question_answer

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)  \[40\sqrt{2}\,km\,{{h}^{-1}}\,S-W\]

Correct Answer: C

Solution :

[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 \[-{{\overrightarrow{v}}_{T}}\] 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