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 S_{n-1} be the distance travelled by the object in (n-1) seconds

let S_{n} be the distance travelled by the article in n seconds

so, the distance travelled in n^{th }second will be

D_{n} = S_{n} - S_{n-1 }(1)

we also know that

s = ut + (1/2)at^{2}

so,

S_{n} = un + (1/2)an^{2 }(2)

S_{n-1} = u(n-1) + (1/2)a(n-1)2 (3)

so, (1) becomes

D_{n} = un + (1/2)an^{2} - u(n-1) + (1/2)a(n-1)^{2}

or

D_{n} = un = (1/2)an^{2} - un + u - (1/2)an^{2} + 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]**

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