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A small block slides without friction down an inclined plane starting from rest. Let $S_n$ be the distance travelled from time $t- n-1$ to $t=n$. Then $\large\frac{S_n}{S_{n_1}}$ is

$\begin{array}{1 1} \large\frac{2n-1}{2n} \\ \large\frac{2n+1}{2n-1} \\ \large\frac{2n-1}{2n+1} \\\large\frac{2n}{2n+1}\end{array} $

1 Answer

For distance $S_n$ travelled from $(n-1)$ to n,
$S_n= 0 [n-n-1)]+\large\frac{1}{2} $$a[n^2-(n-1)^2]$
$S_n = \large\frac{1}{2}$$a [n^2-(n^2+1-2n)]$
$S_n= \large\frac{a}{2}$$(2n-1)$
Again , $ S_{n+1}= \large\frac{1}{2}$$a[(n+1)^2-n^2]$
$\qquad= \large\frac{a}{2} $$[2n+1]$
$\large\frac{S_n}{S_{n+1}}=\large\frac{\frac {a}{2} (2n-1)}{\frac{a}{2} (2n+1)}$
$\qquad= \large\frac{2n-1}{2n+1}$
answered Aug 14, 2014 by meena.p

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