Question

How to drive the 3rd equation of motion?

Answer 1

hey mate! here ur answer:)Third Equation

v2 = u2 + 2aS

where,

v = final velocity,

S = displacement,

u = initial velocity,

a = acceleration (must be constant),

This equation DOES NOT relate to final velocity.

Analytical Proof

We know that,

a = dvdt

v = dSdt

Taking their ratio, we get,

av = dvdS

Hence,

a = vdvdS

Cross multiplying ‘dS’ and integrating both sides,

S∫0 a . dS = v∫u v . dv ⇒ aS = v2 - u22

v2 = u2 + 2aS

Graphical Proof

Following is a v − t graph displaying constant acceleration. (Slope of the curve is constant)



At t = 0 seconds, the particle’s velocity is u m/s.

At t = t seconds, the particle’s velocity is v m/s.

Area under the curve of v − t graph gives displacement.

Now,

S = Area of Rectangle + Area of Triangle

S = ut + 12(v - u)(t)

(Substituting t = (v - u)a, from first equation of motion)

S = u(v - ua) + 12(v - u)(v - ua)

Rearranging the terms,

v2 = u2 + 2aS

hope it helps☺
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