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)`

Answer

To solve the problem, we need to find the ratio \( \frac{S_n}{S_{n+1}} \) where \( S_n \) is the distance traveled by the block from time \( t = n-1 \) to \( t = n \) and \( S_{n+1} \) is the distance traveled from time \( t = n \) to \( t = n+1 \).

### Step 1: Determine the expression for \( S_n \)

The distance traveled by the block on the inclined plane can be expressed using the equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
Since the block starts from rest, \( u = 0 \). The acceleration \( a \) along the incline is given by \( a = g \sin \theta \), where \( g \) is the acceleration due to gravity and \( \theta \) is the angle of the incline.

Thus, the distance \( S_n \) traveled from \( t = n-1 \) to \( t = n \) is:
\[ S_n = \frac{1}{2} a (t_n^2 - t_{n-1}^2) = \frac{1}{2} g \sin \theta \left( n^2 - (n-1)^2 \right) \]
Calculating \( n^2 - (n-1)^2 \):
\[ n^2 - (n-1)^2 = n^2 - (n^2 - 2n + 1) = 2n - 1 \]
Thus, we have:
\[ S_n = \frac{1}{2} g \sin \theta (2n - 1) \]

### Step 2: Determine the expression for \( S_{n+1} \)

Now, we calculate \( S_{n+1} \) which is the distance traveled from \( t = n \) to \( t = n+1 \):
\[ S_{n+1} = \frac{1}{2} g \sin \theta \left( (n+1)^2 - n^2 \right) \]
Calculating \( (n+1)^2 - n^2 \):
\[ (n+1)^2 - n^2 = (n^2 + 2n + 1) - n^2 = 2n + 1 \]
Thus, we have:
\[ S_{n+1} = \frac{1}{2} g \sin \theta (2n + 1) \]

### Step 3: Find the ratio \( \frac{S_n}{S_{n+1}} \)

Now we can find the ratio:
\[ \frac{S_n}{S_{n+1}} = \frac{\frac{1}{2} g \sin \theta (2n - 1)}{\frac{1}{2} g \sin \theta (2n + 1)} \]
The \( \frac{1}{2} g \sin \theta \) cancels out:
\[ \frac{S_n}{S_{n+1}} = \frac{2n - 1}{2n + 1} \]

### Final Answer

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

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