All you say is so wrong!

A constant acceleration is defined as the first derivative of velocity with respect to time or as the second derivative of displacement with respect to time. In my example we are interested on the first definition, because I want to make you reflect on what happens when the velocity changes from a positive value (first state of motion), to 0 m/s (second state of motion), then to a negative value (third state of motion).

If you do the integral of the first derivative you will get the following equation:

v = v0 + a*t

Where

t: time

v: velocity at time t

v0: initial velocity

a: acceleration

Which is the equation of a straight line. For a body that moves vertically upwards under the influence of gravity, “a” is a negative constant. As “a” plays the role of the slope of the line in our equation, velocity “v” is always decreasing starting from its initial value “v0”. It is 0 m/s when

v0 = a*t

That is to say

v = 0 when t = v0/a

At this very instant the body is at rest (v = 0 m/s)

For any other time different from t = v0/a, **no matter how close it is from this value**, the body is already moving (either upwards [v > 0m/s] for a time less than “v0/a”, or downwards [v < 0m/s] for a time greater than “v0/a”).

Therefore, from a mathematical point of view, the change from a positive velocity to 0m/s, and from here to a negative velocity happens in absolutely no time.

Therefore, it is not true that “All change involves time”.

Therefore, your conclusion is false.