68(1002). Two speed skaters started simultaneously on a 10000 m distance. The first, moving at a speed exceeding the second's by $\frac{1}{3} \mathrm{m} / \mathrm{s}$, completed each lap 2 seconds faster than the second and after 10 minutes overtook the second by half a lap. Find the speeds of these skaters and calculate the time it took each of them to complete the entire distance.
Solution. Let the speed of the second skater be $v$ meters per second, then the speed of the first is $\left(v+\frac{1}{3}\right)$ meters per second. Since the first skater overtakes the second by half a lap in 10 minutes (600 seconds), half the circumference of the track is 200 meters $\left(\frac{1}{3} \cdot 600\right)$, i.e., the circumference of the track is 400 meters.

Since the first skater completed each lap 2 seconds faster than the second, we have $2=\frac{400}{v}-\frac{400}{v+\frac{1}{3}}$, from which $3 v^{2}+v-200=0$.

The condition of the problem is satisfied by one (positive) root of the equation: $v=8(m / s)$.

We have: the speed of the second skater -8 m/s, the speed of the first skater $-8 \frac{1}{3} \mathrm{m} / \mathrm{s}$, the time it took the first skater to complete the entire distance:

$$
10000: 8 \frac{1}{3}=1200 \mathrm{~s}=20 \text { min, }
$$

the time it took the second skater to complete the entire distance:

$$
10000: 8=1250 \mathrm{~s}=20 \text { min } 50 \mathrm{~s} \text {. }
$$

Answer: $8 \frac{1}{3} \mathrm{m} / \mathrm{s}, 8 \mathrm{~m} / \mathrm{s} ; 20$ min, 20 min 50 s.