13.396 According to the schedule, a communications officer should first depart from point A, and after 6 hours - a second communications officer at such a speed to catch up with the first one 180 km from point A. However, at the moment of departure, the first communications officer received an order to travel at a speed a km/h greater than originally planned. The second communications officer, however, was not allowed to increase the speed, so to complete the task accurately, he had to depart from point A 3 hours earlier than planned. How long will each communications officer be on the road?
Solution. Let $x$ and $y$ be the initial speeds of the first and second messengers, and $z$ be the time the first messenger was on the road. According to the initial plan, $\frac{180}{x}-\frac{180}{y}=6$. In reality, the speed became $x+a$, and the second messenger was on the road for $z-3$ hours.

$$
\begin{aligned}
& \text { Hence } z(x+a)=(z-3) y=180 \Rightarrow x=\frac{180-a z}{z}, y=\frac{180}{z-3} \Rightarrow \\
& \Rightarrow \frac{180 z}{180-a z}-(z-3)=6 \Rightarrow z=\frac{-3 a+3 \sqrt{a^{2}+240 a}}{2 a}
\end{aligned}
$$

and $z-3=\frac{-9 a+3 \sqrt{a^{2}+240 a}}{2 a}$ must be greater than 0, i.e.

$\sqrt{a^{2}+240 a}>3 a \Leftrightarrow a<30$.

Answer: $\frac{-3 a+3 \sqrt{a^{2}+240 a}}{2 a} \text{ hours}; \frac{-9 a+3 \sqrt{a^{2}+240 a}}{2 a} \text{ hours}$, where $a<30$.