3. Two wagons start simultaneously from places $A$ and $B$ towards each other. Each of them, when they reach the opposite place, immediately turns back. The wagons first met $50 \mathrm{~km}$ from place $A$, and the second time they met $30 \mathrm{~km}$ from place $B$. How long is the track between places $A$ and $B$? (It is assumed that the wagons move at constant speeds.)
Solution. Let the length of the track from $A$ to $B$ be $s$. Let $V_{a}$ denote the speed of the train starting from point $A$, and let $t$ be the time from the start until the first meeting with the other train. During time $t$, the first train has traveled $50 \mathrm{~km}$, so $V_{a} t=50$. The two trains together have traveled the distance from $A$ to $B$ once (on the left diagram, the path traveled by the first train is $A C$, the path traveled by the second train is $B C$, and together they have traveled $A C+B C=A B=s$). By the next meeting, the two trains together have traveled a distance of $3 s$, see the right diagram, the first train has traveled $A B+B D$, and the second train has traveled the path $B A+A D$. Since $B A+A D=s=A B$, we get $A B+B D+B A+A B=3 s$. According to the condition, the speeds of the trains are uniform, so it takes three times as much time to travel three times the distance. Therefore, by the second meeting, the first train has traveled a distance of $3 V_{a} t$, that is, $3 V_{a} t=s+30$. If we substitute $V_{a} t=50$, we get $150=s+30$, hence $s=120$. Therefore, the track between points $A$ and $B$ is $120 \mathrm{~km}$ long.
![](https://cdn.mathpix.com/cropped/2024_06_05_f58d16fda9be15138274g-4.jpg?height=120&width=1092&top_left_y=1746&top_left_x=367)