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# Average acceleration over interval

AP.CALC:
CHA‑4 (EU)
,
CHA‑4.C (LO)
,
CHA‑4.C.1 (EK)
As an example of finding a mean value, we find an average acceleration.

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• why can't we just use the first derivative of S(t)?
we are take derivative two times and take the antiderivetive once.
• You can do exactly that and it will be the same answer. It was likely just an example to make a point.
• why cant we simply take the the average of acceleration function at [1,2], ?
• Technically, you can calculate the average as the way you mentioned, but with very high level of inaccuracy.
Your Idea would works like a charm if the acceleration function we are dealing with was a linear function because what you are literally doing on applying your idea is that you construct a line between the start and endpoint of the interval [1,2] and take the middle point of it which gives you an overestimate result.
`(0.75+12)/2=6.375`

The reason for what happened is that the function changes continuously "how fast it changes with respect to any change in t" so building up a curve.
You need a more accurate tool that consider -nearly- every super small change in function and for that calculus comes in handy.
• At , how do we know that the acceleration is the derivative of the velocity function?
• By definition: Velocity is the rate of change in position, and therefore is directly computed from the derivative of position ( given that the rate of change of a function is by definition a derivative of that function ); then, by extension, since the definition of acceleration is the rate of change in velocity, it is therefore computed from the derivative of velocity.
• Can't we just use V(t) because it's antiderivative of A(t) already.
And 1 (or any other constants) in V(t) will cancel each other anyway.
• I first done function average to s(t) and i got 7/2 as an answer. Then i looked into the video & i realized i should have differentiated the function twice but you got the same asnwer! why is this?
• We could solve this problem without taking the second derivative, and then integrating it. If we use the mean value theorem from differential calculus, it tells us that if a function f(x) is continuous and differentiable over an interval [a,b], then there must be at least one x value in that interval for which f'(x) = ( f(b) - f(a) ) / ( b - a). In other words, if we want to find the mean value of the acceleration function, we just have to find the average slope of the velocity function between a and b. So we would only need to take the first derivative, and evaluate it at 2 and 1, subtract the two values ( f(2) - f(1) ), and divide by 2-1 = 1.

But the way Sal did it, just proves that finding the area and dividing by (b - a) works, and that it's essentially the same process as above. When taking the indefinite integral, you are finding the original function, except that we don't know the y-intercept, which does not matter because it will get canceled when we subtract f(b) - f(a).
• So we took the integral, doesn't that mean we have found the average velocity? I thought we were looking for average acceleration...? Please help :3
• The average value of any integrable function on an interval is the definite integral of that same function on the interval, divided by the length of the interval.

So the average acceleration on a time interval is the definite integral of the acceleration on the time interval divided by the length of the time interval.

Another way to see this is that the definite integral of the acceleration is the change in velocity (i.e. the final velocity minus the initial velocity), and the change in velocity divided by the length of the time interval is the average acceleration on the interval.
• I haven't learned all that after but I still need to do basically the same question with s(t)=t^4-3t^2 over the interval [0,2]. I need to know how to do cc2 part 2 calculator included for a test which is titled Rates of Change and I am finding nothing on the questions they are asking
• I don't understand.. If we want to find the average acceleration then why are we using velocity function? we should use acceleration function. What is going on with the formula of a_avg?
(1 vote)
• Well, we could use both functions. For acceleration function, you just need to input the x value and get y value which is the acceleration. When you are using velocity function, you have to use derivatives to find out the acceleration at a certain point which is an instantaneous rate of change of velocity.
Hope this helps! If you have any questions or need help, please ask! :)