Question

a body starts from rest and moves with a uniform acceleration the ratio of distance covered in the nth second to the distance covered in n second is

Answer 1

Distance covered in nth sec is given by : $u+ \frac{a}{2} (2n-1)$
Distance covered in n sec : $un+ \frac{a}{2} (n)^{2}$
Since both starting from rest. u=0
Ratio= (a)(2n-1)/2: an²/2
= (2n-1):n²

Answer 2

The ratio of the distances is $\bold{2 n-1 : n^{2}}$

Solution:

To determine the distance, we need to use the below equation of motion,

$s=u t+\frac{1}{2} a t^{2}$

Distance covered in nth second is given by the formula given below:

$\Rightarrow u+\frac{a}{2}(2 n-1)$

Distance covered in n seconds is given by the formula given below:

$\Rightarrow u n+\frac{a}{2}(n)^{2}$

We have to find out the ratio of the distance covered in the nth second to the distance covered in n second.

Therefore, the ratio is given as:

$\Rightarrow \frac{u+\frac{a}{2}(2 n-1)}{u n+\frac{a}{2}(n)^{2}}$

Since, both are starting at rest, put, u = 0

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

$\Rightarrow \frac{2 n-1}{n^{2}}$