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

Explanation

To find the ratio of the distance covered by a body in the nth second to that covered in n seconds, we can use the equations of motion under constant acceleration.

### Step-by-Step Solution:

1. **Understanding the Motion**:
The body starts from rest, meaning its initial velocity \( u = 0 \). The body moves with constant acceleration \( a \).

2. **Distance Covered in nth Second**:
The formula for the distance covered in the nth second is given by:
\[ S_n = u + \frac{1}{2} a (2n - 1) \]
Since \( u = 0 \), this simplifies to:
\[ S_n = \frac{1}{2} a (2n - 1) \]

3. **Total Distance Covered in n Seconds**:
The total distance covered in n seconds can be calculated using the formula:
\[ S_n = u n + \frac{1}{2} a n^2 \]
Again, since \( u = 0 \), this simplifies to:
\[ S_n = \frac{1}{2} a n^2 \]

4. **Finding the Ratio**:
Now, we need to find the ratio of the distance covered in the nth second to the total distance covered in n seconds:
\[ \text{Ratio} = \frac{S_n \text{ (nth second)}}{S_n \text{ (n seconds)}} = \frac{\frac{1}{2} a (2n - 1)}{\frac{1}{2} a n^2} \]
The \( \frac{1}{2} a \) cancels out:
\[ \text{Ratio} = \frac{2n - 1}{n^2} \]

5. **Final Result**:
Therefore, the ratio of the distance covered by the body in the nth second to that covered in n seconds is:
\[ \frac{2n - 1}{n^2} \]