# Answer: 6.0 or 6.67

## Solution and Grading Criteria:

For $t=5$ min, the first pedestrian walked: $S_{1}=v_{1} t=7 \cdot 5 \cdot 60=2100$ meters

(5 points)

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

either $S_{2}=2100-200-100=1800$ meters, in which case his speed is:

$v_{2}=\frac{S_{2}}{t}=\frac{1800}{5 \cdot 60}=6.00 \mathrm{~m} / \mathrm{s}$

or $S_{2}=2100-200+100=2000$ meters, in which case his speed is:

$v_{2}=\frac{S_{2}}{t}=\frac{2000}{5 \cdot 60} \approx 6.67 \mathrm{~m} / \mathrm{s}$

#
# Problem No. 5 (15 points)

Two people are moving in the same direction. At the initial moment, the distance between them is $S_{0}=200 \mathrm{~m}$. The speed of the first, faster, pedestrian is $v_{1}=7 \mathrm{~m} / \mathrm{c}$. Determine the speed $v_{2}$ of the second pedestrian, given that after $t=5$ minutes, the distance between them was $S=100$ m.