Question

A bird is flying towards north with a velocity $40km{h}^{-1}$ and a train is moving with velocity $40km{h}^{-1}$ towards east. What is the velocity of the bird noted by a man in the train?

• A. $40\sqrt{2}km{h}^{-1}N-E$
• B. $40\sqrt{2}km{h}^{-1}S-E$
• C. $40\sqrt{2}km{h}^{-1}N-W$
• D. $40\sqrt{2}km{h}^{-1}S-W$

Answer 1

Answer: (C)

To find the relative velocity of bird w.r.t. train, superimpose velocity $-{\vec{v}}_T$ on both the objects. Now as a result of it, the train is at rest, while the bird possesses two velocities, ${\vec{v}}_B$ towards north and $-{\vec{v}}_T$ along west

$\left|{\vec{v}}_{BT}\right|=\sqrt{{\left|{\vec{v}}_B\right|}^2+{\left|-{\vec{v}}_T\right|}^2}$ [By formula, $\theta =90^{\circ }]$
$=\sqrt{40^2+40^2}=40\sqrt{2}km{h}^{-1}$ north-west