13.215. A motorcyclist is moving away from point $A$ along a highway at a constant speed of $a$ km/h. After 30 minutes, a second motorcyclist starts from the same point at a constant speed of $1.25 a$ km/h. How many minutes after the start of the first motorcyclist was a third motorcyclist dispatched from the same point, if it is known that he developed a speed of $1.5 a$ km/h and caught up with the first motorcyclist at the same time as the second motorcyclist?
Solution.

Let $t$ be the time the third racer drove before catching up with the first. This means he drove $(1.5 a \cdot t)$ km. The first racer drove for $t_{1}$ hours.

So $t_{1} \cdot a=t \cdot 1.5 a$.

The second racer drove $\left(t_{1}-\frac{1}{2}\right)$ hours and covered $1.25 a\left(t_{1}-\frac{1}{2}\right)$ km.

Solving the system $\left\{\begin{array}{l}a \cdot t_{1}=1.5 a t, \\ 1.25 a\left(t_{1}-\frac{1}{2}\right)=1.5 a t,\end{array}\right.$ we find $t=\frac{5}{3} \text{ hours}; t_{1}=2.5 \text{ hours}$.

Therefore, the third racer started after the first racer's start by $t_{1}-t=\frac{5}{6} \text{ hours}=50$ minutes.

Answer: after 50 minutes.