A pedestrian set off from city $A$ towards city $B$ in the south. A cyclist started after him a little later. At 13:00, the cyclist caught up with the pedestrian and immediately turned back. Upon returning to $A$, the cyclist turned around again and thus arrived in $B$ at the same time as the pedestrian, at 16:00.

How many times faster is the cyclist's speed compared to the pedestrian's?
Assuming both movements are uniform. Thus, the pedestrian covers 3/4 of the AB distance between 13:00 and 16:00, while the cyclist, after turning back, first covers 1/4 of the AB distance, and then follows the pedestrian, covering the entire AB segment.

Therefore, the cyclist covers 5/4 of the AB distance in 3 hours (from 13:00 to 16:00), while the pedestrian covers 3/4 of the AB distance in the same time. The ratio of the two speeds is thus \(\frac{5}{4} AB : \frac{3}{4} AB = 5:3\).

Note. The plotting of the distance-time or speed-time graphs of the movements (Figures 1/a and 1/b) can provide very useful assistance in solving such problems.

![](https://cdn.mathpix.com/cropped/2024_05_02_52e685cd9c6c75a3d8c6g-1.jpg?height=740&width=935&top_left_y=519&top_left_x=579)

Figure 1/a

If the cyclist starts from point A at times \(I_{1}\) and \(I_{2}\), then the triangles \(O I_{1} T_{1}\) and \(O I_{2} T_{2}\) are clearly similar (Figure 1/a) and the similarity ratio is: \(O T_{1} : O T_{2} = 1 : 4\). On the other hand, if the cyclist starts \(t\) hours after 12:00, i.e., \(O I_{1} = t\), then \(O I_{2} = 4t\); however, \(O I_{2} = 2 - t\), from which \(t = \frac{2}{5}\). The cyclist thus catches up with the pedestrian for the first time in \(1 - t = \frac{3}{5}\) hours, and the ratio of their speeds is \(5:3\).

![](https://cdn.mathpix.com/cropped/2024_05_02_52e685cd9c6c75a3d8c6g-1.jpg?height=464&width=979&top_left_y=1611&top_left_x=562)

Figure 1/b

If we start from the speed-time graph (Figure 1/b), then the area of the shaded rectangles represents the distance covered by the cyclist from 13:00, which is \(3 \cdot v_{k}\). This is 5/4 of the pedestrian's total 4-hour distance, i.e.,

\[
3 v_{k} = \frac{5}{4} \cdot 4 \cdot v_{gy}, \quad \text{from which} \quad v_{k} : v_{gy} = 5:3
\]