## Task 2 - 010932

Kurt is riding a tram along a long straight street. Suddenly, he sees his friend walking in the opposite direction on this street at the same level. After one minute, the tram stops. Kurt gets off and runs after his friend at twice the speed of his friend, but only at a quarter of the average speed of the tram.

How many minutes will it take for him to catch up? How did you arrive at your result?
Let $v_{F}, v_{S}$ and $v_{K}$ be the velocities of the friend, the tram, and Kurt, respectively. Initially, the friend and the tram move apart with the relative velocity $v_{S}+v_{F}$.

At the time of getting off after $t=1 \mathrm{~min}$, both have traveled the distance $s=\frac{v_{S}+v_{F}}{t}$ apart. Subsequently, Kurt catches up with his friend at the relative velocity $v_{K}-v_{F}$.

According to the problem statement, $v_{K}=2 v_{F}=\frac{1}{4} v_{S}$. Therefore, the time required to catch up is:

$$
t^{\prime}=\frac{s}{v_{K}-v_{F}}=\frac{v_{S}+v_{F}}{v_{K}-v_{F}} t=\frac{8 v_{F}+v_{F}}{2 v_{F}-v_{F}} t=9 t=9 \mathrm{~min}
$$