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]