As shown in the picture, the tourist ship is located at point A, heading 60? The sailing speed is 200m/min; Dolphin is located at point B, swimming due east, swimming speed 100 √ 3m/min. AB is 2000 meters apart.

If the "dolphin" is taken as the reference object (thinking that it is still), the dolphin looks at the movement of the ship:

According to the velocity vector diagram, V-boat is relative to dolphin =V-boat -V-dolphin, and the red solid arrow in the diagram is the synthetic motion direction (including velocity) of the ship relative to dolphin.

That is to say, in the eyes of dolphins, a tourist ship at point A, 2000m away, will start from point A and come to it with a red dashed trajectory parallel to the red "V relative speed" vector in the figure (the coming speed is "V relative", that is, the length of the red arrow vector in the speed vector triangle in the figure).

According to the relative trajectory diagram, the nearest distance between the fleet and the dolphins is S (perpendicular to the red relative trajectory from point B):

The detailed solution process of is as follows.

S=AB×sin∠4

( 1)ab = 2000m；

(2)

Because the relative vector of V is parallel to the relative trajectory of red, ∠4=∠5(∠5=∠4).

Because 1 = 60? , then ∠2+∠5=30? (∠2=30? -∠5, then ∠2=30? -∠4)。

In the velocity vector triangle, we know from the sine theorem:

Will ∠5=∠4 and ∠2=30? -∠4 Substitute into sine theorem and find sin∠4.

To sum up, we can get the s value.