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 swimmer starts swimming due north, what will be his resultant velocity (magnitude and direction) ?

$(b)$ If he wants to start from point A on south bank and reach opposite point $B$ on north bank,

       $(i)$ Which direction should he swim ?
       $(ii)$ What will be his resultant speed ?

$(c)$ From two different cases as mentioned in $(a)$ and $(b)$ above, in which case will he reach opposite bank in shorter time ?

Given, Speed of the river $\left(\mathrm{V}_{r}\right)=3 \mathrm{~m} / \mathrm{s}$ (east)

Speed of swimmer $\left(\mathrm{V}_{s}\right)=4 \mathrm{~m} / \mathrm{s}$ (east)

$(a)$ When swimmer starts swimming due north, then his resultant velocity,

$\mathrm{V}=\sqrt{\mathrm{V}_{\mathrm{r}}^{2}+\mathrm{V}_{s}^{2}}=\sqrt{(3)^{2}+(4)^{2}}$

$=\sqrt{9+16}=\sqrt{25}=5 \mathrm{~m} / \mathrm{s}$

$\tan \theta =\frac{\mathrm{V}_{r}}{\mathrm{~V}_{s}}=\frac{3}{4}$

$\tan \theta =0.75$

$\theta &=\tan ^{-1}(0.75)=36^{\circ} 54^{\prime}$

$(b)$ To reach opposite points $B$, the swimmer should swim at an angle $\theta$ of north. Resultant speed of the swimmer,

$\mathrm{V}=\sqrt{\mathrm{V}_{s}^{2}-\mathrm{V}_{r}^{2}}=\sqrt{(4)^{2}-(3)^{2}}$

$=\sqrt{16-9}=\sqrt{7} \mathrm{~m} / \mathrm{s}$

$\tan \theta=\frac{\mathrm{V}_{r}}{\mathrm{~V}}=\frac{3}{\sqrt{7}}$

$\Rightarrow \theta=\tan ^{-1}\left(\frac{3}{\sqrt{7}}\right)$ of north