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)`

Answer

To solve the problem step by step, we can use the equations of motion under constant acceleration. Here's the detailed solution:

### Step 1: Understand the motion
The particle starts from rest under the influence of a constant force. This means that the initial velocity (u) is 0, and the particle experiences a constant acceleration (a).

### Step 2: Write the equation for distance covered
The distance covered by an object under constant acceleration can be expressed using the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the particle starts from rest, \( u = 0 \). Therefore, the equation simplifies to:
\[ s = \frac{1}{2} a t^2 \]

### Step 3: Calculate distance for the first 10 seconds
Let the distance covered in the first 10 seconds be \( s_1 \):
\[ s_1 = \frac{1}{2} a (10)^2 \]
\[ s_1 = \frac{1}{2} a \cdot 100 \]
\[ s_1 = 50a \] (Equation 1)

### Step 4: Calculate distance for the first 20 seconds
Now, let the distance covered in the first 20 seconds be \( s_2 \):
\[ s_2 = \frac{1}{2} a (20)^2 \]
\[ s_2 = \frac{1}{2} a \cdot 400 \]
\[ s_2 = 200a \] (Equation 2)

### Step 5: Relate \( s_1 \) and \( s_2 \)
Now, we can find the relationship between \( s_1 \) and \( s_2 \) by dividing Equation 1 by Equation 2:
\[ \frac{s_1}{s_2} = \frac{50a}{200a} \]
The \( a \) cancels out:
\[ \frac{s_1}{s_2} = \frac{50}{200} = \frac{1}{4} \]

### Step 6: Express \( s_2 \) in terms of \( s_1 \)
From the above relation, we can express \( s_2 \) in terms of \( s_1 \):
\[ s_2 = 4s_1 \]

### Conclusion
Thus, the distance covered in the first 20 seconds is four times the distance covered in the first 10 seconds:
\[ s_2 = 4s_1 \]