A particle starts its motion from rest under the action of a constant force. If the distance covered in first `10 s` is `s_1` and the covered in the first `20 s` is `s_2`, then.

Options

A. `s_2 = 2 s_1`
B. `s_2 = 3 s_1`
C. `s_2 = 4 s_1`
D. `s_2 = s_1`

Explanation

To solve the problem, we will use the equations of motion under constant acceleration.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
The particle starts from rest, which means its initial velocity \( u = 0 \).

2. **Using the Equation of Motion**:
The distance covered by an object under constant acceleration can be given by the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the initial velocity \( u = 0 \), the equation simplifies to:
\[ s = \frac{1}{2} a t^2 \]

3. **Calculating Distance for the First 10 Seconds**:
For the first 10 seconds, we can substitute \( t = 10 \) seconds into the equation:
\[ s_1 = \frac{1}{2} a (10)^2 = \frac{1}{2} a \cdot 100 = 50a \]

4. **Calculating Distance for the First 20 Seconds**:
For the first 20 seconds, we substitute \( t = 20 \) seconds into the equation:
\[ s_2 = \frac{1}{2} a (20)^2 = \frac{1}{2} a \cdot 400 = 200a \]

5. **Finding the Relation Between \( s_1 \) and \( s_2 \)**:
Now we have:
\[ s_1 = 50a \quad \text{and} \quad s_2 = 200a \]
To find the relation between \( s_1 \) and \( s_2 \), we can express \( s_2 \) in terms of \( s_1 \):
\[ s_2 = 4s_1 \]

### Conclusion:
The relation between the distances covered is:
\[ s_2 = 4s_1 \]