13.255. Three swimmers have to swim a lane 50 m long, immediately turn around, and return to the starting point. The first starts first, the second starts 5 s later, and the third starts another 5 s later. At some point in time, before reaching the end of the lane, the swimmers found themselves at the same distance from the start. The third swimmer, having reached the end of the lane and turned back, met the second 4 m from the end of the lane, and the first - 7 m from the end of the lane. Find the speed of the third swimmer.
## Solution.

Let $t$ be the time it takes for the third swimmer to catch up with the first two; $v_{1}$, $v_{2}, v_{3}$ be the speeds of the swimmers. Then $v_{1}(t+10)=v_{2}(t+5)=v_{3} t$, from which $\frac{v_{2}}{v_{3}-v_{2}}=\frac{2 v_{1}}{v_{3}-v_{1}}(1)$

According to the problem, $\left\{\begin{array}{l}\frac{50+4}{v_{3}}=\frac{50-4}{v_{2}}-5, \\ \frac{50+7}{v_{3}}=\frac{50-7}{v_{1}}-10 .\end{array}\right.$

Using equation (1), we solve the system and find $v_{3}=\frac{22}{15}$ (m/s).

Answer: $\frac{22}{15} \mathrm{m} / \mathrm{s}$.