If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:5:07

Average acceleration over interval

CHA‑4 (EU)
CHA‑4.C (LO)
CHA‑4.C.1 (EK)

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.