Question

A body starts from rest and moves with a uniform acceleration. Find the ratio of the distance covered in the ${\mathrm{n}}^{\text{th}}$ sec to the distance covered in $nsec$.

Answer 1

Step 1: Given

Initial velocity $u=0m/s$

Step 2: Equation of motion

Distance covered in $n^{th}$sec. is given by

$s_n=u+\frac{a}{2}(2n-1)$ or $s_n=0+\frac{a}{2}(2n-1)\dots \left(1\right)$

So the distance covered in $n$ seconds is

$s=ut+\frac{1}{2}a{t}^2=0+\frac{1}{2}a{n}^2$

Step 3: Rato of the distance covered,

The ratio of the distance covered in the ${\mathrm{n}}^{\text{th}}$ sec to the distance covered in $nsec$.

Therefore,
$$\begin{gathered} \frac{s_n}{s}=\frac{\frac{a}{2}(2n-1)}{\left(a{n}^2/2\right)} \\ \frac{s_n}{s}=\frac{2}{n}-\frac{1}{n^2} \end{gathered}$$

Hence the ratio of the distance covered in the ${\mathrm{n}}^{\text{th}}$ sec to the distance covered in $nsec$ is $\frac{s_n}{s}=\frac{2}{n}-\frac{1}{n^2}$.