Question

A man who can swim at a speed $v$ relative to the water wants to cross a river of width $d$, flowing with a speed $u$. The point opposite him across the river is A. Consider the following statements:
(a) He can reach the point A in time $\frac{d}{v}$
(b) He can reach the point A in time $\frac{d}{\sqrt{v^2-u^2}}$
(c)The minimum time in which he can cross the river is $\frac{d}{v}$
(d) He cannot reach at A if $u>v$

• A. a, b & c are correct
• B. b, c & d are correct
• C. c, d & a are correct
• D. only b & a are correct

Answer 1

The correct option is B b, c & d are correct

if man starts swimming perpendicular to the flow aiming towards point A then he will reach a point C, $u.\frac{d}{v}$ meter ahead of point A. Because component of velocity of flow will always be in the direction towards the flow.

but maximum component perpendicular to the flow direction is his own velocity. so minimum time to cross the river is $\frac{d}{v}$.

if he wants to reach point A then he must swim at some angle with perpendicular to the flow so that component parallel to flow cancel out the velocity of flow. and component perpendicular to the flow helps him crossing the river.

so if he swims at angle $(90+\theta )$ with the flow then $vSin\theta =u$

$Sin\theta =\frac{u}{v}$ then $Cos\theta =\frac{\sqrt{v^2-u^2}}{v}$

So component perpendicular to the flow is $vCos\theta$.

time taken is $\frac{d}{\frac{v.\sqrt{v^2-u^2}}{v}}$ $=>T=\frac{d}{\sqrt{v^2-u^2}}$.

so statement a is wrong and b,c,d are correct.