$A$ place, from which a car starts, and $B$ place, 80 km away from it, from which a cyclist starts, both set off at the same time, heading towards each other. After one hour of travel, they are still $24 \mathrm{~km}$ apart. At this point, the cyclist rests for 1 hour, and thus they meet 90 minutes after the start.

$1^{0}$. What are their speeds? Where do they meet?

$2^{0}$. Also, illustrate the solution with a graph!
I. Solution. $1^{0}$. The car and the cyclist cover $80-24=56 \mathrm{~km}$ in one hour. If the car travels $x$ km and the cyclist $y \mathrm{~km}$ in one hour, then

$$
x+y=56 \ldots
$$

At a distance of $24 \mathrm{~km}$, the car travels for $\frac{1}{2}$ hour, and the cyclist for $\frac{1}{4}$ hour until they meet, so

$$
\frac{x}{2}+\frac{y}{4}=24 \quad \text { or } \quad 2 x+y=96 \ldots
$$

Subtracting the terms of (1) from those of (2): $x=40$ and thus $y=16$.

The car's speed is 40, and the cyclist's speed is $16 \mathrm{~km}$ per hour.

The car covers 60 km in $\frac{3}{2}$ hours, and the cyclist covers $20 \mathrm{~km}$ in $\frac{5}{4}$ hours.

The meeting point is $60 \mathrm{~km}$ away from $A$ and $20 \mathrm{~km}$ away from $B$.

$2^{0}$. In our Cartesian coordinate system, the x-axis represents the time axis; we start measuring time when the travelers set off. One minute of time corresponds to $0.5 \mathrm{~mm}$. The y-axis serves to measure the distance from $A$; $1 \mathrm{~km}$ of distance corresponds to $0.5 \mathrm{~mm}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_181515d27f17db093bbcg-1.jpg?height=610&width=710&top_left_y=890&top_left_x=705)

The car starts from $A$ (the origin of the coordinate system); after $30^{\prime}$ it is at $A_{1}$, after $60^{\prime}$ at $A_{2}$, and after $90^{\prime}$ at $A_{3}$.

$B$ is on the y-axis, $80 \mathrm{~km}$ away from $A$ $(=4 \mathrm{~cm})$. The cyclist starting from $B$ is at $B_{1}$ after $30^{\prime}$, $80-8=72 \mathrm{~km}$ away from the x-axis $(=3.6 \mathrm{~cm})$. After $60^{\prime}$, the cyclist is at $B_{2}$, $80-16=64 \mathrm{~km}$ away from the x-axis $(=3.2 \mathrm{~cm})$. The cyclist rests here, and the distance from the x-axis does not change for $15^{\prime}$. Therefore, the segment $B_{2} B_{3}$ of the path graph is parallel to the x-axis. In the next $15^{\prime}$, the cyclist's distance from the x-axis decreases by $4 \mathrm{~km}$, and thus the path graph goes from $B_{3}$ to $A_{3}$. The two graphs intersect at $A_{3}$, $60 \mathrm{~km}$ away from $A$ and $20 \mathrm{~km}$ away from $B$.

(Due to uniform motion, the path graph is a straight line.)

Béla Szalma (Érseki g. V. o. Bp. II.)

II. Solution. The car and the cyclist cover $56 \mathrm{~km}$ in one hour. Traveling similarly, they should have covered $28 \mathrm{~km}$ in the next half hour. They actually only covered $24 \mathrm{~km}$. The $4 \mathrm{~km}$ shortfall is due to the cyclist's $1 / 4$ hour rest. Therefore, the cyclist travels $16 \mathrm{~km}$ per hour, and the car travels $56-16=40 \mathrm{~km}$ per hour.

Their meeting point is $40 \times 1.5=60 \mathrm{~km}$ away from $A$.