A body starts from rest and moves with constant acceleration. The ratio of distance covered by the body in `nth` second to that covered in `n` second is.

Options

A. `2/(n^(2))-1/n`
B. `2/(n^(2))+1/n`
C. `2/n-1/(n^(2))`
D. `2/n+1/(n^(2))`

Explanation

To solve the problem, we need to find the ratio of the distance covered by the body in the nth second to the distance covered in the first n seconds.

### Step-by-Step Solution:

1. **Understanding the Variables**:
- Let \( U \) be the initial velocity (which is 0 since the body starts from rest).
- Let \( A \) be the constant acceleration.
- Let \( n \) be the specific second we are interested in.

2. **Distance Covered in nth Second**:
The distance covered in the nth second, denoted as \( S_n \), can be calculated using the formula:
\[ S_n = U + \frac{A}{2} (2n - 1) \]
Since the body starts from rest, \( U = 0 \):
\[ S_n = \frac{A}{2} (2n - 1) \]

3. **Distance Covered in First n Seconds**:
The total distance covered in the first n seconds, denoted as \( S \), is given by the formula:
\[ S = U n + \frac{A}{2} n^2 \]
Again, since \( U = 0 \):
\[ S = \frac{A}{2} n^2 \]

4. **Finding the Ratio**:
Now, we need to find the ratio \( \frac{S_n}{S} \):
\[ \frac{S_n}{S} = \frac{\frac{A}{2} (2n - 1)}{\frac{A}{2} n^2} \]
The \( \frac{A}{2} \) cancels out:
\[ \frac{S_n}{S} = \frac{2n - 1}{n^2} \]

5. **Final Expression**:
Thus, the ratio of the distance covered by the body in the nth second to that covered in the first n seconds is:
\[ \frac{S_n}{S} = \frac{2n - 1}{n^2} \]

### Conclusion:
The final answer is:
\[ \frac{S_n}{S} = \frac{2n - 1}{n^2} \]