## Task 3 - 140613

A cyclist rode from $A$ to $B$ on a road at a constant speed. He started in $A$ at 6:00 AM and covered $12 \mathrm{~km}$ every hour. A second cyclist, who also rode the same route from $A$ to $B$ at a constant speed, started on the same day at 7:00 AM in $A$ and covered $15 \mathrm{~km}$ every hour. He met the first cyclist at the same time in $B$.

a) At what time did the second cyclist catch up with the first one?

b) How long is the route from $A$ to $B$?
The first cyclist needed $t$ hours for the route, the second $t-1$ hours, as he started one hour later. The distances $=$ speed $\cdot$ time are the same for both, i.e., with $v$ in $\frac{\mathrm{km}}{\mathrm{h}}$

$$
t \cdot 12=(t-1) \cdot 15 \quad \rightarrow \quad 12 t=15 t-15 \quad \rightarrow \quad 3 t=15 \quad \rightarrow \quad t=5
$$

The first cyclist for 5 hours, the second 4 hours, meaning he catches up after 4 hours. The distance is thus $5 \cdot 12=4 \cdot 15=60 \mathrm{~km}$.