Question

Derive the third equation of motion [2as = v-2 - u-2] from velocity-time group [under uniform acceleration]?
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Answer 1

Third equation of motion is v^2 = u^2 + 2 as. It gives the velocity acquired by the body in travelling a distance ន.

From the second equation of motion we have :

$s = ut + \frac{1}{2}at {}^{2}$

And from the first equation of motion we have :

$v = u + at$

This and be rearranged and written as :

$at = v - u \\ or \\ t = \frac{v - u}{a}$

putting this value of t in equation ( 1 ) , we get :

s = u(v - u) / a + 1/2a(v - u/ a) ^2

s = uv - u^2 / a + a ( v^2 + u^2 - 2uv)/ 2a^2

s = vu - u^2 / a + v^2 + u^2 - 2uv / 2a

s = 2uv - 2u^2 + v^2 + u^2 - 2uv / 2a

2as = v^2 - u^2

v^2 = u^2 + 2as