Question

The velocity of current in 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 $u$. If the distance through which the boat crossing the river is carried away by the current, is $\frac{bc{v}_0}{2u}$, where $c$ is the width of the river. Find $b$.

Answer 1

Since we are given that the $v_0$ is the river velocity at the middle which is maximum and at the banks it is zero.
Given that $|\overrightarrow{v_{br}}|=v_y=\frac{dy}{dt}=u$ .....(i)
$|{\overrightarrow{v}}_r|=v_x=\frac{dx}{dt}=\left(\frac{2v_0}{c}\right)y$ .....(ii)
From Eqs. (i) and (ii) we have, $\frac{dy}{dx}=\frac{uc}{2v_{0}y}$
or ${\int }_0^{y}ydy=\frac{uc}{2v_0}{\int }_0^{x}dx$ or $y^2=\frac{ucx}{v_0}$

At $y=\frac{c}{2},x=\frac{c{v}_0}{4u}$ or $x_{net}=2x=\frac{c{v}_0}{2u}$

$b=1$