## Task 4 - 080824

A truck traveling at a constant speed $v_{1}$ is overtaken by a car traveling at a constant speed $v_{2}$, 1 hour 25 minutes after the truck started. The car left from the same location 30 minutes later but had a speed 25 km/h greater than the truck.

a) Calculate $v_{1}$ and $v_{2}$!

b) What is the length $s$ of the distance traveled by both vehicles up to the point of overtaking?
At time $t_{i}=85 \mathrm{~min}$, the truck covers the same distance that the car took $t_{2}=55 \mathrm{~min}$ to cover.

If the speed of the truck is denoted by $v_{1}=x \frac{\mathrm{km}}{\mathrm{h}}$, then the speed of the car is $v_{2}=(x+25) \frac{\mathrm{km}}{\mathrm{h}}$. Since $s=t_{1} \cdot v_{1}=t_{2} \cdot v_{2}$, it follows that: $x \cdot \frac{85}{60}=(x+25) \cdot \frac{55}{60}$.

From this, we obtain $x=\frac{5 \cdot 55}{60}=45 \frac{5}{6}$.

a) The speed of the truck was $45 \frac{5}{6} \frac{\mathrm{km}}{\mathrm{h}}$, and that of the car was $70 \frac{5}{6} \frac{\mathrm{km}}{\mathrm{h}}$.

b) Because

$$
\frac{5 \cdot 55}{60} \cdot \frac{85}{60}=64 \frac{67}{72} \approx 64.9
$$

the distance traveled is $s \approx 64.9 \mathrm{~km}$.

These values indeed satisfy all the conditions of the problem, as a check shows.

Solutions of the 2nd Round 1968 adopted from [5]

### 5.10.3 3rd Round 1968, Class 8