Ship A is 10km due west of ship B. Ship A is heading directly north at a speed of 30kmph while ship B is heading in a direction 600 west of north at a speed 20kmph. Their closest distance of approach will be …….

Explanation

`bar(V_(A)) = 30jbar(V_(B))=20Sin60^(0)+20Cos60(hat(j))`
`=-10sqrt3hat(i)+10hat(j)`
`bar(V_(BA))=bar(V_(B))-bar(V_(A)) = -20hat(j)-10sqrt(3)hat(i)`
If `phi` is the angle made by `bar(V_(BA))` with X axis
`Tanphi=(20)/(10sqrt(3))=(2)/(sqrt(3))`
and `Sinphi=(2)/(sqrt(7))`
From `DeltaABC, ("x")/(10)=(2)/(sqrt(7))`
` "x"=(20)/(sqrt(7))=7.56Km`