## Task 3 - 220813

A NVA marching column is $3.2 \mathrm{~km}$ long. A regulation post rides from the end of the marching column, catches up with the head of the marching column after a 5.6 km ride, immediately continues at the same speed for exactly 6 minutes, and then reaches his comrade who is standing at the next road intersection to block the oncoming traffic. Here, he waits for the marching column, which has maintained its average speed throughout the entire time.

a) How do the average speeds of the regulation post and the marching column relate to each other?

b) How many minutes in total does the regulation post have to wait at the intersection for the head of the marching column?
a) The average speeds of the regulation post and the marching column are denoted by $v_{R}$ and $v_{M}$, respectively.

They are in the same ratio as the distances traveled in the same time. The initial distance $s_{0}=5.6 \mathrm{~km}$ traveled by the regulation post consists of the length $s_{1}=3.2 \mathrm{~km}$ of the marching column and the distance $s_{2}$ traveled by the marching column during the overtaking by the regulation post. This distance is therefore $s_{2}=s_{0}-s_{1}=2.4 \mathrm{~km}$. Thus, we have

$$
v_{R}: v_{M}=s_{0}: s_{2}=5.6: 2.4=7: 3
$$

b) The time at which the regulation post leaves the front of the marching column is denoted by $T_{0}$. From this point onward, the regulation post travels an additional distance $s=v_{R} \cdot 6$ min.

The front of the marching column must cover the same distance $s=v_{M} \cdot t$ from the time $T_{0}$ until it reaches the intersection. Therefore, we have

$$
t=\frac{v_{R}}{v_{M}} \cdot 6 \mathrm{~min}
$$

from a) this is $t=\frac{7}{3} \cdot 6 \mathrm{~min}=14 \mathrm{~min}$. This time $t=14 \mathrm{~min}$ consists of the 6 min driving time from the time $T_{0}$ and the waiting time of the regulation post at the intersection. Therefore, the waiting time is 8 min.