Question

A river is flowing due east with a speed $3\frac{m}{s}$. A swimmer can swim in still water at a speed of $4\frac{m}{s}$ (Figure).

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

(2) If he wants to start from point $A$ on south bank and reach opposite point $B$ on north bank,
(a) which direction should he swim?
(b) what will be his resultant speed?

(3) $\left(i\right)$ If swimmer starts swimming due north, what will be his resultant velocity (magnitude and direction)?
$\left(ii\right)$ If he wants to start from point $A$ on south reach opposite point $B$ on north bank, In which case will he reach opposite bank in time?

Answer 1

(1) Formula used:
${\overrightarrow{v}}_m={\overrightarrow{v}}_{mR}+{\overrightarrow{v}}_R$

Given,
Velocity of river ${\overrightarrow{v}}_R=3\frac{m}{s}\hat{i}$
Velocity of man in still water ${\overrightarrow{v}}_{mR}=4\frac{m}{s}$
If swimmer starts swimming due north in still water, ${\overrightarrow{v}}_{mr}=4\frac{m}{s}\hat{j}$


${\overrightarrow{v}}_m={\overrightarrow{v}}_{mR}+{\overrightarrow{v}}_R$
${\overrightarrow{v}}_m=3\hat{i}+4\hat{j}$
$v_m=\sqrt{3^2+4^2}=5\frac{m}{s}$
Direction of swimmer:
$\tan \theta =\frac{3}{4}$
$\theta ={37}^{\circ }$ of north

Final Answer: $5\frac{m}{s}$ at ${37}^{\circ }$ to $N$

(2)(a) Given,
Velocity of river ${\overrightarrow{v}}_R=3\frac{m}{s}\hat{i}$
Velocity of man in still water, ${\overrightarrow{v}}_{mR}=4\frac{m}{s}$
If he wants to reach on the point directly opposite to him (points $B$), the swimmer should swim at an angle $\theta$ of north.
Direction of ${\overrightarrow{v}}_m$ is along north direction.


From diagram
${\overrightarrow{v}}_m={\overrightarrow{v}}_{mR}+{\overrightarrow{v}}_R$
$v_m=\sqrt{4^2-3^2}=\sqrt{7}\frac{m}{s}$


(2)(b) Direction in which swimmer should start swimming
$\tan \theta =\frac{v_m}{v_R}=\frac{3}{\sqrt{7}}$
$\theta =\left(\frac{3}{\sqrt{7}}\right)$ of north


(3) Let width of river be $d$.
Time to reach from $A$ to $B$ in first case $\left(i\right)$:
$t_1=\frac{d}{v_{mR}}$
$v_{mR}=4\frac{m}{s}$
$t_1=\frac{d}{4}$
Time to reach from $A$ to $B$ in second case $\left(ii\right)$ :
$t_2=\frac{d}{v_{mR}\cos \theta }$
$t_2=\frac{d}{v_m}$
$v_m=v_{mR}\cos \theta =\sqrt{7}\frac{m}{s}$
$t_2=\frac{d}{\sqrt{7}}s$
$t_1<t_2$

Final Answer : in case $\left(i\right)$ he reaches the opposite bank in shortest time.