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## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition) > Unit 12

Lesson 2: AP Calculus AB 2015 free response- 2015 AP Calculus AB/BC 1ab
- 2015 AP Calculus AB/BC 1c
- 2015 AP Calculus AB/BC 1d
- 2015 AP Calculus AB 2a
- 2015 AP Calculus AB 2b
- 2015 AP Calculus 2c
- 2015 AP Calculus AB/BC 3a
- 2015 AP Calculus AB/BC 3b
- 2015 AP Calculus AB/BC 3cd
- 2015 AP Calculus AB/BC 4ab
- 2015 AP Calculus AB/BC 4cd
- 2015 AP Calculus AB 5a
- 2015 AP Calculus AB 5b
- 2015 AP Calculus AB 5c
- 2015 AP Calculus AB 5d
- 2015 AP Calculus AB 6a
- 2015 AP Calculus AB 6b
- 2015 AP Calculus AB 6c

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# 2015 AP Calculus AB/BC 3cd

Accelerations and average velocity from velocity function.

## Want to join the conversation?

- Why do we integrate B(t) between the intervals of 0 to 10? Why can we not substitute 0 and 10 into the function of B in order to find the average velocity?(1 vote)
- That would only work if the function was perfectly linear.(4 votes)

- Why don't we integrate the function we found in part C between 0 to 10 to find average velocity? Why does it have to be B(t)?(1 vote)
- Why do we integrate B(t) between the intervals of 0 to 10? Why can we not substitute 0 and 10 into the function of B in order to find the average velocity?(1 vote)
- Why do we integrate B(t) between the intervals of 0 to 10? Why can we not substitute 0 and 10 into the function of B in order to find the average velocity?(0 votes)
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## Video transcript

- [Voiceover] Bob is riding his
bicycle along the same path. For zero is less than or equal to t is less than or equal to 10. Bob's velocity is modeled
by B of t is equal to t to the third minus
six t squared plus 300, where t is measured in minutes
and B of t is measured in meters per minute. Find Bob's acceleration
at time t equals five. Well acceleration, this is a velocity
function right over here, so the acceleration is
going to be derivative of the velocity function
with respect to time. What is the rate of change of
velocity with respect to time? That's acceleration. So we really just want to evaluate this, Bob's acceleration at t equals five, that's going to be B prime of five. So let's first figure
out what B prime of t is. B prime of t is equal to, we'll take the derivative here. It's pretty straightforward. Just use the Power Rule. So it's going to be three
t squared minus 12t, two times negative six
is 12 or negative 12, and then the derivative
of 300 doesn't change with respect to time, so it's just a zero. And so B prime of five is
going to be equal to three times five squared minus 12 times five which is equal to 75 minus 60 which is equal to 15, and the units here, this is acceleration. So this is going to be, his velocity was in meters per minute, and so this is going meters per minute, per minute because remember, time is in terms of minutes. So we could write as meters, and we write it out meters per minute, per minute which is the same
thing as meters per minute, meters per minute squared. All right. Let's do the next part. Based on the model B from part c, find Bob's average velocity
during the interval from zero is less than or equal to t is less than or equal to 10. And if the notion of average velocity or average value of a function
is completely foreign to you, I encourage you to watch the videos on Khan Academy on find
the average of a function. But just to kind of cut to the chase, the average velocity, the average velocity, is going to be the area
under the velocity curve divided by our change in time. So the area under the velocity
curve from t equals zero to t equals 10 of B of t of B of t, dt, divided by our change in time. So it's going to be divided by, well you're going from zero to 10. So 10 minus zero is
going to be equal to 10. And if you wanted the
intuition here it's like well, if you know the area of
something and if you wanted to find its average height, you could just divide by its width. And that's what we're doing here. If we know the area of something, we want to figure out its average height, and so you divide by its width. That's I guess a very
high-level intuition for where this expression came from. And so this is going to be
equal to one tenth times the integral from zero to 10, and B of t is t to the
third power minus 6t squared plus 300 dt, and so this is going to
be equal to one over 10. Take the anti-derivative here, so this is going to be t
to the fourth over four, t to the fourth over four, and then this is going to be, if we increase the exponent here by what is t to the third, and then you divide by three. So it's negative six divided by three is negative two, t to the third, and then plus 300t, 300t, And I'm going to evaluate it. I am going to evaluate it
at 10 and subtract from that and evaluate it at zero. And so this is going to be equal to, this is going to be equal to one tenth, that same one tenth there, and when you evaluate all this at 10 what'd everybody get? Let's see 10 to the fourth power is, 10 to the fourth power is
10,000 divided by four is 2,500, 2,500, and then minus two times 10 to the third. So it's two times a thousand. So minus 2,000. And then 300 times 10, well that's plus 3,000, and then you subtract all
of this evaluated at zero which is just going to be zero. So this going to be equal to, 2,500 minus 2,000 is 500 plus 3,000. This is all, simplifies to 3,500, or 3,500, and then you divide it by 10. This is going to be 350, 350, and it's an average
velocity meters per 350, meters per minute. And we are done.