14. Person A and Person B are doing special training on the same elliptical track: They start from the same point at the same time, running in opposite directions. After each completes the first lap and returns to the starting point, they immediately turn around and accelerate for the second lap. During the first lap, B's speed is $\frac{2}{3}$ of A's speed. For the second lap, A increases their speed by $\frac{1}{3}$, and B increases their speed by $\frac{1}{5}$. It is known that the second meeting point of A and B is 190 meters away from the first meeting point. What is the length of this elliptical track in meters?
14.【Solution】

Let's draw two diagrams (top figure), and assume that the initial speed of A is a, so the speed of B is $\frac{2}{3} \mathrm{a}$. Let the length of the track be $\mathrm{L}$, then the first meeting point of A and B, measured from the starting point in the direction of A's advance, is $\frac{3}{5} \mathrm{~L}$. When A completes the first lap, B has run $\frac{2}{3} L$. After B runs the remaining $\frac{1}{3} L$, A has turned back and is running at a speed of $\mathrm{a}\left(1+\frac{1}{3}\right)=\frac{4}{3} \mathrm{a}$. By the time B completes the first lap, A has turned back and run $\frac{2}{3} L$, at which point B turns back and runs at a speed of $\frac{2}{3} \mathrm{a}\left(1+\frac{1}{5}\right)=\frac{4}{5} \mathrm{a}$. From this point on, the speed ratio of A to B is $\frac{4}{3} \mathrm{a} \div \frac{4}{5} \mathrm{a}=\frac{5}{3}$, or $5: 3$. Therefore, at the second meeting, A has run $\frac{5}{8}$ of the remaining $\frac{L}{3}$, and B has run $\frac{3}{8}$ of it, meaning the second meeting point is $\frac{3}{8} \times \frac{L}{3} = \frac{L}{8}$ from the starting point. The distance between the two meeting points is $\left(\frac{3}{5}-\frac{1}{8}\right) \mathrm{L}=190$ (meters), i.e., $\frac{19}{40} L = 190$ (meters),
$$
\mathrm{L}=400 \text { (meters) }
$$

Answer: The length of the track is 400 meters.