A ship `A` streams due north at `16 km h^-1` and a ship `B` due west at `12 km h^-1`. At a certain instant `B` is `10 km` north east of `A`. Find the
(a) magnitude of velocity of `A relative to B`.
(b) nearest distance of approach of ships.

Answer

(a) `20 km h^-1` ;
(b) `sqrt(2) km`

Explanation

`vec v_(A//B) =12 i + 16 hat j, tan prop`
=`(16)/(12) = (4)/(3)`
(a) `vec v_(A//B) = sqrt((12)^2 + (16)^2)`
=`20 km h^-1`
(b) `DB = EB sin prop = (OB - OE) sin prop`
=`(AB cos 45^@ - (OA)/(tan prop)) sin prop`
=`[(10)/(sqrt(2)) -AB cos 45^@ xx (3)/(4)] xx (4)/(5)`
=`(10)/(sqrt(2))[(1)/(4)] xx (4)/(5) =sqrt(2) km`.
.