## Task 2 - 140822

Four trucks $A, B, C$, and $D$ travel the same route. If $A$ travels at an average speed of $56 \frac{km}{h}$ and $B$ at $40 \frac{km}{h}$, then $A$ takes exactly 2 hours less than $B$ to complete this route.

At what average speed would $C$ need to travel if $D$ were to start 4 hours earlier than $C$, travel at an average speed of $35 \frac{\mathrm{km}}{\mathrm{h}}$, and arrive at the common destination at the same time as $C$?
Assuming that $A$ has covered the distance mentioned in the problem in $t$ hours. Then $B$ needed $(t+2)$ hours for the same distance. Therefore, $56 t=40(t+2)$, from which $t=5$ is obtained.

Thus, $A$ covered the distance with an average speed of $56 \mathrm{~km} \cdot \mathrm{h}^{-1}$ in 5 hours. The distance was therefore $280 \mathrm{~km}$ long.

Since $280: 35=8$, $D$ would need exactly 8 hours to cover this distance at an average speed of $35 \mathrm{~km} \cdot \mathrm{h}^{-1}$. As $C$ is supposed to depart 4 hours later than $D$, he would need to cover the entire distance in exactly 4 hours if he wants to arrive at the destination at the same time as $D$. Since $280: 4=70$, he would need to maintain an average speed of $70 \mathrm{~km} \cdot \mathrm{h}^{-1}$.