Question

A man wants to swim across a river from A to B and back from B to A always following line AB. The distance S between points A and B is S. The velocity of the river current v is constant over the entire width of the river. The line AB makes an angle $\alpha$ with the direction of current. Let the man travel with velocity u and at angle $\beta$ to the line AB. Find time 't' by which he will be back.

• A. $t=\frac{S\left(2ucos\beta \right)}{v^{2}co{s}^2\beta -u^{2}co{s}^2\alpha }$
• B. $t=\frac{S\left(2usin\alpha \right)}{u^{2}si{n}^2\beta -v^{2}si{n}^2\alpha }$
• C. $t=\frac{S\left(20sin\alpha \right)}{v^{2}si{n}^2\beta -u^{2}si{n}^2\alpha }$
• D. $t=\frac{S\left(2ucos\beta \right)}{u^{2}co{s}^2\beta -v^{2}co{s}^2\alpha }$

Answer 1

The correct option is D

$t=\frac{S\left(2ucos\beta \right)}{u^{2}co{s}^2\beta -v^{2}co{s}^2\alpha }$

In this problem, we choose axis along AB and normal to it. Since the man moves along AB, the vertical velocity components of current and man must cancel out, i.e.,$usin\beta =vsin\alpha$. When the man moves from A to B , his resultant velocity along AB;=$(ucos\beta +vcos\alpha$)
Hence,S = $(ucos\beta +vcos\alpha$)$t_1$
While for motion from B to A S = (u cos$\beta$ - v cos $\alpha$)$t_2$

From the condition of the problem,$t_1+t_2=t$

$\frac{S}{ucos\beta +vcos\alpha }+\frac{S}{ucos\beta -vcos\alpha }$
= t;S$\left[\frac{ucos\beta -vcos\alpha +ucos\beta +vcos\alpha }{u^{2}co{s}^2\beta -v^{2}co{s}^2\alpha }\right]=t$
= $\frac{S\left(2ucos\beta \right)}{u^{2}co{s}^2\beta -v^{2}co{s}^2\alpha }=t$