A river of width `omega` is flowing with a uniform velocity v. A boat starts moving from point P also With velocity v relative to the river. The direction of resultant velocity is always perpendicular to the line joining boat and the fixed point R. Point Q is on the opposite side of the river. P, Q and R are in a straight line. If `PQ=QR=omega,` find (a) the trajectory of the boat, (b) the drifting of the boat and (c) the time taken by the boat to cross the river.

Answer

A, B, C

Explanation

(a) Since, the resultant velocity is always
perpendicular to the line joining boat and R, the boat
is moving in a circle of radius `2 omega` and centre at R.

(b) Drifting =`QS=sqrt(4omega^2-omega^2)=sqrt3 omega`.
(c) Suppose at any arbitrary time, the boat is at
point B.

`V_("net")=2V cos theta`
`(d theta)/(dt)=(V_("net"))/(2omega)=(v cos theta)/omega`
or `omega/vsec theta d theta=dt`
`:. int_0^t dt=omega/v int_0^(60^@) sec theta d theta`
`:. omega/v[in(sec theta+tan theta]_0^(60^@)`
or `t=(1.317 omega)/v`