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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 oat 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.

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$y^{2} = \frac{u c x}{v_{0}}$

$x_{n e t} = 2 x = \frac{c v_{0}}{2 u}$

$y^{2} = \frac{u c x}{v_{0}}$

$x_{n e t} = 2 x = \frac{c v_{0}}{4 u}$

$y^{2} = \frac{2 u c x}{v_{0}}$

$x_{n e t} = 2 x = \frac{c v_{0}}{2 u}$

$y^{2} = \frac{5 u c x}{v_{0}}$

$x_{n e t} = 2 x = \frac{c v_{0}}{2 u}$

answer is A.

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Detailed Solution

Given that $|v_{b r}| = v_{y} = \frac{d y}{d t} = u . . . . . . . (i)$

$| v_{r} | = v_{x} = \frac{d x}{d t} = (\frac{2 v_{0}}{c}) y . . . . . . (i i)$
From Eqs. (i) and (ii) we have,

$\frac{d y}{d x} = \frac{u c}{2 v_{0} y}$

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

or $\int_{0}^{y} y d y = \frac{u c}{2 v_{0}} \int_{0}^{x} d x$
or $y^{2} = \frac{u c x}{v_{0}}$

At $y = \frac{c}{2}, x = \frac{c v_{0}}{4 u}$
or $x_{n e t} = 2 x = \frac{c v_{0}}{2 u}$

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