13.406 Three motorcyclists travel the same section AB of the road at constant but different speeds. First, the first motorcyclist passed point A, and 5 s later, in the same direction, the second and third motorcyclists. After some time, the third motorcyclist overtook the first, and 10 s later, the second overtook him. How long will it take the first motorcyclist to travel the distance $\mathbf{A B}$, if the second traveled this distance in 1 min, and the third - 3240 s?
Solution. Let $z$ be the distance between A and B, $x$ be the time (sec) it takes for the first motorcyclist to cover this distance. Then $\frac{z}{x}, \frac{z}{60}, \frac{z}{40}$ are the speeds of the motorcyclists, $\frac{5 z}{x}$ is the difference in distances between them at the moment the second and third pass point A. From this, $\frac{5 z}{x}:\left(\frac{z}{40}-\frac{z}{x}\right)$ is the time it takes for the third to catch up with the first; $\frac{5 z}{x}\left\{\left(\frac{z}{60}-\frac{z}{x}\right)\right.$ is the time it takes for the second to catch up with the first (from the moment the second and third pass point A).

In the end, we have: $\frac{5}{x}:\left(\frac{1}{60}-\frac{1}{x}\right)-\frac{5}{x}:\left(\frac{1}{40}-\frac{1}{x}\right)=10 \Rightarrow x^{2}-110 x+2400=0 \Rightarrow x=80$ ( $x=30$ does not fit, as $x>60$ ).

Answer: 80 sec.