A particle experiences constant acceleration for 20 s after starting from rest. If it travels a distance `S_1` in the first 10 s and distance `S_2` in the next 10 s, the relation between `S_1 and S_2` is :

Options

A. `S_2 = 3S_1`
B. `S_1 = 3S_2`
C. `S_2 = 2S_1`
D. `S_1 = 10S_2`

Explanation

To solve the problem, we need to find the relationship between the distances \( S_1 \) and \( S_2 \) traveled by a particle under constant acceleration over two intervals of time.

### Step-by-step Solution:

1. **Understanding the Motion**:
The particle starts from rest and experiences constant acceleration. The total time of motion is 20 seconds, divided into two intervals of 10 seconds each.

2. **Distance in the First Interval (\( S_1 \))**:
The formula for distance traveled under constant acceleration is given by:
\[ S = ut + \frac{1}{2} a t^2 \]
Since the particle starts from rest, the initial velocity \( u = 0 \). For the first 10 seconds (\( t_1 = 10 \) s):
\[ S_1 = 0 \cdot 10 + \frac{1}{2} a (10^2) = \frac{1}{2} a (100) = 50a \]

3. **Distance in the Second Interval (\( S_2 \))**:
For the second interval, we need to consider the total time of 20 seconds. The distance traveled during the second 10 seconds can be calculated using the same formula, but we need to consider the total time \( t_2 = 20 \) s:
\[ S = ut + \frac{1}{2} a t^2 = 0 \cdot 20 + \frac{1}{2} a (20^2) = \frac{1}{2} a (400) = 200a \]
However, this distance includes the distance traveled in the first 10 seconds. Therefore, the distance \( S_2 \) for the second interval is:
\[ S_2 = \text{Total distance in 20 s} - S_1 = 200a - 50a = 150a \]

4. **Finding the Relation Between \( S_1 \) and \( S_2 \)**:
Now we have:
\[ S_1 = 50a \quad \text{and} \quad S_2 = 150a \]
To find the ratio of \( S_1 \) to \( S_2 \):
\[ \frac{S_1}{S_2} = \frac{50a}{150a} = \frac{50}{150} = \frac{1}{3} \]
Thus, we can express \( S_2 \) in terms of \( S_1 \):
\[ S_2 = 3S_1 \]

### Conclusion:
The relationship between \( S_1 \) and \( S_2 \) is:
\[ S_2 = 3S_1 \]