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

Explanation

To solve the problem, we need to find the ratio \( \frac{S_n}{S_{n+1}} \), where \( S_n \) is the distance traveled by a particle sliding down a frictionless inclined plane from time \( t = n-1 \) seconds to \( t = n \) seconds.

### Step-by-Step Solution:

1. **Understanding the Motion**:
The particle is sliding down a frictionless inclined plane, which means it is undergoing uniformly accelerated motion due to gravity. The initial velocity \( u \) is 0 (since it starts from rest), and the acceleration \( a \) can be expressed as \( g \sin(\theta) \), where \( \theta \) is the angle of the incline.

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

3. **Distance for (n+1)th Second**:
Similarly, the distance traveled in the (n+1)th second, \( S_{n+1} \), is given by:
\[ S_{n+1} = u + \frac{a}{2} (2(n+1) - 1) \]
Again, substituting \( u = 0 \):
\[ S_{n+1} = \frac{a}{2} (2(n+1) - 1) = \frac{a}{2} (2n + 2 - 1) = \frac{a}{2} (2n + 1) \]

4. **Finding the Ratio**:
Now, we can find the ratio \( \frac{S_n}{S_{n+1}} \):
\[ \frac{S_n}{S_{n+1}} = \frac{\frac{a}{2} (2n - 1)}{\frac{a}{2} (2n + 1)} \]
The \( \frac{a}{2} \) terms cancel out:
\[ \frac{S_n}{S_{n+1}} = \frac{2n - 1}{2n + 1} \]

5. **Conclusion**:
Therefore, the ratio \( \frac{S_n}{S_{n+1}} \) is:
\[ \frac{S_n}{S_{n+1}} = \frac{2n - 1}{2n + 1} \]