13.270. Two cruise ships are due to arrive in Odessa with a 1-hour gap. Both ships are traveling at the same speed, but circumstances have led to the first ship being delayed by \(t_{1}\) minutes, and the second by \(t_{2}\) minutes. Upon receiving a radio instruction to arrive on time, both captains simultaneously increased the speeds of their ships: the first by \(v_{1}\) km/h, and the second by \(v_{2}\) km/h, as a result of which both ships arrived in Odessa exactly on schedule. What speed were the ships traveling at before receiving the radio signal?

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1181.jpg?height=536&width=561&top_left_y=55&top_left_x=414)

Fig. 13.11
Solution.

Graphical representation of the conditions of the problem in Fig. 13.11.

We obtain $S_{1}=\left(v+v_{1}\right) T, S_{2}=\left(v_{1}+v_{2}\right)(T+1)=v\left(T+1+\frac{t_{2}}{60}\right)$ Simplifying and eliminating $T$, we have $v=\frac{60 v_{1} v_{2}}{v_{1} t_{2}-v_{2} t_{1}}$ km/h.

Answer: $\frac{60 v_{1} v_{2}}{v_{1} t_{2}-v_{2} t_{1}}$ km/h.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1181.jpg?height=472&width=424&top_left_y=1207&top_left_x=125)

Fig. 13.12