$A$ river is flowing due east with a speed $3\, ms^{-1}$. $A$ swimmer can swim in still water at a speed of $4\, ms^{-1}$ (figure).
$(a)$ If the swimmer starts swimming due north,what will be his resultant velocity (magnitude and direction)?
$(b)$ If he wants to start from point $A$ on the south bank and reach the opposite point $B$ on the north bank,
$(i)$ In which direction should he swim?
$(ii)$ What will be his resultant speed?
$(c)$ From the two different cases as mentioned in $(a)$ and $(b)$ above,in which case will he reach the opposite bank in a shorter time?

(N/A) Given: Speed of the river $V_r = 3\, ms^{-1}$ (east),Speed of the swimmer in still water $V_s = 4\, ms^{-1}$.
$(a)$ When the swimmer swims due north,the resultant velocity $V$ is the vector sum of $V_r$ and $V_s$.
$V = \sqrt{V_r^2 + V_s^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5\, ms^{-1}$.
The direction $\theta$ with respect to the north is given by $\tan \theta = \frac{V_r}{V_s} = \frac{3}{4} = 0.75$.
$\theta = \tan^{-1}(0.75) \approx 36.87^{\circ}$ east of north.
$(b)$ To reach point $B$ directly opposite to $A$,the swimmer must swim at an angle $\theta$ upstream (west of north) such that the horizontal component of his velocity cancels the river's velocity.
$(i)$ $\sin \theta = \frac{V_r}{V_s} = \frac{3}{4} = 0.75 \Rightarrow \theta = \sin^{-1}(0.75) \approx 48.6^{\circ}$ west of north.
$(ii)$ The resultant speed $V$ is the vertical component: $V = \sqrt{V_s^2 - V_r^2} = \sqrt{4^2 - 3^2} = \sqrt{16 - 9} = \sqrt{7} \approx 2.65\, ms^{-1}$.
$(c)$ Let $d$ be the width of the river. Time taken in case $(a)$ is $t_a = \frac{d}{V_s} = \frac{d}{4}$. Time taken in case $(b)$ is $t_b = \frac{d}{V} = \frac{d}{\sqrt{7}}$. Since $\sqrt{7} < 4$,$t_b > t_a$. Thus,the swimmer reaches the opposite bank in a shorter time in case $(a)$.