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`

Explanation

To solve the problem, we need to find the distances \( x_1 \), \( x_2 \), and \( x_3 \) traveled by a body under uniform acceleration over three intervals of 10 seconds each. The body starts from rest, so its initial velocity \( u = 0 \).

### Step-by-Step Solution:

1. **Identify the formula for distance under uniform acceleration**:
The distance traveled by an object under uniform acceleration can be calculated using the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the body starts from rest, \( u = 0 \), so the equation simplifies to:
\[ s = \frac{1}{2} a t^2 \]

2. **Calculate \( x_1 \)** (distance traveled in the first 10 seconds):
For the first 10 seconds, \( t = 10 \) s:
\[ x_1 = \frac{1}{2} a (10)^2 = \frac{1}{2} a \cdot 100 = 50a \]

3. **Calculate \( x_2 \)** (distance traveled in the next 10 seconds):
The distance traveled in the first 20 seconds can be calculated as:
\[ x_1 + x_2 = \frac{1}{2} a (20)^2 = \frac{1}{2} a \cdot 400 = 200a \]
Therefore, to find \( x_2 \):
\[ x_2 = (x_1 + x_2) - x_1 = 200a - 50a = 150a \]

4. **Calculate \( x_3 \)** (distance traveled in the last 10 seconds):
The total distance traveled in 30 seconds is:
\[ x_1 + x_2 + x_3 = \frac{1}{2} a (30)^2 = \frac{1}{2} a \cdot 900 = 450a \]
Therefore, to find \( x_3 \):
\[ x_3 = (x_1 + x_2 + x_3) - (x_1 + x_2) = 450a - 200a = 250a \]

5. **Find the ratio \( x_1 : x_2 : x_3 \)**:
Now we have:
\[ x_1 = 50a, \quad x_2 = 150a, \quad x_3 = 250a \]
The ratio can be expressed as:
\[ x_1 : x_2 : x_3 = 50a : 150a : 250a \]
Dividing each term by \( 50a \):
\[ 1 : 3 : 5 \]

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

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