Question

A block is kept on the floor of an elevator at rest. The elevator starts descending at acceleration 12 m/$s^2$. Find displacement of the block during the first 0.2 s after the start.

Answer 1

As the elevator is descending with an acceleration of 12 m/$s^2$ (>g) so the block will lose contact, hence will fall freely.
$s=ut+\frac{1}{2}g{t}^2$
$\Rightarrow s=\frac{1}{2}\times 10\times \frac{2}{10}\times \frac{2}{10}=0.2m$.
Hence the displacement of the block is 0.2 m.