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?

40 \sqrt{2} km/h N - E

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40 \sqrt{2} km/h S - E

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40 \sqrt{2} km/h N - W

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40 \sqrt{2} km/h S - W

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Solution

The correct option is C $40 \sqrt{2} km/h N - W$
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 $V_{B} = 40 k m / h$ due north
$V_{T} = 40 k m / h$ due east
$\Rightarrow - V_{T} = - 40 k m / h$ due west
Thus magnitude of relative velocity, $∣ V_{B / T} ∣= \sqrt{(40)^{2} + (- 40)^{2}} = 40 \sqrt{2} k m / h$

Since $V_{B}$ is in north direction and $- V_{T}$ will be in west direction, hence there resultant should be in north-west direction.

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