A pedestrian set out from point $A$ to point $B$. At the same time, a cyclist set out from $B$ to $A$. After one hour, the pedestrian found himself exactly halfway between point $A$ and the cyclist. Another 15 minutes later, they met, and each continued on their way.

How much time did the pedestrian spend on the journey from $A$ to $B$? (The speeds of the pedestrian and the cyclist are constant.)

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Let one hour after the start of the movement, the pedestrian be at point $C$, and the cyclist at point $D$. Then the segment $A C$ is equal to the segment $C D$ (see figure). Let them meet at point $E$ after 15 minutes.

![](https://cdn.mathpix.com/cropped/2024_05_06_f36000480089c1ccffcbg-25.jpg?height=309&width=849&top_left_y=1395&top_left_x=610)

The pedestrian spent 60 minutes on the path $A C$, so he also spent 60 minutes on the path $C D$. Therefore, the path $E D$ took the pedestrian $60-15=45$ minutes, and the cyclist 15 minutes. Consequently, the pedestrian's speed is 3 times less than the cyclist's speed.

The cyclist traveled the path $B D$ in one hour, so it took the pedestrian 3 hours to travel this path. The pedestrian spent 2 hours on the path $A D$, so the total time for the entire journey was 5 hours.

## Answer

5 hours.