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

Options

A. The average speed of the particle while passing the windows satisfy the relation `v_(av1)gtv_(av2)gtv_(av3)`.
B. The time taken by the particle to cross the windows satisfies the relation `t_1`lt`t_2`lt`t_3`.
C. The magnitude of the acceleration of the particle while crossing the windows, satisfies the relation `a_1 = a_2 !=a_3`.
D. The change in the speed of the particle, while crossing the windows, would satisfy the relation `Deltav_1`lt`Deltav_2`lt`Deltav_3`.

Answer

A, B, D

Explanation

a,b,d. 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 distance is increasing as the particle is going
up. Hence, `t_1`lt`t_2`lt`t_3`.
As `v_(av) = Distance/Time , we have `v_(av) prop 1/Time`
So `v_(av1)`gt`v_(av2)`gt`v_(av3)`
Acceleration throughout the motion remains same from equation.
`vecv = vecu + vec(at), Deltav prop t`
So `Deltav_1` lt`Deltav_2` lt`Deltav_3`.