Question

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

Answer 1

s= ut +½ at²

s= 0 + ½ * -12* (0.2)²

s= -6*0.04

s=-0.024

Displacement can be -ve where as distance can never be -ve

# -12 because elevator is descending