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

40 \sqrt{2} k m h^{- 1} N - E

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40 \sqrt{2} k m h^{- 1} S - E

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40 \sqrt{2} k m h^{- 1} N - W

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40 \sqrt{2} k m h^{- 1} S - W

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Solution

The correct option is D $40 \sqrt{2} k m h^{- 1} N - W$
To find the relative velocity of bird w.r.t. train, superimpose velocity $- {\to V}_{T}$ on both the object. Now as a result of it, the train is at rest, while bird possesses two velocities, ${\to V}_{B}$ towards north and $- {\to V}_{T}$ along west.
$| {\to V}_{B T} | = \sqrt{| {\to V}_{B} |^{2} + | {\to V}_{T} |^{2}}$ [By formula, $θ = 90^{\circ}]$
$= \sqrt{40^{2} + 40^{2}} = 40 \sqrt{2} k m h^{- 1}$ north-west.

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A bird is flying towards north with a velocity 40 kmh−1 and a train is moving with velocity 40 kmh−1 towards east. What is the velocity of the bird noted by a man in the train?

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