Question

A boat is moving towards east with velocity $4\text{m/s}$ with respect to river flowing towards north with velocity $2\text{m/s}$ and the wind is blowing towards north with velocity $6\text{m/s}$. The direction of the flag blown over by the wind hoisted on the boat is

• A. North-west
• B. South-east
• C. ${\tan }^{-1}\frac{1}{2}$ with east
  
• D. North

Answer 1

The correct option is A North-west

Let the $+ve-x$ axis along the east direction and $+ve-y$ axis along the north direction.

The velocity of boat w.r.t river is given as,
$v_{br}=4\hat{i}\text{m/s}$

Velocity of river w.r.t ground,
$v_{rg}=2\hat{j}\text{m/s}$

Velocity of wind w.r.t ground is,
$v_{wg}=6\hat{j}\text{m/s}$

The direction of flag hoisted on the boat will be along the direction of velocity of wind w.r.t boat.

Applying the concept of relative velocity,
$\Rightarrow {\overrightarrow{v}}_{wb}={\overrightarrow{v}}_{wg}-{\overrightarrow{v}}_{bg}$

Since ${\overrightarrow{v}}_{br}={\overrightarrow{v}}_{bg}-{\overrightarrow{v}}_{rg}$

$\Rightarrow {\overrightarrow{v}}_{wb}={\overrightarrow{v}}_{wg}-({\overrightarrow{v}}_{br}+{\overrightarrow{v}}_{rg})$

${\overrightarrow{v}}_{wb}=6\hat{j}+(-2\hat{j})+(-4\hat{i})$

${\overrightarrow{v}}_{wb}=4\hat{j}-4\hat{i}$

Thus, $\tan \theta =\left|\frac{v_y}{v_x}\right|=1$

$\Rightarrow \theta ={45}^{\circ }$

On observing the component of velocity for $v_{wb}$ it is clear that the direction will be along north-west.


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