118(987). A swimmer is swimming against the current of a river and meets an empty boat floating downstream. Continuing to swim against the current for another $t$ minutes after the moment of meeting, he then turns back and catches up with the boat $s$ meters from the meeting point. Find the speed of the river current.
Solution. Let the speed of the swimmer in still water be $v_{1}$ meters per minute, and the speed of the boat (the speed of the river current) be $v_{2}$ meters per minute. Then the swimmer's speed downstream is $\left(v_{1}+v_{2}\right)$ meters per minute, and his speed upstream is $\left(v_{1}-v_{2}\right)$ meters per minute.

In $t$ minutes after the moment of meeting, the swimmer swam upstream a distance of $\left(v_{1}-v_{2}\right) t$ meters, and downstream a distance of $\left(v_{1}-v_{2}\right) t+s$ meters. Downstream, the swimmer swam for $\frac{\left(v_{1}-v_{2}\right) t+s}{v_{1}+v_{2}}$ minutes, and in total (from the moment of meeting the boat to the moment when he caught up with the boat) $\frac{\left(v_{1}-v_{2}\right) t+s}{v_{1}+v_{2}}+t$ minutes, which is equal to the time the boat traveled $\left(\frac{s}{v_{2}}\right)$.

We have the equation

$$
\frac{\left(v_{1}-v_{2}\right) t+s}{v_{1}+v_{2}}+t=\frac{s}{v_{2}}
$$

from which

$$
\begin{gathered}
\frac{v_{1} t-v_{2} t+s+v_{1} t+v_{2} t}{v_{1}+v_{2}}=\frac{s}{v_{2}} \\
\left(s+2 v_{1} t\right) v_{2}=s\left(v_{1}+v_{2}\right) \\
s v_{2}+2 v_{1} v_{2} t=s v_{1}+s v_{2} \\
v_{2}=\frac{s v_{1}}{2 v_{1} t}=\frac{s}{2 t}
\end{gathered}
$$

Answer: the speed of the river current is $\frac{s}{2 t}$ meters per minute.