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# Average acceleration over interval

As an example of finding a mean value, we find an average acceleration.

Video transcript

Let's say that we have a particle that's travelling in one dimension and it's
positioned as a function of time is given as t to the
third power plus 2 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 1 to t is equal to 2. What is this, what is this going to be
equal to? So assuming you've given a go at it, and
the first thing you might have realized is that we're trying to take the average
value of a function that we don't know
explicitly yet. We know the position function, but not the
acceleration function. But luckily, we also know that the
acceleration function is the derivative with respect to
time of the velocity function which is the
derivative with respect to time of the position
function. So the acceleration function is the second
derivative of this. And the we have to find its average value
over this interval. So let's do that. Let's take the derivative of this twice. But before we do it, let me just even
rewrite this. So it's gonna be a little bit easier to
differentiate it. So if we just take each of these two terms
in numerator and divide them by t-squared. We're gonna get t to 3rd divided by
t-squared is just t. And then 2 divided by t square, we can
write that as plus 2 t to the negative 2 power. And now lets, lets take the derivative. So the velocity function as a, as a, the
velocity as a function of time. The derivative of this with respect to
time, so it's going to be derivative of t with respect to t is
1. Derivative of 2 t to the negative 2, let's
see negative 2 times positive 2 is negative 4 t to the, and we just
decremented the exponent here. t to the negative, negative 3 power. And now to find the acceleration as a
function of time, we just find and take the derivative of this
with respect of time. So acceleration as a function of time is
equal to, actually since I've already used that color for the average,
let me do a different color now. So acceleration as a function of time is just the derivative of this with respect
to t. So derivative of a constant with respect
to time, well it's not changing, so it's
zero. And then over here negative 3 times
negative 4 is positive 12 times t to the let's decrement that
exponent to the negative 4 power. Now to find the average value, all we have
to do now, average, average value is essentially
take the definite integral of this. Over the interval And divide that by the
width of the interval. So or we could say, we could take, we
could divide by the width of the interval, one over two
minus one. And this all simplifies to one. Times the definite integral over the
interval. So one to two, of a of t, which is, so that's gonna be 12 t to the negative four
power dt. So what is this simplified to? Once again this is 1 over 1, that's just
going to be one. We take the anti derivative of this, we
actually, well let me just, this is going to be equal to the
anti derivative of this, so we're gonna go t to the negative 3 power but then we
divide by negative 3, so n anti derivative of this is going to
be if we don't take the. Well, an anti-derivative is going to be
negative 4, t to the negative 3 power, and we saw that over here, obviously, if you were
really just taking an indefinite role, we would have
to put some constant here, but in the definite rule,
even if we put a constant here, it would get can, if we, if we, assuming the same constant,
it would get cancelled out when you actually
do the calculation. But the anti-derivative of this, we
increment the exponent. And then we divide by that new exponent. So 12 divided by negative 3 is negative 4. And we're going to evaluate that from at 2
and at 1. So this is going to be equal to. When we evaluated it at 2 at the upper
bound of our interval, it's gonna be negative 4 times 2
to the negative 3 power. So it's negative 4 times what is that 2,
that's 1 over 2 to the 3rd over times one-eighth is one
way to think about that. And then we're going to have minus this
evaluated one. So minus negative 4 times t to the
negative 3. Well t to the negative 3 is just 1 so it's
going to be negative 4 times 1. And this is going to be equal to, we're
really in the home stretch now, this is equal to, this part
right over here is, negative one-half. So this is negative one-half. And this part right over here is positive
four. So, positive four, minus one-half, we
could wri, either write that as three and a half. Or if we wanted to write it as an improper fraction, we could write this as
seven halves. So the average value of our acceleration
over this interval is seven halves. If the position was given in meters and
time was in seconds. And this would be seven halves meters per
second squared is the average exceleration between time and one
second and time and two seconds.