4. A train $110 \mathrm{~m}$ long is moving at a speed of $\frac{25}{3} \mathrm{~m} / \mathrm{s}$. On its way to the track at $09.10 \mathrm{~h}$, it encountered a pedestrian walking in the same direction and passed him in $15 \mathrm{sec}$. At $09.16 \mathrm{~h}$, it met a pedestrian walking towards it and passed him in $12 \mathrm{sec}$. At what time did the pedestrians meet?
Solution. Let the first pedestrian have walked $x$ meters in 15 seconds. Then

$$
\begin{gathered}
110+x=\frac{25}{3} \cdot 15 \\
x=15
\end{gathered}
$$

This means the first pedestrian walked 15 meters in 15 seconds, so his speed is $1 \mathrm{~m} / \mathrm{s}$.

Let the second pedestrian have walked $y$ meters in 12 seconds. Then

$$
\begin{gathered}
110-y=\frac{25}{3} \cdot 12 \\
y=10
\end{gathered}
$$

This means the second pedestrian walked $10 \mathrm{~m}$ in 12 seconds, so his speed is $\frac{5}{6} \mathrm{~m} / \mathrm{s}$.

The distance between the meeting point of the first pedestrian and the second pedestrian is $360 \cdot \frac{25}{3}=3000 \mathrm{~m}$. Let the pedestrians meet after $t$ seconds from the first pedestrian's meeting. Then

$$
\begin{gathered}
t \cdot 1+(t-360) \cdot \frac{5}{6}=3000 \\
t+\frac{5}{6} t-300=3000 \\
t=1800
\end{gathered}
$$

This means the pedestrians met after 30 minutes, or at 9:40 AM.

## Second Year