How can we derive the distance travelled in nth second by calculus method? Please derive it...

It's quite simple...

 

let us assume that the initial velocity of the object be 'u' and it moves under constant acceleration 'a'.

Now,

let Sn-1 be the distance travelled by the object in (n-1) seconds

let Sn be the distance travelled by the article in n seconds

 

so, the distance travelled in nth second will be

Dn = Sn - Sn-1 (1)

 

we also know that

s = ut + (1/2)at2

 

so,

Sn = un + (1/2)an2 (2)

Sn-1 = u(n-1) + (1/2)a(n-1)2 (3)

 

so, (1) becomes

Dn = un + (1/2)an2 - u(n-1) + (1/2)a(n-1)2

or

Dn = un = (1/2)an2 - un + u - (1/2)an2 + an - (a/2)

 

thus, by solving further, we get

the distance travelled in nth second of uniformly accelerated motion

D n = u + (a/2)[2n - 1]

  • 12
What are you looking for?