A particle started with initial velocity u and moving with uniform acceleration a. change of velocity it will gain in the $n^{th}$second of its journey is 

Method: (1)

Let initial velocity = u

Acceleration $=a$

The final velocity after t = n seconds is equal to ${\text{v}}_{\text{1}}$

We can have $v_1=u+an\to \left(1\right)$

The final velocity of the body after t = (n – 1) seconds is equal to ${\text{v}}_{\text{2}}$

We can get ${\text{v}}_{\text{2}}=\text{u}+a(n-1)$

So the change in velocity in$n^{th}$second

= velocity after n seconds – velocity after (n – 1) seconds

= u + an – (u + a(n – 1))

= u + an – u – an + a

$\therefore$The change in velocity in the $n^{th}$ second $=a$

Method: (2)

We can simply work it out as given below

We know that body is given to be travelling with uniform acceleration $=a$

It means in every one second duration it will attain change of velocity equal to $a$

So obviously it will gain change of velocity in the $n^{th}$second also equal to$a$.