## Task 3 - 230813

On a $22.5 \mathrm{~km}$ long tram line, trains are to run in both directions from 8:00 AM to 4:00 PM at ten-minute intervals, starting with the departure time exactly at 8:00 AM at both terminal stops. The average speed of the trains is $18 \frac{\mathrm{km}}{\mathrm{h}}$. Each train, upon arrival at a terminal stop, should take a break before its departure, which is more than 10 minutes but less than 20 minutes.

a) When will the train that departed at 8:00 AM from one terminal stop leave this terminal stop for the second time?

b) How many trains are sufficient to maintain the described traffic flow?

c) How much time elapses for a train traveling from one terminal stop to the other, on average, from one encounter with an oncoming train to the next encounter with an oncoming train?
a) Since the distance between the two terminal stops is $s=22.5 \mathrm{~km}$ and is traveled at a speed of $v=18 \frac{\mathrm{km}}{\mathrm{h}}$, the train requires the time

$$
t=\frac{s}{v}=\frac{22.5}{18} h=1.25 h=1 h 14 \mathrm{~min}
$$

The train that departed from one terminal stop at 8:00 AM will therefore reach the other terminal stop at 9:15 AM.

From there, it must depart at a time that is a whole multiple of 10 minutes due to the break regulation, so at 9:30 AM. Accordingly, it must depart from the first terminal stop one hour later, at 11:00 AM.
b) From the second terminal stop, trains must be available for departure times of 8:00, 8:10, 8:20, 8:30, 8:40, 8:50, 9:00, 9:10, and 9:20, which amounts to 9 trains.

An equal number will be required for departure from the first terminal stop at the same times. From 9:30 AM onwards, the continuation of the planned schedule with the already listed trains is possible. Therefore, a total of 18 trains are sufficient.

c) Let the first-mentioned train be $Z_{0}$; the first and second trains it encounters be $Z_{1}$ and $Z_{2}$, respectively.

During the journey, $Z_{1}$ and $Z_{2}$ maintain a constant distance from each other; this is also the distance between $Z_{0}$ and $Z_{2}$ at the time $Z_{0}$ and $Z_{1}$ meet. Since train $Z_{2}$ arrives at a location 10 minutes later than $Z_{1}$, it takes exactly 10 minutes to cover this distance.

Due to the movement of trains $Z_{0}$ and $Z_{2}$ towards each other, their distance decreases at twice the speed of the travel speed of each individual train. Therefore, this distance is reduced to 0 in half the time it would take a single train to cover this distance. This means:

Until the time of the encounter between $Z_{0}$ and $Z_{2}$, half of 10 minutes, which is 5 minutes, elapse.