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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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