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 )

Step-by-step solution

The acceleration (downward) of the elevator is 12 ~m / s ^2 which is greater than gravitational acceleration g . So, the block will act as a freely falling body. From the motion of the block aligned s & =u t+ 1 2 g t^2 & =0 t+ 1 2 10 (0.2)^2 & = 1 2 10 0.2 0.2 & =20 ~cm aligned