I didn't understand the 3 equation of motion
Answers
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 =
dv
dt
v =
dS
dt
Taking their ratio, we get,
a
v
=
dv
dS
Hence,
a = v
dv
dS
Cross multiplying ‘dS’ and integrating both sides,
S∫0 a . dS = v∫u v . dv ⇒ aS =
v2 - u2
2
v2 = u2 + 2aS
Graphical Proof
Following is a v − t graph displaying constant acceleration. (Slope of the curve is constant)
Curve 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 +
1
2
(v - u)(t)
(Substituting t =
(v - u)
a
, from first equation of motion)
S = u(
v - u
a
) +
1
2
(v - u)(
v - u
a
)
Rearranging the terms,
v2 = u2 + 2aS
Hope that it was useful
equation???? which equation your talking about it's not equation it's derivation