88. The Chase (I). Ship $P$ has spotted ship $Q$, which is sailing in a direction perpendicular to $P Q$, maintaining its course. Ship $P$ is chasing $Q$, always heading directly towards $Q$; the speed of both ships is the same at any moment (but can vary over time). Without calculations, it is clear that $P$ is sailing along a curved path; if the chase lasts long enough, the trajectory of the pursuing ship and the trajectory of the fleeing ship will eventually become almost identical. What will then be the distance $P Q$, if initially it was 10 nautical miles?
88. Let $\alpha$ denote the instantaneous angle between the direction $P Q$ and the path of ship $Q$ (Fig. 163), and $v$ - the speed of ships $P$ and $Q$ at that moment. The mutual approach of the ships is influenced by the speed $v$ of ship $P$, directed towards $Q$, and the component $v \cos \alpha$ of the speed of ship $Q$, both in the same direction. Therefore, the speed of approach of the ships is $v(1-\cos \alpha)$.

The projection $S$ of point $P$ on the path of ship $Q$ moves along this path with a speed of $v \cos \alpha$, while ship $Q$ moves away with a speed of $v$, so the distance $S Q$ increases at a speed of $v(1-\cos \alpha)$. Since the distance $P Q$ decreases, as we noted earlier, at the same speed, the sum $P Q + S Q$ is constant, and therefore equals 10 miles, as it did at the initial moment.

After an infinitely long

![](https://cdn.mathpix.com/cropped/2024_05_21_5d77f4829f88503a49eag-161.jpg?height=542&width=742&top_left_y=457&top_left_x=360)

Fig. 163. time, $P$ will coincide with $S$ and will be equal to $P Q + S Q = 2 P Q = 10$ miles, and $P Q = 5$ miles.