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.
2as(t) = v^2-u^2 is a relation bweteen distance travelled s(t) , initial velocity vu and final velocity v.
Let the distance travelled by a body be only a function of time or we can represent the distance as d = f(t).
Now speed is the distance travelled in a unit time. The instantaneous speed is the derivative of the function that represents distance.
Or the speed, s(t) = f'(t)
Acceleration is the change in speed divide by the time taken. The instantaneous acceleration is the derivative of the function representing speed or s'(t).
As s(t) = f’(t), acceleration can also be represented as the second derivative of the function representing distance travelled or f''(t).