[ Motion problems ]

A cyclist rode from point A to point B, where he stayed for 30 minutes, and then returned to A. On the way to B, he overtook a pedestrian, and 2 hours later met him on the return trip. The pedestrian arrived in B at the same time the cyclist returned to A. How much time did it take the pedestrian to travel from A to B, if his speed is four times less than the cyclist's speed?
The first method. The distance that the pedestrian covers in 2 hours, we will take as a unit. Then the cyclist covers this same distance in 30 minutes. From the moment of their first meeting, the pedestrian has walked one unit, while the cyclist has traveled three units (he rested for half an hour in point B). Therefore, the distance from the point of their second meeting to point B is one unit. Then, after the second meeting, the pedestrian will walk another unit, while the cyclist will travel 4 units in the same time. Therefore, the distance between A and B is 5 units, so it took the pedestrian 10 hours to travel from A to B.

The second method. Let $v$ km/h be the speed of the pedestrian, then the speed of the cyclist is $4v$ km/h. Let the first meeting of the cyclist and the pedestrian occur at a distance of $x$ km from point B. Then in 2 hours, which passed until the second meeting, the pedestrian walked $2v$ km and was at a distance of $x-2v$ km from point B, while the cyclist traveled a distance of $1.5 \times 4v = 6v$ (km) and was at a distance of $6v-x$ km from point B.

Therefore, $x-2v=6v-x$, from which $x=4v$. This means that the point of the first meeting is at a distance from point B that the cyclist covers in one hour, and the point of the second meeting is at a distance that the cyclist covers in half an hour. Since 2 hours after the second meeting, the pedestrian will be at point B, and the cyclist will be at point A, it took the cyclist 2.5 hours to travel from B to A. Therefore, it will take the pedestrian 4 times longer, that is, 10 hours, to travel from A to B.

## Answer

10 hours.