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?

Explanation

Given that `v_(br)`=4km/h and `v_(r)=2km//h`
`theta=sin^(-1)((v_(r))/(v_(br)))=sin^(-1)((2)/(4))`
`sin^(-1)((1)/(2))=30^(@)`

Hence, to reach the point directly oppostie to starting point he shou ld heat the boat at an angle of `30^(@)` with AB or `90^(@)+30^(@)-120^(@)` withe the river flow.
(b) Time taken by the boatman to cross the river w=width of river =4km
`v_(br)=4km//hr`
and `theta=30^(@)`
`therefore t=(r)/(4cos 30^(@))=(2)/sqrt(3)h`
(c) For shortest time `theta=0^(@)`
and `t_("min")=(w)/(v_(br)cos 0^(@))=(4)/(4)=1h`
Hence, he shou ld his boat perpendicu lar to the river current for crossing the river in shortest time and this shortest time is 1h.
(d) `t=t_(CD)+t_(DC)`
or `t=(CD)/(v_(br)-v_(r))+(DC)/(v_(br)+v_(r))`
`=(2)/(4-2)+(@)/(4+2)=1+(1)/(3)=(4)/(3)h`