A particle experiences constant acceleration for 20 s after starting from rest. If it travels a distance `X_(1)`, in the first 10 s and distance `X_(2)`, in the remaining 10s, then which of the following is true?

Options

A. `X_(1) = 2X_(2)`
B. `(X_(1) = X_(2)`
C. `X_(1) = 3X_(2)`
D. None of these

Answer

To solve the problem of a particle experiencing constant acceleration for 20 seconds after starting from rest, we can break it down into steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
The particle starts from rest and accelerates uniformly for a total of 20 seconds. We need to find the distances traveled in the first 10 seconds (denoted as \( X_1 \)) and the next 10 seconds (denoted as \( X_2 \)).

2. **Using the Equation of Motion**:
The equation for distance traveled under constant acceleration is given by:
\[ X = ut + \frac{1}{2} a t^2 \]
where \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.

3. **Calculating \( X_1 \)**:
For the first 10 seconds:
- Initial velocity \( u = 0 \) (starts from rest)
- Time \( t = 10 \) seconds
- Distance \( X_1 \) can be calculated as:
\[ X_1 = 0 \cdot 10 + \frac{1}{2} a (10)^2 = \frac{1}{2} a \cdot 100 = 50a \]

4. **Calculating \( X_2 \)**:
For the next 10 seconds (from 10 to 20 seconds):
- The initial velocity at \( t = 10 \) seconds is \( v = u + at = 0 + a \cdot 10 = 10a \).
- The total time for this segment is also 10 seconds.
- The distance \( X_2 \) can be calculated as:
\[ X_2 = (10a) \cdot 10 + \frac{1}{2} a (10)^2 = 100a + \frac{1}{2} a \cdot 100 = 100a + 50a = 150a \]

5. **Finding the Relationship Between \( X_1 \) and \( X_2 \)**:
Now we have:
- \( X_1 = 50a \)
- \( X_2 = 150a \)

To find the ratio \( \frac{X_1}{X_2} \):
\[ \frac{X_1}{X_2} = \frac{50a}{150a} = \frac{1}{3} \]
This implies:
\[ 3X_1 = X_2 \]

### Conclusion:
The relationship between the distances traveled in the first 10 seconds and the next 10 seconds is:
\[ 3X_1 = X_2 \]