Question

A bird is flying towards north with a velocity 40 km/h and a train is moving with a velocity 40 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 $40\sqrt{2}\text{km/h}N-W$

Relative velocity of bird with respect to train, $V_{B/T}=V_B-V_T$
where $V_B$ and $V_T$ are the velocity of bird and train with respect to ground.
Since $V_B=40km/h$due north
$V_T=40km/h$due east
$\Rightarrow -V_T=-40km/h$due west
Thus magnitude of relative velocity, $\mid V_{B/T}\mid =\sqrt{(40{)}^2+(-40{)}^2}=40\sqrt{2}km/h$

Since $V_B$ is in north direction and $-V_T$ will be in west direction, hence there resultant should be in north-west direction.