# Derivation of equation of motion

1
by varun4

2015-09-22T20:42:13+05:30

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We know that
a= v-u / t
v-u = at
therefore v = u +at ; This is our first eq of motion
velocity = distance traveled / time taken v =  s/t => s= vt

If u is the initial velocity of an uniformly accelerating body and v is its velocity after a time t, then since the acceleration is UNIFORM, we can find the average velocity of the body as follows average velocity = (u+v)/2 Now, the distance s, traveled in the time t by the body is given by distance traveled = average velocity x time s = [(u+v)/2]t From eq 1 we have v=u+at, substituting this in the above equation for v, we get s = [(u+u+at)/2]t => s = [(2u+at)/2]t => s = [(u + (1/2)at)]t               => s =  ut + (1/2)at2 ; This is our 2nd eq of motion

We start with squaring eq 1 Thus we have v2 = (u+at)2                  =>    v2 = u2 + a2t2 + 2uat            =>    v2 = u2 + 2uat + a2t2              =>    v2 = u2 + 2a(ut + (1/2)at2) now, using eq 2  we have =>   v2 = u2 + 2as
That was the third eq of motion.

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