## Task 1 - 260731

Mr. Anders drove past a gas station (A) on the highway at a speed of $100 \frac{\mathrm{km}}{\mathrm{h}}$. After driving an additional distance of $175 \mathrm{~km}$, Mr. Anders had to switch the fuel valve to reserve.

Since the next gas station (B) is still $45 \mathrm{~km}$ away from this point on the highway, Mr. Anders reduced his speed to $60 \frac{\mathrm{km}}{\mathrm{h}}$ to consume less fuel.

At what average speed did Mr. Anders cover the distance between A and B? (The short braking distance, during which the speed was reduced from $100 \frac{\mathrm{km}}{\mathrm{h}}$ to $60 \frac{\mathrm{km}}{\mathrm{h}}$, should not be considered in the calculation, as it only slightly affects the average speed.)
The distance from A to the point where the speed was reduced is $s_{1}=175$ $\mathrm{km}$, the distance from this point to B is $s_{2}=45 \mathrm{~km}$, so the total distance from A to B is $s=s_{1}+s_{2}=220 \mathrm{~km}$. (1)

The speed at which the first segment was covered is $v_{1}=100 \frac{\mathrm{km}}{\mathrm{h}}$. Since this speed was constant, the travel time for this segment is

$$
t_{1}=\frac{s_{1}}{v_{1}}=\frac{175}{100} \mathrm{~h}=\frac{7}{4} \mathrm{~h}
$$

The speed at which the second segment was covered is $v_{2}=60 \frac{\mathrm{km}}{\mathrm{h}}$. Since this speed was also constant, the travel time for this segment is

$$
t_{2}=\frac{s_{2}}{v_{2}}=\frac{45}{60} \mathrm{~h}=\frac{3}{4} \mathrm{~h}
$$

Therefore, the total travel time from A to B is $t=t_{1}+t_{2}=\frac{5}{2} \mathrm{~h}$ (2). From (1) and (2), the required average speed is

$$
v=\frac{s}{t}=\frac{220 \mathrm{~km}}{\frac{5}{2} \mathrm{~h}}=88 \frac{\mathrm{km}}{\mathrm{h}}
$$