Question

A man swims across a river with speed of $5\text{km/hr}$ with respect to the river in positive $y$ direction, while a boat goes upstream with speed $12\text{km/hr}$ with respect to the river in negative x direction. How fast and in which direction does the man appear to go to the boatman ? Given that the velocity of flowing water is $2\text{km/hr}$ in positive $x$ direction.

• A. $13\text{km/hr}$ in a direction making angle ${\tan }^{-1}\left(\frac{5}{12}\right)$ with x-axis
• B. $13\text{km/hr}$ in a direction making angle ${\tan }^{-1}\left(\frac{5}{12}\right)$ with y-axis
• C. $13\text{km/hr}$ in a direction making angle ${\tan }^{-1}\left(\frac{3}{4}\right)$ with x-axis
• D. $13\text{km/hr}$ in a direction making angle ${\tan }^{-1}\left(\frac{3}{4}\right)$ with y-axis

Answer 1

The correct option is A $13\text{km/hr}$ in a direction making angle ${\tan }^{-1}\left(\frac{5}{12}\right)$ with x-axis

Given that velocity of man w.r.t. the boat is $5\hat{j}$
$v_{mr}=5\hat{j}$
Also velocity of the boat w.r.t the river $v_{br}=-12\hat{i}$
And velocity of the river = $2\hat{i}$
Hence velocity of man w.r.t. the boat is
$v_{mr}-v_{br}=5\hat{j}+12\hat{i}$

Magnitude of the resultant $=\sqrt{5^2+{12}^2}=13\text{km/hr}$
Direction with x axis :
$tan\theta =\frac{v_y}{v_x}=\frac{5}{12}$
$\theta =ta{n}^{-1}\frac{5}{12}$