A body starts from rest and is uniformly accelerated for `30` s. The distance travelled in the first `10` s is `x_(1)`, next `10` s is `x_(2)` and the last `10` s is `x_(3)`. Then `x_(1):x_(2):x_(3)` is the same as :-

Options

A. `1:2:4`
B. `1:2:5`
C. `1:3:5`
D. `1:3:9`

Answer

To solve the problem, we need to find the distances \( x_1 \), \( x_2 \), and \( x_3 \) traveled by a body that starts from rest and is uniformly accelerated over three intervals of 10 seconds each.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Initial velocity \( u = 0 \) (the body starts from rest).
- Let the uniform acceleration be \( a \).
- The total time of acceleration is \( 30 \) seconds.

2. **Calculate \( x_1 \)**:
- The distance traveled in the first \( 10 \) seconds can be calculated using the formula:
\[ x_1 = u t + \frac{1}{2} a t^2 \]
- Substituting \( u = 0 \) and \( t = 10 \) seconds:
\[ x_1 = 0 \cdot 10 + \frac{1}{2} a (10)^2 = \frac{1}{2} a \cdot 100 = 50a \]

3. **Calculate \( x_2 \)**:
- The distance traveled in the next \( 10 \) seconds (from \( t = 10 \) to \( t = 20 \)) can be calculated using the total distance formula up to \( t = 20 \) seconds and subtracting \( x_1 \):
\[ x_1 + x_2 = \frac{1}{2} a (20)^2 = \frac{1}{2} a \cdot 400 = 200a \]
- Thus, we have:
\[ x_2 = (x_1 + x_2) - x_1 = 200a - 50a = 150a \]

4. **Calculate \( x_3 \)**:
- The distance traveled in the last \( 10 \) seconds (from \( t = 20 \) to \( t = 30 \)) can be calculated similarly:
\[ x_1 + x_2 + x_3 = \frac{1}{2} a (30)^2 = \frac{1}{2} a \cdot 900 = 450a \]
- Therefore, we find \( x_3 \) as follows:
\[ x_3 = (x_1 + x_2 + x_3) - (x_1 + x_2) = 450a - 200a = 250a \]

5. **Determine the Ratios**:
- Now we have:
\[ x_1 = 50a, \quad x_2 = 150a, \quad x_3 = 250a \]
- The ratio \( x_1 : x_2 : x_3 \) is:
\[ x_1 : x_2 : x_3 = 50a : 150a : 250a \]
- Simplifying this ratio:
\[ = 50 : 150 : 250 = 1 : 3 : 5 \]

### Final Answer:
The ratio \( x_1 : x_2 : x_3 \) is \( 1 : 3 : 5 \).