Question

A river is flowing with a speed of 1 km/hr. A swimmer wants to go to point 'C' starting from 'A'. He swims with a speed of 5 km/hr at an angle $\theta$ w.r.t the river flow. If AB = BC = 400 m. At what angle with river bank should swimmer swim?

• A. $\theta ={37}^{\circ }$
• B. $\theta ={53}^{\circ }$
• C. $\theta ={45}^{\circ }$
• D. $\theta ={60}^{\circ }$

Answer 1

The correct option is B

$\theta ={53}^{\circ }$

$v_R=1\hat{i}$

$v_{MR}=5\cos \theta \hat{i}+5\sin \theta \hat{j}$

We know that velocity of man w.r.t. ground should point in the direction ${45}^{\circ }$ to the bank as AB = BC

$v_{MR}=\frac{v}{\sqrt{2}}\hat{i}+\frac{v}{\sqrt{2}}\hat{j}$

$v_{MR}=v_M-v_R$

$5cos\theta \hat{i}+5sin\theta \hat{j}=\frac{v}{\sqrt{2}}\hat{i}+\frac{v}{\sqrt{2}}\hat{j}-1\hat{i}$

$5cos\theta \hat{i}+5sin\theta \hat{j}=(\frac{v}{\sqrt{2}}-1)\hat{i}+\frac{v}{\sqrt{2}}\hat{j}$

$\Rightarrow 5cos\theta =\frac{v}{\sqrt{2}}-1$ --------------(i)

$5sin\theta =\frac{v}{\sqrt{2}}$ ----------------(ii)

Substituting (ii) in (i)

$5cos\theta =5sin\theta -1$

On squaring both sides we get

$25co{s}^2\theta =25si{n}^2\theta +1-10sin\theta$

$25-25si{n}^2\theta =25si{n}^2\theta +1-10sin\theta$

$(\because co{s}^2\theta =1-si{n}^2\theta )$

$50si{n}^2\theta -10sin\theta -24=0$

$sin\theta =\frac{10\pm \sqrt{100+4800}}{100}$

= $\frac{10\pm \sqrt{4900}}{100}$

$sin\theta =\frac{4}{5}$

$\theta ={53}^{\circ }$