1. A bus, starting at 6:00 AM, travels between points $A$ and $B$ at a constant speed, and upon reaching point $A$ or $B$, it immediately turns around. Petya on a moped and Vasya on a bicycle simultaneously set off from point $A$ to point $B$ at 6:00 AM, moving at constant speeds. It is known that the bus, during its second trip from point $A$ to point $B$, caught up with Vasya at 1:30 PM and arrived at point $B$ at 3:00 PM, simultaneously with Petya. Determine the ratio of Vasya's and Petya's speeds. Specify the time at which the bus, on its first trip from point $B$ to point $A$, caught up with Vasya.

(12 points)
Solution. We solve graphically. Let's represent all movements in the coordinates $t$ (time) and $S$ (distance). The trajectory of Petya's movement from point $A$ to point $B$ is the segment $A_{1} B_{2}$. The trajectory of Vasya's movement from point $A$ to point $B$ is the segment $A_{1} B^{3}$.

The bus traveled the distance $A B$ three times from 6 to 15 hours. The bus takes 3 hours to travel the distance $A B$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9a8ac98263d65b486910g-1.jpg?height=463&width=904&top_left_y=796&top_left_x=979)
Vasya caught up with the bus halfway between points $A$ and $B$. Therefore, Vasya spent 15 hours on the journey from $A$ to $B$, and Petya spent 9 hours. The ratio of the speeds of Vasya and Petya is 3:5. Let's determine at what hour the bus, on its first trip from point $B$ to point $A$, caught up with Vasya. Let this time be $x$. The abscissa of the intersection of the medians of the triangle $A_{1} B_{2} A_{2}$ is equal to $x$. Therefore, $\frac{x-6}{13.5-x}=\frac{2}{1}, x=11$.

Answer: 3:5, at 11 o'clock.