Cullen Banks

2023-03-03

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?

Elliot Watson

Beginner2023-03-04Added 1 answers

Distance travelled in t sec $={S}_{t}$

$\therefore {S}_{t}=u+at-a/2$

as u=0

$\frac{{S}_{n}}{{S}_{n+1}}=\frac{an-a/2}{a(n+1)-a/2}=\frac{a(2n-1)\times 2}{2a(2n+1)}\phantom{\rule{0ex}{0ex}}=\frac{2n-1}{2n+1}=\frac{5}{7}\phantom{\rule{0ex}{0ex}}\Rightarrow 14-7=10n+5\phantom{\rule{0ex}{0ex}}\Rightarrow 4n=12\phantom{\rule{0ex}{0ex}}\Rightarrow n=3$

$\therefore {S}_{t}=u+at-a/2$

as u=0

$\frac{{S}_{n}}{{S}_{n+1}}=\frac{an-a/2}{a(n+1)-a/2}=\frac{a(2n-1)\times 2}{2a(2n+1)}\phantom{\rule{0ex}{0ex}}=\frac{2n-1}{2n+1}=\frac{5}{7}\phantom{\rule{0ex}{0ex}}\Rightarrow 14-7=10n+5\phantom{\rule{0ex}{0ex}}\Rightarrow 4n=12\phantom{\rule{0ex}{0ex}}\Rightarrow n=3$