The current velocity of river grows in proportion to the distance from its bank and reaches the maximum value `v_0` in the middle. Near the banks the velocity is zero. A boat is moving along the river in such a manner that the boatman rows his boat always perpendicular to the current. The speed of the boat in still water is u. Find the distance through which the boat crossing the river will be carried away by the current, if the width of the river is c. Also determine the trajectory of the boat.

Options

A. The distance carried by the boat is `X_(max)=(2cu)/(v_(0))`
B. The distance carried by the boat is `X_(max)=v_(0)/(2cu)`
C. The trajectary of the boat is `y^(2)=(v_(0)c)/ux`
D. The trajectary of the boat is `y^(2)=(v_(0)c)/ux`

Answer

B, D

Explanation

Given that `|vecv_(br)|=v_(y)=(dy)/(dt)=u`..(i)
`|vecv_(br)|=v_(x)=(dx)/(dt)=((2v_(0))/c)y`..(ii)
From eq.(i) and (ii) we have,
`(dy)/(dx)=(uc)/(2v_(0)y)y^(2)=(ucx)/(v_(0))`
At `y=c/2,x=(cv_(0))/(4u) or x_("net")=2x=(cv_(0))/(2u)`