13.401 Two speed skaters started simultaneously on a running track for a distance of $s$ m. When the winner reached the finish line, the other had one more lap to go. Determine the length of the running track, given that the winner, running each lap $a$ s faster than the loser, completed the distance in $t$ min.
Solution. Let $x$ - be the length of the running track (or circle). According to the condition, $\frac{s}{60 t}$ - is the speed (m/sec) of the first athlete. Hence, $\frac{x}{s / 60 t}+a$ - is the time it takes for the second athlete to run one lap. Since the number of laps to the finish is $\frac{s}{x}$, then $\left(\frac{s}{x}-1\right)\left(\frac{60 x t}{s}+a\right)-$ is the time it takes for the second athlete to run $\frac{s}{x}-1$ laps. According to the condition, in the same time, i.e., in $60 t$ seconds, the first athlete finished the distance. Therefore, $\left(\frac{s}{x}-1\right)\left(\frac{60 x t}{s}+a\right)=60 t \Rightarrow 60 t x^{2}+s a x-s^{2} a=0 \Rightarrow x=\frac{-s a+\sqrt{s^{2} a^{2}+240 t a s^{2}}}{120 t}$. Answer: $\frac{s}{120 t}\left(\sqrt{a^{2}+240 a t}-a\right)$ m.