# A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12 m/s2. Find the displacement of the block during the first 0.2 s after the start. Take g = 10 m/s2.

Given:

The two bodies are separated because the elevator is moving downward with an acceleration of 12 ms-2 (>g) and the body moves with acceleration, g = 10 ms-2 [Freely falling body]

Now, for the block: g = 10 m/s �, u = 0, t = 0.2 s

So, the distance travelled by the block is given by

=5×0.04 =0.2 m =20 cm

The displacement of the body is 20 cm during the first 0.2 s

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Pseudo Force52 mins
Challenging Quiz on Newton Laws of Motion Part 364 mins
Pully Problems | Getting JEE Ready68 mins
Check your understanding of linear momentum22 mins
Why edges are raised for Curved roads? - Banking of roads54 mins
Into the world of springs - Spring force48 mins
Basics strengthener - Newton Laws of motion61 mins
Understand Newton's 2nd law Exhaustively34 mins
Elaborative Lecture on Constraint Motion59 mins
Escaping the reality - Concept of Pseudo Force54 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses