3$\cdot$34 A person is sailing a sailboat downstream along a river and encounters a speedboat coming upstream. He asks the speedboat driver: "Is there a ship coming behind you?" The speedboat driver replies: "I passed a ship half an hour ago." The sailboat continues to sail for another half hour and meets a ship coming upstream. It is known that the ship's speed in still water is twice the sailboat's speed in still water. Therefore, the speedboat's speed in still water is
(A) 4 times.
(B) 5 times.
(C) 6 times.
(D) Cannot be determined due to the unknown current speed.
("Zu Chongzhi Cup" Junior High School Mathematics Invitational, 1991)
[Solution] Let the still water speeds of the motorboat, the steamer, and the sailboat be $v, v_{1}, v_{2}$, respectively. According to the problem, $v_{1} = 2 v_{2}$. Let the speed of the current be $u$.
The distance between the motorboat and the steamer in the half hour from when the motorboat meets the steamer to when it meets the sailboat is
$$
\frac{1}{2}(v-u)-\frac{1}{2}\left(v_{1}-u\right) .
$$

This is exactly the distance traveled by the sailboat and the steamer in the half hour after they meet, so we have
$$
\frac{1}{2}(v-u)-\frac{1}{2}\left(v_{1}-u\right)=\frac{1}{2}\left(v_{2}+u\right)+\frac{1}{2}\left(v_{1}-u\right) \text {, }
$$

and given $v_{1}=2 v_{2}$, we get $v=5 v_{2}$.
Therefore, the answer is $(B)$.