13.446 From the same point, two pedestrians set out simultaneously in the same direction along a straight section of the highway at constant but different speeds. After 2 hours, the distance between them was $s$ km. After that, the pedestrians started walking faster and spent 10 minutes less on each kilometer of the way. Another 2 hours later, the distance between them became $3 s$ km. Find the distances traveled by the pedestrians in the first two hours of movement.
Solution. Let $x$ be the distance (km) traveled by the first pedestrian in the first 2 hours. Then $\frac{120}{x}$ is the number of minutes spent by him to travel 1 km.

Similarly, $\frac{120}{x+s}$ is the number of minutes spent by the second pedestrian to travel 1 km. After increasing their speed, the first and second pedestrian began to spend $\frac{120}{x}-10$ and $\frac{120}{x+s}-10$ minutes, respectively, to travel 1 km. Therefore, in the subsequent two hours, each of them traveled $\frac{120}{\frac{120}{x}-10}$ and $\frac{120}{\frac{120}{x+s}-10}$ km, respectively.

$$
\begin{aligned}
& \text { Hence } \frac{120}{\frac{120}{x+s}-10}-\frac{120}{\frac{120}{x}-10}=2 s \Rightarrow x^{2}+(s-24) x+72-12 s=0 \Rightarrow \\
\Rightarrow x= & \frac{24-s \pm \sqrt{s^{2}+288}}{2}
\end{aligned}
$$

If $x=\frac{24-s+\sqrt{s^{2}+288}}{2}$, then $x>12$ and $\frac{120}{x}<10$, which is impossible.

Answer: $\frac{24-s-\sqrt{s^{2}+288}}{2}$ km, $\frac{s+24-\sqrt{s^{2}+288}}{2}$ km.