## Task 4 - V10734

A group of Young Pioneers hikes from the town A located in the valley to the mountain B, which is $9 \mathrm{~km}$ away. They set off from A at 8:00 AM and arrive in B at 12:00 PM.

The next day, the group hikes the same path back. They start at 8:30 AM and arrive in A at 11:00 AM.

Is there a point on the way from A to B where the group is at the same time on both days? Justify your answer!
There is such a point in time.

On the way from A to B, the group has a speed of $v_{1}=\frac{9 \mathrm{~km}}{4 \mathrm{~h}}$, and on the way from B to A, a speed of $v_{2}=\frac{9 \mathrm{~km}}{2.5 \mathrm{~h}}=\frac{18 \mathrm{~km}}{5 \mathrm{~h}}$.

From A to B, the group is exactly $s=t_{1} \cdot v_{1}=\frac{9}{4} t_{1} \mathrm{~km}$ away from A after a travel time of $t_{1}$. From B to A, the distance from B after $t_{2}$ is exactly $s_{2}=t_{2} \cdot v_{2}=\frac{18}{5} t_{2} \mathrm{~km}$. Since the total distance is $9 \mathrm{~km}$, it follows that $s_{1}=9 \mathrm{~km}-s_{2}$. Since the group starts half an hour earlier on the second day, it also follows that $t_{2}=t_{1}-0.5$ h. Therefore, the equation for the point with the same distance from A and the same time is

$$
\frac{9}{4} t_{1}=9-\left(t_{1}-0.5\right) \frac{18}{5}
$$

The solution to the equation is $t_{1}=\frac{24}{13} \mathrm{~h}$, which means the time is 9:51 AM. At this moment, the pioneers are $4.15 \mathrm{~km}$ away from A on each day. The verification confirms the result.