## 177. Math Puzzle $2 / 80$

A moped rider drives seven kilometers at a speed of $v_{1}=30 \mathrm{~km} / \mathrm{h}$ and then four kilometers at $v_{2}=40 \mathrm{~km} / \mathrm{h}$.

What is his average speed?
The total distance $s_{G}$ of the moped rider consists of $s_{1}=7 \mathrm{~km}$ and $s_{2}=4 \mathrm{~km}$, thus $s_{G}=11 \mathrm{~km}$.

To calculate the average speed $\bar{v}$, the total time $t_{G}$ must first be determined: $v=\frac{s}{t}$, from which it follows that $t=\frac{s}{v}$.

$$
t_{1}=\frac{7 \mathrm{~km} \cdot \mathrm{h}}{30 \mathrm{~km}}=\frac{7}{30} \mathrm{~h} \quad, \quad t_{2}=\frac{4 \mathrm{~km} \cdot \mathrm{h}}{40 \mathrm{~km}}=\frac{3}{30} \mathrm{~h}
$$

$t_{G}=t_{1}+t_{2}=\frac{10}{30} \mathrm{~h}=\frac{1}{3} \mathrm{~h}$

For the average speed, we now get:

$$
\bar{v}=\frac{s_{G}}{t_{G}}=33 \frac{\mathrm{km}}{\mathrm{h}}
$$

The moped rider thus travels at an average speed $\bar{v}$ of $33 \mathrm{~km} / \mathrm{h}$.