Question

Why is distance travelled by body in nth second 2n-1?

Answer 1

how can you say it is 2n-1

Answer 2

Given,

let Initial velocity  of an  object  is  u and  acceleration is  a.

Let the distance traveled in n second  from initial position is  $S_n$ .

According  second  equation of  motion,

Distance traveled in  t second,

$S=ut+\frac{1}{2}at^{2}$

Distance traveled  in  n second,

$S_n=un+\frac{1}{2}an^{2}...........i)$

Distance traveled in (n-1) second,

$S_{n-1}=u(n-1)+\frac{1}{2}a(n-1)^{2}...........ii)$

Hence,

Distance traveled in nth second,

$S_{n^{th}}=S_n-S_{n-1}\\\\\;\;\;\;\;\;=un+\frac{1}{2}an^{2}-[u(n-1)+\frac{1}{2}a(n-1)^{2}]\\\\=un+\frac{1}{2}an^{2}-un+u-\frac{1}{2}a(n-1)^{2}\\\\=un+\frac{1}{2}an^{2}-un+u-\frac{1}{2}a(n^{2}+1-2n)\\\\=u+\frac{1}a(n^{2}-n^{2} -1+2n)\\ \\S_{n^{th} }=u+\frac{1}{2}a(2n-1)$