Question

A bird is flying due east with a velocity of 4 m/s. The wind starts to blow with a velocity of 3 m/s due north. What is the magnitude of relative velocity of bird w.r.t. wind? Find out the angle it makes with the x-axis?

• A. 7m/s,$ta{n}^{-1}\left(\frac{3}{4}\right)$
• B. 7 m/s, $ta{n}^{-1}\left(\frac{4}{3}\right)$
• C. 5 m/s, $ta{n}^{-1}\left(\frac{3}{4}\right)$
• D. 5 m/s,$ta{n}^{-1}\left(\frac{4}{3}\right)$

Answer 1

The correct option is C

5 m/s, $ta{n}^{-1}\left(\frac{3}{4}\right)$

Method - 1:

Whenever two bodies are moving at right angle to each other,

then their relative velocity is obtained by taking square root of

the sum of square root of the sum of squares of respective velocities.

Here Velocity of bird = $V_B=4m{s}^{-1}$ towards east.

Velocity of wind = $v_w=3m{s}^{-1}$ towards north, Velocity of bird

w.r.t wind = $v_{bw}=?$

${\overrightarrow{v}}_{b,w}={\overrightarrow{v}}_b-{\overrightarrow{v}}_w={\overrightarrow{v}}_b+(-{\overrightarrow{v}}_w)$

So,${\overrightarrow{v}}_{b,w}=\sqrt{v_b{}^2+v_w{}^2}=\sqrt{(3{)}^2+(4{)}^2}=\sqrt{25}=5m{s}^{-1}$

To find direction we need to find angle $\beta$ is angle between relative velocity and velocity of bird.

$tan\beta =\frac{3sin{90}^{\circ }}{4+3cos{90}^{\circ }}=\frac{3}{4}$

[$\therefore ={90}^{\circ }$, angle between them w.r.t east i.e., $v_b$ and $v_w$].B = $ta{n}^{-1}\left(\frac{3}{4}\right)$

Method - 2:

${\overrightarrow{v}}_{b,w}={\overrightarrow{v}}_b-{\overrightarrow{v}}_w={\overrightarrow{v}}_b+(-{\overrightarrow{v}}_w)=4\hat{i}+(-3\hat{j})\left(\frac{m}{s}\right)=4\hat{i}-3\hat{j}\left(m/s\right)$

$|{\overrightarrow{v}}_{b,w}|=\sqrt{(4{)}^2+(3{)}^2}=5m/s$

Here the direction of the relative velocity of the bird is $|tan\beta |=\frac{3}{4}\Rightarrow \beta =$

$ta{n}^{-1}\left(\frac{3}{4}\right)$

Here the relative velocity of the bird with respect to wind is 5 m/s and in the direction $ta{n}^{-1}\left(\frac{3}{4}\right)$ from east towards south.