The flow velocity of a river increases linearly with the distance (r) from its bank and has its maximum value $v_{0}$ in the middle of the river. The velocity near the bank is zero. A boat which can move with speed u in still water moves in the river in such a way that it is always perpendicular to the flow of current. Find

(i) The distance along the bank through which boat is carried away by the flow current, when the boat crosses the river.
(ii) The equation of trajectory for the coordinate system shown. Assume that the swimmer starts from origin.

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Solution

Since the velocity of the river is perpendicular to the velocity of the boat, it will not change the time by which boat will reach to the other bank. Now total time taken by the boat is $\frac{d}{u}$ , time taken by the boat to reach the middle of the river is $\frac{d}{2 u}$ . Now in this time the x-component of the velocity will change from 0 to $v_{0}$ hence the acceleration is $\frac{v_{0}}{\frac{d}{2 u}} = \frac{2 u v_{0}}{d}$ . No the x distance travelled by boat to reach the centre of the river is $x_{\frac{1}{2}} = \frac{1}{2} a_{x} t^{2} = \frac{1}{2} \frac{2 u v_{0}}{d} {(\frac{d}{2 u})}^{2} = \frac{v_{0} d}{4 u}$ . Hence the total distance travelled is $\frac{v_{0} d}{4 u}$ .
Now we have
$y = u t \Rightarrow t = \frac{y}{u}$
Now we have
$x = \frac{1}{2} a_{x} t^{2} = \frac{1}{2} \frac{2 u v_{0}}{d} {(\frac{y}{u})}^{2} = \frac{v_{0}}{u d} y^{2} f o r y \leq \frac{d}{2}$
x=v_0t -\frac{1}{2}a_x t^2 =\frac{v_0}{u}y- \frac{v_0}{ud} ~~for~y> \frac{d}{2}\)

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