Question

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 $t=n$. If $\frac{S_n}{S_{n+1}}$ is found to be 5/7, then value of $n$ is

Answer 1

Distance travelled in $t^{th}$ sec $=S_t$
$\therefore S_t=u+at-a/2$
as $u=0$,

$\frac{S_n}{S_{n+1}}=\frac{an-a/2}{a(n+1)-a/2}=\frac{a(2n-1)\times 2}{2a(2n+1)}=\frac{2n-1}{2n+1}=\frac{5}{7}$
$\Rightarrow 14n-7=10n+5$
$\Rightarrow 4n=12$
$\Rightarrow n=3$