## Task 2 - 040712

A freight train traveled $35 \frac{3}{4} \mathrm{~km}$ in the first hour and in the following $2 \frac{1}{2}$ hours, it covered an additional 92.7 $\mathrm{km}$. For the return trip on the same route, it took three hours and 12 minutes. Calculate the average speed for the entire trip! Round to one decimal place!
The total distance is: $\left(35 \frac{3}{4}+92.7\right) \mathrm{km} \cdot 2=256.9 \mathrm{~km}$.

The total travel time is: $(1+2.5+3.2) \mathrm{h}=6.7 \mathrm{~h}$.

The average speed is thus $v=\frac{s}{t}=\frac{256.9 \mathrm{~km}}{6.7 \mathrm{~h}} \approx 38.3 \frac{\mathrm{km}}{\mathrm{h}}$.