Problem No. 5 (15 points)

Two people are moving in the same direction. At the initial moment, the distance between them is $S_{0}=100 \text{ m}$. The speed of the first, faster, pedestrian is $v_{1}=8 \text{ m} / \text{s}$. Determine the speed $v_{2}$ of the second, if it is known that after $t=5$ min the distance between them was $S=50 \text{ m}$.
Answer: 7.5 or 7.83

## Solution and Evaluation Criteria:

For $t=5$ min, the first pedestrian walked: $S_{1}=v_{1} t=8 \cdot 5 \cdot 60=2400$ meters

Therefore, there are two possible answers. In this time, the second pedestrian walked:

or $S_{2}=2400-100-50=2250$ meters, in this case his speed:

$v_{2}=\frac{S_{2}}{t}=\frac{2250}{5 \cdot 60}=7.50 \mathrm{~m} / \mathrm{s}$

or $S_{2}=2400-100+50=2350$ meters, in this case his speed:

$v_{2}=\frac{S_{2}}{t}=\frac{2350}{5 \cdot 60} \approx 7.83 \mathrm{~m} / \mathrm{s}$