Question

a block is kept on the floor of an elevator at rest .the elevator starts descending with an acceleration of 12m/s2 . find the displacement of the block during the first 0.2s after the start

Answer 1

a=12my kithi uY88yri/s2 s=0.2s

Answer 2

Given:

Acceleration of elevator = 12 $m/s^2$

Since, the acceleration of the elevator downwards is greater than the gravitational force, and the block gets separated and will behave like a free fall body with acceleration equal to g which is 9.8 $m/s^2$

Where u = 0 and given t = 0.2 sec

Hence the displacement of block during first 0.2 seconds is = ut + ½ $at^2$

Displacement = (0 x 0.2) + ½ x 9.8 x $(0.2)^2$

Displacement = 0 + 4.9 x (0.04) = 0.196 meter = 19.6 centimeter.