Question

A bird is flying towards north with a velocity $40\text{km/h}$ and a train is moving with a velocity $40\text{km/h}$ towards east. What is the velocity of the bird noted by a man in the train ?

• A. $40\sqrt{2}\text{km/h N-E}$
• B. $40\sqrt{2}\text{km/h S-E}$
• C. $40\sqrt{2}\text{km/h N-W}$
• D. $40\sqrt{2}\text{km/h S-W}$

Answer 1

The correct option is C $40\sqrt{2}\text{km/h N-W}$

Given:
Velocity of bird, $V_B=40\text{km/h}$
Velocity of train, $V_T=40\text{km/h}$

Both velocity of bird and train as mention in question is shown below.

From the concept of relative motion, velocity of bird w.r.t. train will be

${\overrightarrow{V}}_{BT}=\overrightarrow{V_B}-\overrightarrow{V_T}$

Where,
$\overrightarrow{V_B}=$ Velocity of bird w.r.t. ground
$\overrightarrow{V_T}=$ Velocity of train w.r.t. ground

So, above expression can be written as

${\overrightarrow{V}}_{BT}=\overrightarrow{V_B}+(-\overrightarrow{V_T})$

So, the final diagram on the basis of expression will be

So, the magnitude velocity of bird w.r.t. train is given by

$|{\overrightarrow{V}}_{BT}|=\sqrt{|\overrightarrow{V_B}{|}^2+|-\overrightarrow{V_T}{|}^2}$

$\text{[By formula,}\theta ={90}^o]$

$|{\overrightarrow{V}}_{BT}|=\sqrt{{40}^2+{40}^2}=40\sqrt{2}\text{km/hr}$

So, ${\overrightarrow{V}}_{BT}=40\sqrt{2}\text{km/hr North-West}$

Hence, option $\left(c\right)$ is correct.

Why this question ? Relative position ${\overrightarrow{r}}_{AB}=\overrightarrow{r_A}-\overrightarrow{r_B}$ Relative velocity, ${\overrightarrow{V}}_{AB}=\overrightarrow{V_A}-\overrightarrow{V_B}$