrian and the truck met after $9 / 5$ hours from the start of the movement. Converting this time into hours and minutes, we get that the pedestrian and the truck met at $15:48$.

## Criteria

The following is the most appropriate single criterion:

7 p. Any complete solution to the problem.

2 6. The system of motion equations is correctly written.