Question

A body starts from rest and moves with an uniform acceleration. The ratio of distance covered in the $n^{th}$ second to the distance covered in $n$ seconds is :

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

Answer 1

The correct option is A $(\frac{2}{n}-\frac{1}{n^2})$

Distance covered in $n$ seconds,
o $\to$ start from rest
$s=ut+\frac{1}{2}a{t}^2$
$s=\frac{1}{2}a{n}^2-----\left(1\right)$

Distance covered in $n^{th}$ second $=s_n-s_{n-1}$

$\Rightarrow s_n-s_{n-1}=\frac{1}{2}a{n}^2-\frac{1}{2}a(n-1{)}^2-----(2)$

So, $\frac{e{q}^n\left(2\right)}{e{q}^n\left(1\right)}=\frac{\frac{1}{2}a{n}^2-\frac{1}{2}a(n-1{)}^2}{\frac{1}{2}a{n}^2}=\frac{2n-1}{n^2}$$=\frac{2}{n}-\frac{1}{n^2}$