A small block slides down on a smooth inclined plane starting from rest at time $t=0$. Let $S_n$ be the distan

Question

A small block slides down on a smooth inclined plane starting from rest at time $t=0$. Let $S_n$ be the distance traveled by the block in the interval $t=n-1$ to $t=n$. Then the ratio $\frac{S_n}{S_{n+1}}$ is:

Accepted Answer

1.  **Identify the Formula:** The distance traveled by a body undergoing uniformly accelerated motion in the $n$-th second (interval from $t = n-1$ to $t = n$) is given by the formula:
    $$S_n = u + \frac{a}{2}(2n - 1)$$
    where $u$ is the initial velocity and $a$ is the acceleration.
2.  **Apply Conditions:** The block starts from rest, so $u = 0$. The acceleration $a$ is constant ($g \sin 	heta$ on an incline). Thus:
    $$S_n = \frac{a}{2}(2n - 1)$$
3.  **Calculate for (n+1)-th Interval:** For the next interval ($t = n$ to $t = n+1$), we substitute $n$ with $(n+1)$:
    $$S_{n+1} = \frac{a}{2}[2(n + 1) - 1] = \frac{a}{2}(2n + 2 - 1) = \frac{a}{2}(2n + 1)$$
4.  **Calculate Ratio:**
    $$\frac{S_n}{S_{n+1}} = \frac{\frac{a}{2}(2n - 1)}{\frac{a}{2}(2n + 1)} = \frac{2n - 1}{2n + 1}$$
5.  **Context:** This is a generalization of Galileo's law of odd numbers found in NCERT .

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