Ship A is 10 km due west of ship B. Ship A heading directly north at a speed of 30 kmph while ship B is heading in a direction `60^(@)` west of north at a speed 20 kmph. Their closest distance of approach will be…….

Explanation

`bar(V_(A)) = 30 j bar(V_(B)) = 20 Sin 60^(@)(hat(i)) + 20 Cos 60 (hat(j))`
`= -10 sqrt(3)hat(i) + 10 hat(j)`
`bar(V_(BA)) = bar(V_(B)) - bar(V_(A)) = -20 hat(j) - 10 sqrt(3)hat(i)` `Tan phi = (20)/(10sqrt(3)) = (2)/(sqrt(3))`
and `Sin phi = (2)/(sqrt(7))`
From `Delta ABC , (x)/(10) = (2)/(sqrt(7))`
`x = (20)/(sqrt(7)) = 7.56 Km`