Could someone explain this equation to me?

[Centurion]

Average rate of change of height with respect to time = [s(a+h) - s(a)]/h. So h represents a small interval around time a. S(a) represents the position of the object at time a. Um. I tried plugging arbitrary numbers into the quation to make it so it makes more sense to me.
Then the more general equation for average rate of change:
Average rate of change of f over the interval from a to a+h = [f(a+h) - f(a)]/h]
I don't get this whole concept of a + h; h is a time interval around time a...

 

[Deleted member 4993]

Average rate of change of height with respect to time = [s(a+h) - s(a)]/h. So h represents a small interval around time a. S(a) represents the position of the object at time a. Um. I tried plugging arbitrary numbers into the quation to make it so it makes more sense to me.
Then the more general equation for average rate of change:
Average rate of change of f over the interval from a to a+h = [f(a+h) - f(a)]/h]
I don't get this whole concept of a + h; h is a time interval around time a...

Suppose

f(x) = x² + 3
then

f(a) = a² + 3

and

f(a+h) = (a+h)² + 3 →

f(a+h) - f(a) = [(a+h)² +3] - [a² +3] = [(a² + 2ah + h²) +3] - [a² +3] = 2ah + h²

[f(a+h) - f(a)]/h = [2ah + h²]/h = [h(2a + h)]/h = 2a + h

 

[HallsofIvy]

If the height at t= 5 is f(5)= 4 m and 2 seconds later (t= 5+2= 7) it is f(5+ 2)= 4.1 meters then it has grown f(5+2)- (5)= 4.1- 4= .1 meters in two seconds which is an average rate of growth of .1/2= .05 meters per second. That is (f(5+2)- f(5))/2, or (f(a+h)- f(a))/h.