Question

15. A river is flowing with the speed of 4 m/s.
A swimmer wants to reach just opposite point on the
other bank. If speed of the swimmer in still water is
3 m/s, then the swimmer should swim at angle
(1) 143° with the direction of river flow
(2) 127° with the direction of river flow
(3) 30° with normal of the bank
(4) Swimmer cannot reach just opposite point on
the other bank while swimming​

Answer 1

$\setlength{\unitlength}{1mm}</p><p>\begin{picture}(5,5)</p><p>\put(0,35){\line(1,0){50}}</p><p>\put(0,5){\line(1,0){50}}</p><p>\put(20,14){\vector(1,0){12}}</p><p>\put(20,14){\vector(0,1){9}}</p><p>\put(20,14){\vector(-4,3){12}}</p><p>\put(30,10){ \small{\vec{v_r}} }</p><p>\put(20,25){ \small{\vec{v_s}} }</p><p>\put(8,25){ \small{\vec{v_{sr}}} }</p><p>\put(17,17){ \small{\theta} }</p><p>\end{picture}$

Have a look at the figure. Here $\vec{v_r}$ is the velocity of the river, i.e., 4 m s^(-1), $\vec{v_s}$ is the velocity of the swimmer in still water, i.e., 3 m s^(-1) [if the water was in still then the swimmer could swim perpendicular to the river flow or the river bank, that's why the velocity], and $\vec{v_{sr}}=\vec{v_s}-\vec{v_r}$ is the velocity of the swimmer wrt the river.

Let θ be the angle made by the swimmer with the vertical.

The figure can be rearranged as the following.

$\setlength{\unitlength}{1mm}</p><p>\begin{picture}(5,5)</p><p>\put(0,35){\line(1,0){50}}</p><p>\put(0,5){\line(1,0){50}}</p><p>\put(8,23){\vector(1,0){12}}</p><p>\put(20,14){\vector(0,1){9}}</p><p>\put(20,14){\vector(-4,3){12}}</p><p>\put(13,25){ \small{\vec{v_r}} }</p><p>\put(22,18){ \small{\vec{v_s}} }</p><p>\put(9,16){ \small{\vec{v_{sr}}} }</p><p>\put(17,17){ \small{\theta} }</p><p>\end{picture}$

So,

$\theta=\tan^{-1}\left(\dfrac{|\vec{v_r}|}{|\vec{v_s}|}\right)\\\\\\\theta=\tan^{-1}\left (\dfrac{4}{3}\right)\\\\\\\theta=53^{\circ}$

This angle is made by the swimmer with the vertical, i.e., perpendicular to the flow of the river or the river bank. So the angle made by the swimmer with the river will be 53° + 90° = 143°.

Hence the answer is (1).