A small block slides down on a smooth incline | vedclass

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(B) The distance travelled by an object in the $n^{th}$ second starting from rest is given by the formula $S_n = u + \frac{a}{2}(2n-1)$.Since the bloc

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$\frac{2n-1}{2n+1}$

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(B) The distance travelled by an object in the $n^{th}$ second starting from rest is given by the formula $S_n = u + \frac{a}{2}(2n-1)$.Since the block starts from rest,the initial velocity $u = 0$.Therefore,the distance travelled in the $n^{th}$ interval is $S_n = \frac{a}{2}(2n-1)$.Similarly,the distance travelled in the $(n+1)^{th}$ interval is $S_{n+1} = \frac{a}{2}(2(n+1)-1) = \frac{a}{2}(2n+2-1) = \frac{a}{2}(2n+1)$.Now,calculating the ratio $\frac{S_n}{S_{n+1}}$:$\frac{S_n}{S_{n+1}} = \frac{\frac{a}{2}(2n-1)}{\frac{a}{2}(2n+1)} = \frac{2n-1}{2n+1}$.

$A$ 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 $\frac{S_{n}}{S_{n+1}}$ is

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