Main content

## Calculus 2

### Unit 4: Lesson 2

Straight-line motion- Motion problems with integrals: displacement vs. distance
- Analyzing motion problems: position
- Analyzing motion problems: total distance traveled
- Motion problems (with definite integrals)
- Analyzing motion problems (integral calculus)
- Worked example: motion problems (with definite integrals)
- Motion problems (with integrals)
- Average acceleration over interval

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 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.

## Want to join the conversation?

- why can't we just use the first derivative of S(t)?

we are take derivative two times and take the antiderivetive once.(32 votes)- You can do exactly that and it will be the same answer. It was likely just an example to make a point.(24 votes)

- why cant we simply take the the average of acceleration function at [1,2], ?(6 votes)
- 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.(17 votes)

- At0:49, how do we know that the acceleration is the derivative of the velocity function?(2 votes)
- 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.(13 votes)

- 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.(5 votes) - 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?(2 votes)
- 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).(6 votes)

- 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(2 votes)
- 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.(3 votes)

- I haven't learned all that after2:30but 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(2 votes)
- 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! :)(3 votes)

- Is there a video that explains average acceleration over a period of time using a simpler formula, or no formula at all? For example: Within 8.0 seconds, an object's velocity on the x-axis increases from a constant velocity of 131 km/hr to 184 km/hr. How do I find the average acceleration?(2 votes)
- Are ideas of mean value and average value the same.(1 vote)
- "Mean" and "average" are the same thing.(2 votes)

## Video transcript

- [Voiceover] Let's say
that we have a particle that's traveling in one
dimension, and its position as a function of time is given as t to the third power plus
two over t-squared. What I would like you to
do is pause this video and figure out what the
average acceleration is of this particle over the
interval, the closed interval, from t is equal to one
to t is equal to two. What is this? What is this going to be equal to? Assuming you've given a
go, and the first thing you might have realized is we're
trying to get the average value of a function that we
don't know explicitly at. We know the position function but not the acceleration function. But luckily, we also know
that the acceleration function is derivative with respective
time of the velocity function which is the derivative with respect to time of the position function. The acceleration function is
the second derivative of this. We have to just find its average
value over this interval. Let's do that. Let's take the derivative of this twice. But before we do that, let
me just even rewrite this. It's just going to be a little bit easier to differentiate it. If we just take each of these
two terms in the numerator and divide them by t
squared, we're going to get t to the third, divided by
t-squared is just the t. Then two divided by T
squared, we could write that as plus two t to the negative two power. Now, let's take the derivative. The velocity function, as
the velocity as the function of time, just the derivative
with this with respect to time. It's going to be derivative
of t with respect to t as one. Derivative of two t to the negative two. Let's see, negative two times
positive two is negative four. T to the, we just decremented
the exponent here, t to the negative three power. Now, to find acceleration as
a function of time, we just find, take the derivative of
this with respect to time. Acceleration as a function
of time is equal to, actually it's already used
at color for the average so let me do a different color now. Acceleration has a
function of time is just the derivative of this with respect to t. Derivative of a constant with respect to time was not changing so it's a zero. Then over here, negative three times negative four is positive
12, times t to the, let's decrement that exponent
to the negative four power. Now to find the average
value, all we have to do now, average value is essentially
take the definitive role of this over the interval, and divide that by the width of the interval. Or we could say, we could take, we can divide by the
width of the interval. One, over two minus one, and this also simplifies to one. Times the definitive role of the interval. One to two of a of t which is, so this could be 12 t to
the negative four power d t. What is this simplified to? Once again, this is one over one. That's just going to be one. We take the antiderivative of this. Let me just, so this
is going to be equal to the antiderivative of this
is, so we go t to the negative three power but we
divide by negative three. An antiderivative of this
is going to be, if we don't take that, an antiderivative
is going to be negative four t to the negative three power,
and we saw that over here. Obviously, if you were really
just taking an in-definitive role, we would have to
put some concept here. But in the definitive role,
if we put a concept here, Assuming the same concept
that we get canceled out when you actually do a
calculation, but the entire derivative of this, we
increment the exponent, and then we divide by that new exponent. Twelve divided by negative
three is negative four. We are going to evaluate
that from two and that one. This is going to be equal to,
when we evaluated it at two at the upper bound of our
intervals, it could be negative four times two to
the negative three powers. It's negative four times, what is that? Two, that's one over to
the third of times 1/8, is one way to think about that. Then we are going to have
minus this evaluated one. Minus negative four times
t to the negative three of one to the negative three is just one. This is going to be
negative four times one. This is going to be equal to, or really in the homestretch now, this is equal to, this part right over here, is negative 1/2. This is negative 1/2, and this part right over here is positive four. Positive four minus 1/2. We could either write
that as three and a half, or if we wanted to write
that as an improper fraction, we could write this as 7/2. The average value of our acceleration over this interval is 7/2. If this position was given a
meters and time was in seconds, then this would be 7/2
meters per seconds squared, is the average acceleration between time in one second and time at two seconds.