Using calculus prove the relation between distance, time, speed and acceleration

We presume s(t) = A+Bt+Ct^2 , where s is the distance
travelled by a particle in time t with a starting velocity u and final velocity v and
acceleration a.

To find the relation between  s,a,u and
v.

At time 0, t= 0.

s(0) =
A+B*0+C*0^2 = 0. So, A =0.

Differentiate both sides of s(t)
= A+Bt+Ct^2 , with respect to time variable t, we
get:

s'(t) = B +2Ct. So when t = 0,  s'(0) = B+2C*0 = u.
So  B  = u.

s"(t) = (B+2Ct)' = (B)' +(2Ct)' = 0+2C. So 2C =
acceleration a.

So, C =
a/2.

Replacing the values for A= 0, B = u and C = a/2 in
the pressumed equation , s(t) = A+BT+Ct^2 , we get:

s(t) =
ut+(1/2)at^2...........(1).

Also B = u and velocity after
time t is v = B+2Ct or

v = u
+2(a/2)t

v =
u+at..............................(2)

So equations (1) and
(2) established the relation between the distance , initial velocity, final velocity,
acceleration and time.

Also s(t) = ut+(1/2)(at^2).
Substitute for t from u+at = v or t = (v-u)/a .

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

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

s(t) = (v^2 =
u^2)/2a.

Or

2as(t) = v^2-u^2 
is a relation bweteen  distance travelled s(t) , initial velocity vu and final velocity
v.