\section*{Task 1 - 041031}

A pedestrian starts walking from A to B, which is \(12.75 \mathrm{~km}\) away, at 9:00 AM with a constant speed.

On the same road, a tram departs from A to B at 9:26 AM. It overtakes the pedestrian at 9:36 AM and, after a 4-minute stop in B, returns. On the way back, it meets the pedestrian at 10:30 AM.

a) How many kilometers do the pedestrian and the tram travel on average per hour?

b) At what distance from A does the tram overtake the pedestrian, and where does it meet him on the return trip?
}

Let \(v_{1}\) be the speed of the pedestrian and \(v_{2}\) the speed of the tram. Let \(s_{1}\) be the distance from A to the first meeting point and \(s_{2}\) the distance from B to the second meeting point. Then we have:

\[
\text { (1): } \quad v_{1}=\frac{s_{1}}{36}=\frac{12.75-s_{2}}{90} \quad ; \quad(2): \quad v_{2}=\frac{s_{1}}{10}=\frac{12.75+s_{2}}{60}
\]

The solutions to this system of linear equations are \(s_{1}=3\) and \(s_{2}=5.25\). Therefore:

\[
v_{1}=\frac{3 \text{ km}}{\frac{3}{5} \text{ h}}=5 \frac{\text{km}}{\text{h}} \quad ; \quad v_{2}=\frac{12.75 \text{ km}+5.25 \text{ km}}{1 \text{ h}}=18 \frac{\text{km}}{\text{h}}
\]

The pedestrian covers an average of \(5 \text{ km}\) per hour, and the tram covers an average of \(18 \text{ km}\) in the same time.

b) The pedestrian is overtaken by the tram \(3 \text{ km}\) away from A and meets it again \(5.25 \text{ km}\) away from B.