a block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of `12 m/s^2`. Find the displacement of the block during the first 0.2 s after the start. Take `g=10 m/s^2`.

Explanation

Because the elevator is moving downward with an acceleration `12 m/s^2 (gtg)`, the body gets separated.
So, body moves with acceleration,
`g=10 m/s^2`
[Freely falling body]`
and the elevator moves with acceeration `12 m/s^2`.
Now, the bock had acceleration,
` g= 10 m/s^2`
`u=0`
`t=0.2 sec`

So teh distance travelled by the block is given by
` S= ut+ 1/2 at^2 `
`= 0+1/2 10(0.2)^2= 5xx0.04`
`=0.2 m = 20 cm`
The displacement of body is 20 cm during first 0.2 sec.