## 6. Math Puzzle 11/65

In a pursuit race, the drivers start one minute apart.

How fast is a driver going if he catches up to the driver in front of him, who is traveling at 36 $\frac{\mathrm{km}}{\mathrm{h}}$, after $6 \mathrm{~km}$?
Given are $\mathrm{t}=1 \mathrm{~min}, \mathrm{~s}=6 \mathrm{~km}, v_{1}=36 \frac{\mathrm{km}}{\mathrm{h}}$. We need to find $v_{2}$.

Solution: $t_{1}=\frac{s}{v_{1}}=10 \mathrm{~min}, t_{2}=10 \mathrm{~min}-1 \mathrm{~min}=9 \mathrm{~min}$ and $v_{2}=\frac{s}{t_{2}}=40 \frac{\mathrm{km}}{\mathrm{h}}$.