Ship A is 10 km due west of ship B. Ship A is heading directly north at a speed of 30 km/h. While ship B is heading in a direction `60^(@)` west of north at a speed of 20 km/h.
(i) Determine the magnitude of the velocity of ship B relative to ship A.
(ii) What will be their distance of closest approach ?

Explanation

(a) `V_(A) = 30` kmph, `V_(B) = 20` kmph, `theta = 60^(@)`
Relative velocity of ship B w.r.to ship A is
`|V_(B) - V_(A)| = sqrt(V_(A)^(2) + V_(B)^(2) - 2V_(A)V_(B) cos theta)`
`= sqrt(30^(2) + 20^(2) - 2 xx 30 xx 20 xx cos 60^(@))`
`= sqrt(900 + 400 - 2 xx 600 xx (1)/(2)) = sqrt(700)`
`= 10sqrt(7)` kmph.
(b) The distance between the ship A and ship B = 10 km.
Ship A moves along north relative to ship B. The disfance of closest approach is `BD = AB sin 45^(@)`
`= 10 xx (1)/(sqrt(2)) = (10)/(sqrt(2))` km.