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`.

Explanation

To solve the problem, we need to analyze the motion of the block inside the descending elevator. Here are the steps to find the displacement of the block during the first 0.2 seconds after the elevator starts descending:

### Step 1: Understand the scenario
The elevator is descending with an acceleration of \( a = 12 \, \text{m/s}^2 \) and the acceleration due to gravity is \( g = 10 \, \text{m/s}^2 \). Since the elevator's acceleration is greater than the acceleration due to gravity, the block will not remain in contact with the floor of the elevator after it starts descending.

### Step 2: Determine the effective acceleration of the block
When the elevator descends with an acceleration greater than gravity, the effective acceleration acting on the block can be calculated as:
\[
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