A block is kept on the floor of an elevator at rest. The elevator starts descending with acceleration of 5 m/s ^2 . The displacement of the block during the first 0.2 s after the start is Take g = 10 m/s ^2

Step-by-step solution

The acceleration of the elevator is a = 5 m/s ^2 downwards. Since the downward acceleration of the elevator is less than the acceleration due to gravity ( a Using the second equation of motion for the displacement of the block with initial velocity u = 0 and time t = 0.2 s : s = ut + 1 2 at^2 s = 0 + 1 2 5 (0.2)^2 s = 2.5 0.04 s = 0.1 m Converting the displacement into centimeters: s = 0.1 100 cm = 10 cm Answer: 10 cm