Question

A man can swim with a 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 are correct?

• A. If he crosses the river in minimum time, $x=\frac{du}{v}$
• B. $x$ cannot be less than $\frac{du}{v}$
• C. For $x$ to be minimum, he has to swim in a direction making an angle of $\frac{\pi }{2}+{\sin }^{–1}\left(\frac{v}{u}\right)$ with the direction of the flow of water.
• D. $x$ will be maximum if he swim in a direction making an angle of $\frac{\pi }{2}-{\sin }^{–1}\left(\frac{v}{u}\right)$ with the direction of the flow of water.

Answer 1

Answer: (A), (C)

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

(A) For minimum time the swimmer will have to start moving perpendicular to the river flow. (vertical component of the swimmer's speed with respect to ground should be $v$)
Then the vertical component of velocity will be $v$ and distance to be covered will be $d$. Hence time to cross the river = $\frac{d}{v}$
The x component of the swimmer's speed in ground frame will u.
Hence drift = $u\times \frac{d}{v}$
The time required for crossing $=t=\left(d/v\right)\cos \theta$ ... (i)
$x=(u-v\sin \theta )t=ud/v\sec \theta -d\tan \theta$ (from equation (i))
For $x$ to be minimum, $\frac{dx}{d\theta }=0$
This gives $\sin \theta =\frac{v}{u}$