Question

A small block slides without friction down an inclined plane starting from rest. Let $S_n$ be the distance travelled from time $t=\left(n-1\right)tot=n$. The $\frac{S_n}{S_{n+1}}$ is.

• A. $\frac{\left(2n-1\right)}{2n}$
• B. $\frac{\left(2n+1\right)}{\left(2n-1\right)}$
• C. $\frac{\left(2n-1\right)}{\left(2n+1\right)}$
• D. $\frac{\left(2n\right)}{\left(2n+1\right)}$

Answer 1

The correct option is C $\frac{\left(2n-1\right)}{\left(2n+1\right)}$

For distance $S_n$ travelled from $(n-1)$ to $n$

$S_n=0[n-n-1]+\frac{1}{2}a[n^2-(n-1{)}^2]$

$S_n=\frac{1}{2}a[n^2-(n^2+1-2n\left)\right]$

$S_n=\frac{a}{2}(2n-1)$

Again , $S_{n+1}=\frac{1}{2}a\left[\right(n+1{)}^2-n^2]$

$=\frac{a}{2}[2n+1]$

$\frac{S_n}{S_{n+1}}=\frac{\frac{a}{2}(2n-1)}{\frac{a}{2}(2n+1)}$

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