13.407 Three people, A, B, and C, approached the shore of a reservoir; A set off to the opposite shore swimming at a speed of v km/h; simultaneously, B and C set off in a motorboat at a speed of $10 v$ km/h. After some time, C decided to cover the remaining distance by swimming at the same speed as A. At the same moment, B turned back to pick up A, who quickly got into the boat and continued the journey together with B. All three arrived at the opposite shore simultaneously. Determine the time of the crossing, given that the width of the reservoir is $b$ km, and the current speed is zero.
Solution. Let $T_{1}$ be the point on the shore where the crossing began, $T_{4}$ - the point where it ended, $T_{2}$ - the point where A met the boat, $T_{3}$ - the point where B disembarked from the boat. Since A swam the segment $T_{1} T_{2}$, and C swam the segment $T_{3} T_{4}$, and they both arrived at $T_{4}$ simultaneously, then $T_{1} T_{2}=T_{3} T_{4}$. Let $x=T_{2} T_{3}$. Then $T_{1} T_{2}=T_{3} T_{4}=\frac{b-x}{2}$, and therefore, the boat traveled a distance of $\frac{b-x}{2}+2 x=\frac{b+3 x}{2}$ before meeting A.

From this, $\frac{b+3 x}{2}: \frac{b-x}{2}=10 v: v \Rightarrow x=\frac{9 b}{13}$.

In total, the boat traveled a distance of $b+2 x=\frac{31 b}{13}$.

Therefore, the time of the crossing is $\frac{31 b}{130 v}$.

Answer: $\frac{31 b}{130 v}$ hours.