Question

A man wants to reach point B on the opposite bank of a river flowing at a speed as shown in figure. What minimum speed relative to water should the man have so that he can reach point B? In which direction should he swim?

Answer 1

A man has to reach point B on the opposite bank of a river at minimum speed

Let the speed of man swimming in water is $v$ m/sec

the speed of water flow is $u$ m/sec.

Relative speed of man with repect to the water= $( u-vcos\theta )$

$tan45$° = $\frac{vsin\theta}{u-vcos\theta}$

$\frac{vsin\theta}{u-vcos\theta}$ = 1

${vsin\theta}={u-vcos\theta}$

$v=\frac{u}{sin\theta+cos\theta}$

$v=\frac{u}{\sqrt{2} (sin\theta+45)}$

$For v should be minimum$

$sin(\theta+ 45 )$ be maximum i.e

$\\sin(\theta+ 45 )=90\\\therefore \theta=45$

$\therefore v_m_i_n=\frac{u}{\sqrt{2}(sin90) }$

$\therefore v_{min}=\frac{u}{\sqrt{2}}$

$The minimum speed relative to water should the man have so that he can reach point B is \frac{u}{\sqrt{2} } m/sec .$

$Man should swim in the opposite direction of flow of water at 45$°

Answer 2

Answer:

A man has to reach point B on the opposite bank of a river at minimum speed

Let the speed of man swimming in water is vv m/sec

the speed of water flow is uu m/sec.

Relative speed of man with repect to the water= ( u-vcos\theta )(u−vcosθ)

tan45tan45 ° = \frac{vsin\theta}{u-vcos\theta}

u−vcosθ

vsinθ

\frac{vsin\theta}{u-vcos\theta}

u−vcosθ

vsinθ

= 1

{vsin\theta}={u-vcos\theta}vsinθ=u−vcosθ

v=\frac{u}{sin\theta+cos\theta}v=

sinθ+cosθ

u

v=\frac{u}{\sqrt{2} (sin\theta+45)}v=

2

(sinθ+45)

u

$

sin(\theta+ 45 )sin(θ+45) be maximum i.e

\begin{lgathered}\\sin(\theta+ 45 )=90\\\therefore \theta=45\end{lgathered}

sin(θ+45)=90

∴θ=45

\therefore v_{min}=\frac{u}{\sqrt{2}}∴v

min

=

2

u

The minimum speed relative to water should the man have so that he can reach point B is $\frac{u}{\sqrt{2} } m/sec .

Man should swim in the opposite direction of flow of water at 45 °