Relative velocity ship problem intuitively

[anurag07]

Homework Statement

Two ships A and B are 4 km apart. A is due west of B. If A moves with uniform velocity of 8 km/hr due east and B moves with a uniform velocity of 6 km/hr due south. Calculate 1) the magnitudeof the velocity of A in relation to B 2) the closest distance apart of A and B

Homework Equations

The first solution is pretty simple but the second one, I don't get it. I know how to just solve it but without any intuition. Can someone help me get an intuitive idea of the solution of 2) ? The solution will be down there.

The Attempt at a Solution

In the example BD is drawn from B to relative velocity of A with respect to B represented by AE. From the given distance between A and B is 4km i.e. AB = 4km. In ABE sin x = BE/AE = 6/10 ( x= Angle EAB, BE = Velocity of B in opposite direction, AE = relative velocity of A in respect to B )
x=36.67
Closest distance apart from A to B is BD.
Now in BDA,
BD = sin x * AB = sin 36.87 * 4 = 2.4 km
This is the closest distance.

 

[phinds]

Posting tiny sideways pics is not a good way to get people to look at them.

 

[anurag07]

Posting tiny sideways pics is not a good way to get people to look at them.

Dont have a good phone. Sorry. And vertical pictures are even tinier.

 

[haruspex]

The idea is to use B's reference frame. In that frame, what are the NS and EW components of A's motion. At what point does B perceive A as being at closest approach?