derive the 3rd equation of motion

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 = Vavx t.................. (ii)

Putting the value of Vavand 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]

Hope this works for you :-)

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(3) Third equation of Motion
v^2 = u^2 +2as

sol.
We know that
V = u + at
=> v-u = at
or t = (v-u)/a ………..eq.(3)

Also we know that
Distance = average velocity X Time
.: s = [(v+u)/2] X [(v-u)/a]
=> s = (v^2 – u^2)/2a

=>2as = v^2 – u^2

or v^2 = u^2 + 2as

This is the third equation of motion.

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