A particle is thrown in vertically in upward direction and passes three equally spaced windows of equal heights. Then

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The average speed of the particle while passing the windows satisfy the relation vav1>vav2>vav3

The time taken by the particle to cross the windows satisfies the relation t1

The magnitude of the acceleration of the particle while crossing the windows, satisfies the relation a1=a2≠a3

The change in the speed of the particle, while crossing the windows, would satisfy the relation Δv1<Δv2<Δv3

answer is A.

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Detailed Solution

As the particle is going up, it is slowing down, i.e., speed is decreasing and hence we can say that time taken by the particle to cover equal distances is increasing as the particle is going up. Hence, t1vav2>vav3Acceleration throughout the motion remains same from equation, v→=u→+at→,Δv∝tSo Δv1<Δv2<Δv3

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