A particle starts sliding down a frictionless inclined plane. If ` S_(n) ` is the distance travelled by it from time ` t = n-1 sec `, to `t = n sec`, the ratio ` (S_(n))/(s_(n+1))` is

Options

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

Explanation

To solve the problem, we need to find the ratio of the distances traveled by a particle sliding down a frictionless inclined plane during two consecutive seconds. The distance traveled in the nth second, denoted as \( S_n \), can be calculated using the formula:

\[ S_n = u + \frac{1}{2} a (2n - 1) \]

Since the particle starts from rest, the initial velocity \( u = 0 \). Therefore, the formula simplifies to:

\[ S_n = \frac{1}{2} a (2n - 1) \]

### Step-by-step Solution:

1. **Calculate \( S_n \)**:
\[ S_n = \frac{1}{2} a (2n - 1) \]

2. **Calculate \( S_{n+1} \)**:
Using the same formula for \( n+1 \):
\[ S_{n+1} = \frac{1}{2} a (2(n + 1) - 1) = \frac{1}{2} a (2n + 2 - 1) = \frac{1}{2} a (2n + 1) \]

3. **Set up the ratio \( \frac{S_n}{S_{n+1}} \)**:
\[ \frac{S_n}{S_{n+1}} = \frac{\frac{1}{2} a (2n - 1)}{\frac{1}{2} a (2n + 1)} \]

4. **Simplify the ratio**:
The \( \frac{1}{2} a \) terms cancel out:
\[ \frac{S_n}{S_{n+1}} = \frac{2n - 1}{2n + 1} \]

### Final Result:
The ratio \( \frac{S_n}{S_{n+1}} \) is:
\[ \frac{S_n}{S_{n+1}} = \frac{2n - 1}{2n + 1} \]