On an $A B=8 \mathrm{~km}$ long road, buses travel in both directions at a speed of $12 \mathrm{~km}$ per hour. The first bus from each location departs at 6 o'clock, and a new bus departs every 10 minutes.

A pedestrian starts from $A$ towards $B$ at $81 / 4$ hours; their speed is $4 \mathrm{~km}$ per hour.

Determine graphically how many oncoming buses the pedestrian encounters, and when and where.
The $X$-axis starting from $A$ will be the time axis, such that the 8 o'clock time point is placed at point $A$, and on this axis, a 10-minute time interval corresponds to $1-1 \mathrm{~cm}$.

The $Y$-axis starting from $A$ will measure the distance $A B=8 \mathrm{~km}$. (1 km corresponds to $1 \mathrm{~cm}$.) The bus covers the distance $A B$ in $\frac{8}{12}=\frac{2}{3}$ hours $=40$ minutes.

![](https://cdn.mathpix.com/cropped/2024_05_02_021dbc5815a02335bcfbg-1.jpg?height=536&width=1182&top_left_y=377&top_left_x=466)

The slanted lines, which are parallel to the one starting from $A$ and $B$ at 8 o'clock, indicate the movement of the cars starting from $A$ and $B$. These cars start from $A$ at $8^{h}, 8^{h} 10^{\prime}, 8^{h} 20^{\prime}$, etc., and arrive at $B$ at $8^{h} 40^{\prime}, 8^{h} 50^{\prime}, 9^{h}$, etc.

The pedestrian starts from $A$ at $8^{h} 15$ (E) and arrives at $B$ at $10^{h} 15$ (F).

Their path is indicated by the line segment $E F$. This line segment intersects 16 lines that start from $B$; among these, the first one starts at $7^{h} 40^{\prime}$ and arrives at $A$ at $8^{h} 20^{\prime}$, and the last one starts at $10^{h} 10^{\prime}$ and arrives at $10^{h} 50^{\prime}$.

The times of the encounters. The car that starts at $7^{h} 40^{\prime}$ from $B$ has already been traveling for 35 minutes when the pedestrian starts. At that moment, the car is $8-\frac{35}{60} \cdot 12=1 \mathrm{~km}$ away from $A$. The distance between the car and the pedestrian decreases by $12+4=16 \mathrm{~km}$ per hour. Therefore, they meet after $\frac{1}{16}$ hours $=3^{\prime} 45^{\prime \prime}$, which is at $8^{h} 18^{\prime} 45^{\prime \prime}$.

The second car, which starts from $B$ at $7^{h} 50^{\prime}$, has been traveling for 25 minutes by the time the pedestrian starts, so it is $8-\frac{25}{60} \cdot 12=3 \mathrm{~km}$ away from $A$ when the pedestrian starts. They meet $\frac{3}{16}$ hours after the pedestrian starts, or $\frac{2}{16}=\frac{1}{8}$ hours after the first encounter.

Each subsequent encounter with a car occurs $\frac{1}{8}$ hours $=7^{\prime} 30^{\prime \prime}$ later. The time of the last, 16th encounter is:

$$
8^{h} 18^{\prime} 45^{\prime \prime}+15\left(7^{\prime} 30^{\prime \prime}\right)=8^{h} 18^{\prime} 45^{\prime \prime}+112^{\prime} 30^{\prime \prime}=10^{h} 11^{\prime} 15^{\prime \prime}
$$

which is before $10^{h} 15^{\prime}$.

The locations of the encounters. The pedestrian covers $250 \mathrm{~m}$ in $\frac{1}{16}$ hours and $500 \mathrm{~m}$ in $\frac{1}{8}$ hours. The pedestrian meets the cars starting from $B$ at a distance of $250+n \cdot 500 \mathrm{~m}$ from $A$, where $n=0,1,2 \ldots 15$.

Pál Koren (Fazekas Mihály r. VI. o. Debrecen.).