A motor boat set out at 11a.m. from a position `-6hati-2hatj` relative to a marker buoy and travel at a steady spee dof magnutdue `sqrt(53)` on a direc course to intercept a ship. The ship maintains a steady velocity vector `3hati+4hatj` and at 12 noon is at a postion `3hati-hatj` form the body. Find (a) the velocity vector of the motor boat, (b) the time of interecption and (c) the position vector of point of interception from the buoy if distances are measured in mkilometeres and speds i kilometre per hour.

Answer

(a) `7vec1+2vecj`; (b) 12.30 p.m.; (c) `(9)/(2)veci+vecj`

Explanation

Position vector of ship after 't' hours from 12 noon is,
`vecr_(s)=vecr_(as)+vecv_(s)t=(3hati-hatj)+(3hati+4hatj)t` .....(i)
Position vector of motor boat at that instant
`vecr_(B)=(-6hati-2hatj)+(ahati+bhatj)(1+t)` ............(ii)
where `ahati+bhatj=vecv_(B)` (velocity of motor boat)
Given `a^(2)+b^(2)=53` ............(iii)
when they intercept `vecr_(s)=vecr_(B)`
Equate coefficients of `hati` and `hatj` and use equation (iii)
we get b=2 and `(34)/(5)`: with `b=(34)/(5)` time 't' comes out to be negative which is not possible so b=2, from this a=7 and `t=(1)/(2)`.
Position vector of point of interception is `(a)/(2)hati+hatj`