A man can row a boat with 4km/h in still water, if he is crossing a river where the current is 2 km/h.
(a) In what direction will his boat be holded, if he wants to reach a point on the other bank, directly opposite to starting point?
(b) If width of the river 4km, how long will the man take to cross the river, with the condition in part (a)?
(c) In what direction shou ld he heat the boat if he wants to cross the river in shorest time and what is this minimum time?
(d) How long will it take him to row 2 km up the stream and then back to his starting point?

Options

A. `120` , `(2)/(sqrt(3))h` , ` 1 hr , 90` , `(4/3) h`
B. `120` , `(2)/(sqrt(5))h` , ` 1 hr , 90` , `(4/3) h`
C. `120` , `(2)/(sqrt(3))h` , ` 1 hr , 90` , `(5/3) h`
D. `150` , `(2)/(sqrt(3))h` , ` 1 hr , 90` , `(4/3) h`

Answer

A

Explanation

(a) Given, that `v_(br)=4 km//h` and `v_(r )=2 km//h`
`therefore theta = sint^(-1)((v_(r ))/(v_(br)))=sin^(-1)((2)/(4))=sin6(-1)((1)/(2))=30^(@)`
Hence, to reach the point directly aopposite to starting point he should head the boat at an angle of `30^(@)` with AB or `90^(@)+30^(@)=120^(@)` with the river flow.
(b) Time taken by the boatman to cross the river
w = width of river = 4 km
`v_(br)=4 km//h`
and `theta = 30^(@)`
`thereofre t = (4)/(4 cos 30^(@))=(2)/(sqrt(3))h`.
(c ) For shortest time `theta = 0^(@)`
and `t_(min)=(w)/(v_(br)cos 0^(@))=(4)/(4)=1h`
Hence, he should head his boat perpendicular to the river current for crossing the river in shortest time and this shortest time is 1 h.
(d) `t=t_(CD)+t_(DC)`

or `t=(CD)/(v_(br)-v_(r ))+(DC)/(v_(br)+v_(r ))=(2)/(4-2)+(2)/(4+2)=1+(1)/(3)=(4)/(3)h`