13.259. A steamship, 2 hours after departure from pier $A$, stops for 1 hour and then continues its journey at a speed equal to 0.8 of its initial speed, as a result of which it is late to pier $B$ by 3.5 hours. If the stop had occurred 180 km further, then under the same conditions, the steamship would have been late to $B$ by 1.5 hours. Find the distance $A B$.
Solution.

Let $AB = x$ km, $v$ km/h be the speed of the steamboat.

According to the problem, $\left\{\begin{array}{l}\frac{x}{v} + 3.5 = 2 + 1 + \frac{x - 2v}{0.8v}, \\ \frac{x}{v} + 1.5 = \frac{2v + 180}{v} + 1 + \frac{x - 2v - 180}{0.8v},\end{array}\right.$ from which $x = 270$ km.

Answer: 270 km.