Question

Derive first equation of motion

Answer 1

You have not mentioned which three equations. But, I guess you are  talking about the three  equations of uniformly accelerated motion. These are
1.  v=u+atv=u+at 
2.  S=ut+12at2S=ut+12at2
3. v2−u2=2aSv2−u2=2aS

Before you learn derivations of these equations you need to have a prerequisite knowledge of Differential Equations(Calculus).

Ok, so lets start deriving them.

Derivation of the first equation:
dvdt=advdt=a
∫dv=∫adt=a∫dt∫dv=∫adt=a∫dt (since a is uniform ie constant)
∫vudv=∫t0dt∫uvdv=∫0tdt
v−u=a(t−0)v−u=a(t−0)
Thus, v=u+atv=u+at

Derivation of the second equation :
dxdt=vdxdt=v
∫x2x1dx=∫t0vdt∫x1x2dx=∫0tvdt
putting value of v from the first equation : v=u+atv=u+at
∫x2x1dx=∫t0(u+at)dt∫x1x2dx=∫0t(u+at)dt
x2−x1=S=ut+12at2x2−x1=S=ut+12at2
Thus, S=ut+12at2S=ut+12at2

Derivation of the third equation :
dxdt=vdxdt=v and dvdt=a→dt=dvadvdt=a→dt=dva
Hence, ∫vdv=∫adx→∫vudv=a∫x2x1dx∫vdv=∫adx→∫uvdv=a∫x1x2dx
v2−u22=a(x2−x1)v2−u22=a(x2−x1)
Thus, v2−u2=2aSv2−u2=2aS

Bingo. All the three equations are derived. :)
Mark as a braineist answer

Answer 2

Hello

 Consider a body initial moving with velocity "Vi". After certain interval of time "t", its velocity becomes "Vf". Now

Change in velocity = Vf - Vi 
OR 
DV =Vf – Vi

Due to change in velocity, an acceleration "a" is produced in the body. Acceleration is given by

a = DV/t

 Putting the value of "DV"

a = (Vf – Vi)/t
at = Vf – Vi
at + Vi =Vf


SECOND EQUATION OF MOTION
OR 
S = Vit + 1/2at2

 Consider a car moving on a straight road with an initial velocity equal to ‘Vi’. After an interval of time ‘t’ its velocity becomes ‘Vf’. Now first we will determine the average velocity of body.

Average velocity = (Initial velocity + final velocity)/2
OR
Vav = (Vi + Vf)/2

but Vf = Vi + atPutting the value of Vf

Vav = (Vi + Vi + at)/2
Vav = (2Vi + at)/2    
Vav = 2Vi/2 + at/2   
Vav = Vi + at/2      
                      Vav = Vi + 1/2at.......................................(i)

we know that

S = Vav x t

Putting the value of ‘Vav’

S = [Vi + 1/2at] t

S = Vit + 1/2at2

THIRD EQUATION OF MOTION
OR 
2aS = Vf2 – Vi2

 Initial velocity, final velocity, acceleration, and distance are related in third equation of motion.Consider a body moving initially with velocity ‘Vi’. After certain interval of time its velocity becomes ‘Vf’. Due to change in velocity, acceleration ‘a’ is produced in the body. Let the body travels a distance of ‘s’ meters. 
According to first equation of motion:

Vf = Vi + at     
OR 
Vf – Vi = at
OR   
                      (Vf – Vi)/a = t....................(i)

Average velocity of body is given by:

Vav = (Initial velocity + Final velocity)/2
                              Vav = (Vi + Vf)/2.................. (ii)

we know that :

              S = Vav x t.................. (ii)

Putting the value of Vav and t from equation (i) and (ii) in equation (iii)

S = { (Vf + Vi)/2} { (Vf – Vi)/a}
2aS = (Vf + Vi)(Vf – Vi)

According to [ (a+b)(a-b)=a2-b2]


2as = vf2-vi2

I hope this may help you