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+3haj`
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 shou ld be perpendicu lar to the relative motion. Let the optimum velocity of motorboat is
`vecv_(B)=ahati+bhatj` Velocity of steamer is
`vecv_(S)=10hati-5hatj`
`vecv_(BS)=vecv_(S)=(a-10)hati+(b+5)hatj`
Now, `vecv_(B)botvecv_(BS)` or `vecv_(B).vecv_(BS)=0`
or `a(a-10)+b(b+5)=0` ...........(i)
But we know that speed of the motorboat is 5m/s. so
`a^(2)+b^(2)=25` ..............(ii)
Solving Eqns. (i) and (ii), we get a=0 or 4
When a=0,b=-5 and a=4,b=3.
Hence either `vecv_(B)=-5hatj` or `vecv_(B)=4hati+3hatj`.
However a diagram shows that when `vecv_(B)=-5hatj`. the motorboat is moving futher aways form the steamer . So,
`vecv_(B)=4hati+3hatj`.
(ii) Now at time `t vecr_(B)=-(1hati+16hatj)+(4hati+3hatj)t`
and `vecr_(S)=(4hati+36hatj)+(10hati-5hatj)t`
If the motorboat can reach the steamer, there is a value of t for which `vecr_(B)=vecr_(S)`
equating the coefficient of `hati`, we get t=-5/2 negative time implies that the motorboat cannot ever react the steamer.