A small block slides down on a smooth inclined plane, starting from rest at time t=0. Let S_n be the distance travelled by the block in the interval t=n−1 to t=n. Then, the ratio S_n/S_n+1 is :

The correct option is 2n - 1/2n + 1.

Step-by-Step Explanation:

Let the block slide down the smooth inclined plane with a constant acceleration $a$ under the influence of gravity. Since the block starts from rest at time $t=0$, its initial velocity is:
$u=0$

The distance traveled by an object undergoing constant acceleration in the $n^{\text{th}}$ second (i.e., in the time interval from $t=n-1$ to $t=n$) is given by the formula:
$S_n=u+\frac{a}{2}(2n-1)$

Substituting $u=0$ into the formula, the distance traveled in the $n^{\text{th}}$ interval is:
$S_n=\frac{a}{2}(2n-1)$

Similarly, the distance traveled in the next interval, which is the ${(n+1)}^{\text{th}}$ second (from $t=n$ to $t=n+1$), is obtained by replacing $n$ with $n+1$:

Simplifying the expression inside the brackets:

Now, we find the ratio of $S_n$ to $S_{n+1}$:

Canceling out the common factor $\frac{a}{2}$ from the numerator and denominator, we get:
$\frac{S_n}{S_{n+1}}=\frac{2n-1}{2n+1}$