Question

If the ratio of swimmer velocity to the river velocity is ‘2’, then the angle at which the swimmer swims
with the water flow to cross the river in minimum time is

Answer 1

Answer:

Correct option is

A

sinθ

The correct option is A.

Letdistance=d

The angle made with the upstream=θ

Therefore,

90−θ=Angle made with the normal stream

Hence,

Shortest Time Method:

Time taken, t=

v

d

Shortest distance method:

Time taken = t'=

vcos(90−θ)

d

=

vsinθ

d

Therefore, required ratio

=

t

′

t

=

v

d

sinθ

v

d

=sinθ

Answer 2

Answer:

The angle at which the swimmer swims in minimum time is 60°.

Explanation:

Given the ratio of swimmer velocity (${v_{s}$) to the river velocity(${v_{s}$), $\frac{v_{s} }{v_{r} } =2$

The swimmer take the minimum time by covering the shortest distance, when the velocity of swimmer with respect to the river velocity is perpendicular to the upstream.

The width of the river is the shortest distance, and let it be d

Angle made by swimmer velocity with the upstream = $\theta$

Relative velocity of the swimmer with respect to river velocity$v_{s/r}=v$

Minimum time taken to travel a distance d with relative velocity $v_{s/r}$ is $t_{min} =\frac{d}{v}$

Time taken to travel the minimum distance is $t=\frac{d}{v sin\theta}$

Taking the ratio of both the times, $\frac{t_{min}}{t} =\frac{ \frac{d}{v}}{\frac{d}{vsin\theta}}$

$\implies \frac{d}{v}}\times {\frac{v sin\theta}{d}}={\frac{v sin\theta}{v}}=sin\theta$

And from the diagram,  $sin\theta=\frac{v_{r}}{v_{s}} =\frac{1}{2}$  

Therefore, $\theta=60^{o}$

Therefore, the angle at which the swimmer swims in minimum time is 60°.