A man can swim with velocity v relative to wa | vedclass

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(C) Let the man swim at an angle $	heta$ with the downstream direction. The velocity components are $v_x = u + v \cos 	heta$ and $v_y = v \sin 	het

Vedclass · Accepted Answer

For $x$ to be minimum,he has to swim in a direction making an angle of $\frac{\pi}{2} + \sin^{-1}(\frac{v}{u})$ with the direction of the flow of water.

Vedclass · Answer

(C) Let the man swim at an angle $	heta$ with the downstream direction. The velocity components are $v_x = u + v \cos 	heta$ and $v_y = v \sin 	heta$.The time taken to cross the river is $t = \frac{d}{v \sin 	heta}$.The downstream drift is $x = v_x t = (u + v \cos 	heta) \frac{d}{v \sin 	heta} = \frac{d}{v} (u \csc 	heta + v \cot 	heta)$.For maximum time,$\sin 	heta$ must be minimum. Since $	heta$ is the angle with the downstream,the range is $0 To minimize $x$,we differentiate $x$ with respect to $	heta$ and set it to zero: $\frac{dx}{d	heta} = \frac{d}{v} (-u \csc 	heta \cot 	heta - v \csc^2 	heta) = 0$.This implies $u \cos 	heta + v = 0$,so $\cos 	heta = -v/u$.Since $	heta$ is the angle with the flow,the angle with the normal to the flow is $\alpha = 	heta - \pi/2$. Using $\cos 	heta = -v/u$,we find $\sin \alpha = v/u$,so $\alpha = \sin^{-1}(v/u)$.The angle with the flow is $	heta = \pi/2 + \sin^{-1}(v/u)$. Thus,option $(c)$ is correct.

$A$ man can swim with velocity $v$ relative to water. He has to cross a river of width $d$ flowing with a velocity $u$ $(u > v)$. The distance through which he is carried downstream by the river is $x$. Which of the following statements is correct?

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