A small block slides without friction down an iclined plane starting form rest. Let `S_(n)` be the distance traveled from time `t = n - 1` to `t = n`. Then `(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 will follow these steps:

### Step 1: Understand the motion of the block
The block slides down an inclined plane under the influence of gravity. The acceleration of the block down the incline is given by \( a = g \sin \theta \), where \( g \) is the acceleration due to gravity and \( \theta \) is the angle of inclination.

### Step 2: Use the formula for distance traveled in nth second
The distance traveled by the block during the nth second can be calculated using the formula:
\[ S_n = u + \frac{a}{2}(2n - 1) \]
Since the block starts from rest, the initial velocity \( u = 0 \). Therefore, the formula simplifies to:
\[ S_n = \frac{a}{2}(2n - 1) \]
Substituting \( a = g \sin \theta \):
\[ S_n = \frac{g \sin \theta}{2}(2n - 1) \]

### Step 3: Calculate the distance traveled in the (n + 1)th second
Using the same formula for \( n + 1 \):
\[ S_{n+1} = \frac{g \sin \theta}{2}(2(n + 1) - 1) \]
This simplifies to:
\[ S_{n+1} = \frac{g \sin \theta}{2}(2n + 2 - 1) = \frac{g \sin \theta}{2}(2n + 1) \]

### Step 4: Find the ratio \( \frac{S_n}{S_{n+1}} \)
Now, we can find the ratio of the distances traveled in the nth and (n + 1)th seconds:
\[ \frac{S_n}{S_{n+1}} = \frac{\frac{g \sin \theta}{2}(2n - 1)}{\frac{g \sin \theta}{2}(2n + 1)} \]
The \( \frac{g \sin \theta}{2} \) terms cancel out:
\[ \frac{S_n}{S_{n+1}} = \frac{2n - 1}{2n + 1} \]

### Final Answer
Thus, the ratio \( \frac{S_n}{S_{n+1}} \) is:
\[ \frac{S_n}{S_{n+1}} = \frac{2n - 1}{2n + 1} \]