Question

A man who can swim at a velocity $v$ relative to water wants to cross a river of width $b$, flowing with a speed $u$.

• A. The minimum time in which he can cross the river is $\frac{b}{v}$
• B. He can reach a point exactly opposite on the bank in time $t=\frac{b}{\sqrt{v^2-u^2}}$ if $v>u$
• C. He cannot reach the point exactly opposite on the bank if $u>v$
• D. He cannot reach the point exactly opposite on the bani if $v>u$

Answer 1

Answer: (A), (B), (C)

The correct options are
A The minimum time in which he can cross the river is $\frac{b}{v}$
B He can reach a point exactly opposite on the bank in time $t=\frac{b}{\sqrt{v^2-u^2}}$ if $v>u$
C He cannot reach the point exactly opposite on the bank if $u>v$

Here, the man swims at a velocity $v$ and $\theta$ is the angle made by his motion with the direction of flow of water.
When we resolve velocity $v$ into components $vsin\theta$ and $vcos\theta$, for minimum time $vsin\theta$ has to be maximum or $\theta ={90}^o$.
$\therefore t_{min}=\frac{b}{v}$
For reaching a point exactly opposite
Net velocity $=\sqrt{v^2-u^2}$ (but $v>u$)
$\therefore t=\frac{b}{\text{net velocity}}$.

Options A, B and C are correct.