7.2. A pedestrian left point $A$ for point $B$, which are 5 km apart. At the same time, a cyclist left point $B$ towards the pedestrian, with a speed twice that of the pedestrian. Upon meeting the pedestrian, he turned around and rode back to $B$. Upon reaching $B$, the cyclist turned again and rode towards the pedestrian, and so on. What distance will the cyclist have traveled by the time the pedestrian arrives at $B$?

## 7.2. Calculations


7.2. A n s w e r: 10 km. We can assume that the cyclist is always moving in one direction (the length of the path does not change from this).

His speed is twice the speed of the pedestrian, so in the time it takes the pedestrian to walk 5 km, the cyclist will travel 10 km.