# Average velocity or average values of functions and calculus

## Main Question or Discussion Point

Hi, I have some questions regarding values of function and average acceleration/velocity and really just want to make sure i am right :)

if a function, let's say acceleration = 2 m/s^2 is a constant
Then the average acceleration = (v2-v1)/(t2-t1) between two points is the slope of a straight line, this straight line is the Velocity function v(t) =2*t. and the slope is 2, so the average acceleration is constantly 2.

The average Speed would then be (X(t2)-X(t1))/(t2-t1) the difference between two points on the distance vs time graph, and the function for this graph would be the integral of velocity function so x(t)= t^2.
However average velocity would not be constant because the magnitude of velocity depends on time, so if time is large, the average velocity would be large. The average velocity would be the slope of the secant connecting two points on the position vs time graph.

Herrer is my question:
1) When we find the average acceleration we do it in terms of the velocity function, and when we find the average velocity we do it in terms of the position function. So average acceleration is NOT the slope of the acceleration function , and average velocity is NOT the slope of the velocity function. So what are we actually finding mathematically speaking? I can see we are finding the secant which connects two points of the integral function. However the function of distance could increase for 1 year, and it if suddently dropped to zero after a year, then the average velocity between the point or origen (t0,x0) and the point where the distance dropped to zero would be zero, so the average velocity value can not be the same as the average function value of the velocity function,right? So is this correct: The average velocity between 2 points is (X(t2)-X(t1))/(t2-t1) which is the same as
integral((v(t),(v(t2),v(t1))/(t2-t1) so it's actually the area under the curve of the velocity function divided the time interval (or the difference in t2-t1), and since the area under the velocity function between two points is the same as the function value of the distance function at the point t2, then the average velocity actuallymeans

(volume of acceleration function between t2 and t1)/(Time difference)

So the larger the volume of a(t) increases per time, the larger the average velocity? and so if the volume drops below zero (negative decelleration) and decellerates until the velocity stops, then the average velocity is zero because the vulime of acceleration is zero between those two times?

However the function of distance could increase for 1 year, and it if suddently dropped to zero after a year, then the average velocity between the point or origen (t0,x0) and the point where the distance dropped to zero would be zero
If the distance drops to zero at a point then you're describing a teleportation, in which case the velocity is undefined. But let's say for argument's sake that the distance function is not discontinuous but simply very quickly reduces from its value at 1 year back to zero after 1 year and 1 second. Then that's a very great distance to cover in one second, corresponds to a momentary very high velocity. So a year of very low velocity plus a second of extremely high velocity, when averaged over time, gives you zero net velocity.

Stephen Tashi