A motor boat has a speed of `5 m//s`.At time `t=0`,its position vector relative to a origin is `(-11hati+16hatj) m`,having the aim of getting as close as possible to a steamer.At time `t=0`,the steamer is at the point `(4hati+36hatj) m` and is moving with constant velocity `(10hati-5hatj) m//s`.Find the direction in which the motorboat must steer

Options

A. `3hati+3hatj`
B. `3hati+4hatj`
C. `4hati+3hatj`
D. `4hati+4hatj`

Answer

C

Explanation

In order to approach as close as possible to steamer, the direction of motion of motorboat should be perpendicular to the relative motion.Let the optimum velocity of motor boat is
`vecupsilon_(B)=a hat i+ b hat j` Velocity of steamer is
`vec upsilon _(S)=10hati-5hatj`
`vec upsilon_(BS)=vec upsilon _(B)-vec upsilon _(S)=(a-10)hati+(b+5)hatj`
`vec upsilon _(B)_|_vec upsilon_(BS)` or `vec upsilon _(B).vec upsilon_(BS)=0`
or `a(a-10)+b(b+5)=0..(1)`
But we know that speed of the motorboat is `5m//s`,so `a^(2)+b^(2)=25..(2)`
Solving (1) and (2), we get `a=0` or `4`
when `a=0,b=-5` and `a=4,b=3` Hence either
`vec upsilon_(B)=-5hatj` or `vec upsilon_(B)=4hati+3hatj`
However a diagram shows that when `vec upsilon_(B)=-5hatj` the motorboat is moving further away from the steamer.So `vec upsilon_(B)=-4hati+3hatj`