A block is kept the floor of an elevator at rest. The elevator starts descending with an acceleration of `12m//s^(2)`. Find the displacement of the block during the first `1s` after the start.

Explanation

`mg-N=ma_(0)`
`N=m(g-a_(0))`
Here `a_(0)gtg`, the block will not be in contact with the floor
For the block`:` `x=(1)/(2)g t^(2)=(1)/(2)xx10xx1^(2)=5m`
In this duration, the elevator will fall a distance
`s'=(1)/(2)a_(0)t^(2)=(1)/(2)xx12xxt^(2)=6m`