Question

Derive the expression for distance traveled by a body in nth second

Answer 1

Let AB be the distance travelled by a body in n seconds and AC be the distance travelled by the body in (n-1) seconds


Thus,the distance travelled in the nth second is CB=Sn

Sn=AB-AC. (2.5)
From
S=ut+1/2an^2

AB=un+1/2an^2. (2.6)

AC=u(n-1)+1/2a(n-1) (2.7)

Substituting (2.6) and (2.7)


Sn=un+1/2an^2-[u(n-1)+- a (n-1)^2
Sn=un+1/2an^2-[un-u+1/2an^2-an-1/2a

Sn=un+1/2an^2-un+u-1/2an^2+an-1/2a
Sn=u +an-1/2a
Sn=u+a(n-1/2)
Sn u+1/2a(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,

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