2. An airplane flew from $A$ to $B$, initially at a speed of $180 \mathrm{~km} / \mathrm{h}$, and when it had $320 \mathrm{~km}$ less to fly than it had already flown, it increased its speed to $250 \mathrm{~km} / \mathrm{h}$. In this way, the average speed of the airplane for the entire journey from $A$ to $B$ was $200 \mathrm{~km} / \mathrm{h}$. Determine the length of the path from $A$ to $B$.
Solution. The time to travel the path from $A$ to $D$ is $\frac{x+320}{180}$, and the time to travel the path from

![](https://cdn.mathpix.com/cropped/2024_06_05_6a3de2303cb9ec4e16a2g-1.jpg?height=135&width=600&top_left_y=1420&top_left_x=880)
$D$ to $B$ is $\frac{x}{250}$. Therefore, we have $2 x+320=200\left(\frac{x+320}{180}+\frac{x}{250}\right)$, from which we get $x=400 \mathrm{~km}$. According to this, the length of the path from $A$ to $B$ is $2 x+320=1120 \mathrm{~km}$.