A body starts from rest and moves with a uniform acceleration. The ratio of the distance covered in the nth sec to the distance covered in n sec is

Options

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

Explanation

To solve the problem of finding the ratio of the distance covered in the n-th second to the distance covered in n seconds for a body starting from rest and moving with uniform acceleration, we can follow these steps:

### Step 1: Understand the equations of motion
The distance covered by a body under uniform acceleration can be described by the equations of motion. For a body starting from rest (initial velocity \( u = 0 \)), the distance \( S_n \) covered in \( n \) seconds is given by:
\[ S_n = ut + \frac{1}{2} A t^2 \]
Substituting \( u = 0 \):
\[ S_n = \frac{1}{2} A n^2 \]

### Step 2: Calculate the distance covered in the n-th second
The distance covered in the n-th second, denoted as \( S_n^{(nth)} \), can be calculated using the formula:
\[ S_n^{(nth)} = u + \frac{1}{2} A (2n - 1) \]
Again substituting \( u = 0 \):
\[ S_n^{(nth)} = \frac{1}{2} A (2n - 1) \]

### Step 3: Set up the ratio
We need to find the ratio of the distance covered in the n-th second to the distance covered in n seconds:
\[ \text{Ratio} = \frac{S_n^{(nth)}}{S_n} = \frac{\frac{1}{2} A (2n - 1)}{\frac{1}{2} A n^2} \]

### Step 4: Simplify the ratio
The \( \frac{1}{2} A \) terms cancel out:
\[ \text{Ratio} = \frac{2n - 1}{n^2} \]

### Step 5: Final expression
Thus, the final expression for the ratio of the distance covered in the n-th second to the distance covered in n seconds is:
\[ \text{Ratio} = \frac{2n - 1}{n^2} \]