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## AP®︎ Calculus AB (2017 edition)

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

Lesson 3: AP Calculus AB 2011 free response- 2011 Calculus AB free response #1a
- 2011 Calculus AB Free Response #1 (b, c, & d)
- 2011 Calculus AB free response #2 (a & b)
- 2011 Calculus AB free response #2 (c & d)
- 2011 Calculus AB free response #3 (a & b)
- 2011 Calculus AB free response #3 (c)
- 2011 Calculus AB free response #4a
- 2011 Calculus AB free response #4b
- 2011 Calculus AB free response #4c
- 2011 Calculus AB free response #4d
- 2011 Calculus AB free response #5a
- 2011 Calculus AB free response #5b
- 2011 Calculus AB free response #5c
- 2011 Calculus AB free response #6a
- 2011 Calculus AB free response #6b
- 2011 Calculus AB free response #6c

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# 2011 Calculus AB free response #1a

Determining whether speed is increasing. Difference between speed and acceleration. Created by Sal Khan.

## Want to join the conversation?

- As someone who isn't familiar with what AP/AB placement is, could someone explain it briefly so I can think of a local comparison?(8 votes)
- AP stands for Advanced Placement, while AB could stand for Calculus AB which represents the first part of the Calculus cumulative course. The AP exam includes an AP Calculus AB exam, where the AB stands quite simply for the 1st section of Calculus taught according to the AP standards.(22 votes)

- At7:01: Why is the speed increasing?(4 votes)
- So unlike velocity, speed is a scalar quantity. That is, it only has a magnitude (not direction). So, at7:01when Sal references the particle's speed and says that it is "increasing", he's doing so with the knowledge that the particle's velocity vector is moving in the negative direction. Try picking a negative number and moving in the negative direction, the magnitude (or absolute value) of the number increases, doesn't it? It is confusing, but this is all he's saying here.

I hope this helps.(16 votes)

- Whatever happened to the video for parts b, c, & d??(13 votes)
- What are the applications of Calculus?(3 votes)
- Calculus has
*many*applications. It has so many applications that I cannot begin to describe them to you. Calculus pervades math (obviously) and the sciences. Even a scientific discipline like biology can use calculus to study rates of diffusion etc. It also has very important implications for the study of physics (which was kind of why it was invented) and engineering. Whenever you play a video game that involves laws of physics (like gravity, magnets...some sort of attractive forces) you can bet that there is calculus involved. The machine may not be doing integrals like you can, but it makes numerical calculations based on the same principles of Riemann sums that you have learned.

The possibilites are limitless!(17 votes)

- At around1:20: What is the magnitude?(5 votes)
- Remember, velocity is a vector quantity, so it has both direction and magnitude. In the example at1:20, the particle said to be moving in the negative direction and with a magnitude of 5.(4 votes)

- Do I use radians or degrees for this problem?(2 votes)
- You should never use degrees in calculus. It is always radians unless for some bizarre reason some problem asks for degrees. Even if you do use degrees, you have to effectively convert them into radians to do the math.(5 votes)

- I'm confused by everything concerning the velocity here. Can anyone re - explain it to me, please?(2 votes)
- Velocity is like having a line and moving on it

moving right is positive

moving left is negative

Your speed is how fast you are moving

Your velocity basicaly tells you where you are going to be in a certain amount of time

It tells you how far and in what direction

If you say that your velocity is -15 fps then your are going left at a speed of 15 fps.

While Saying that your velocity is 15 fps means that you are going right at at a speed of 15 fps.

Same speed, different direction, different velocity.(5 votes)

- Correct me if I am wrong:

If acceleration is positive and velocity is positive, then velocity is increasing and speed is increasing.

If acceleration is positive and velocity is negative, then velocity is increasing but speed is decreasing.

If acceleration is negative and velocity is positive, then velocity is decreasing and speed is decreasing.

If acceleration is negative and velocity is negative, then velocity is decreasing but speed is increasing.(3 votes) - At around1:41, what you're saying is that speed is equal to the absolute value of the velocity?(2 votes)
- yeah, speed is the absolute value (magnitude) of the velocity, whether it be one dimensional or a two dimensional vector or a three dimensional vector(3 votes)

- I am still confused about the speed and velocity thing at7:12. What is that about?(1 vote)
- Speed indicates how fast something is moving.

Velocity indicates how fast something is moving in a certain direction.

Hence, speed is a positive number while velocity is positive or negative depending on the direction.(4 votes)

## Video transcript

Problem 1. For 0 is less than or equal to
t is less than or equal to 6, a particle is moving
along the x-axis. The particle's position x of
t is not explicitly given. The velocity of the
particle is given by v of t is equal to all of this
business right over here. The acceleration of the
particle is given by a of t is equal to all of this
business over here. They actually didn't
have to give us that because the
acceleration is just the derivative of the velocity. And they also gave us-- they
don't tell us the position function, but they tell us
where we start off, x of 0 is equal to 2. Fair enough. Now let's do part
A. Is the speed of the particle increasing
or decreasing at time t equals 5.5? Give a reason for your answer. It looks like they did
something a little sneaky here because they gave us
a velocity function and then they ask about a speed. And you might say, wait,
aren't those the same thing? And I would say, no, they
aren't quite the same thing. Velocity is a magnitude
and a direction. It is a vector quantity. Speed is just a magnitude. It is a scalar quantity. And to see the difference,
you could have a velocity-- and this isn't maybe
particular to this problem because they don't
give us the units-- but you could have a velocity
of negative 5 meters per second. And maybe if we're talking
about on the x-axis, this would mean
we're moving leftward at 5 meters per
second on the x-axis. So the magnitude is
5 meters per second. This is the magnitude. And the direction is specified
by the negative number. And that is the direction. Your velocity could be
negative 5 meters per second, but your speed would just
be 5 meters per second. So your speed is 5
meters per second whether you're going to
the left or to the right. Your velocity, you actually
care whether you're going to the left or the right. So let's just keep
that in mind while we try to solve this problem. So the best way to figure out
whether our rate of change is increasing or decreasing is
to look at the acceleration. Because acceleration
is really just the rate of change of velocity. And then, we can
think a little bit about this velocity
versus speed question. So what is the
acceleration at time 5.5? Get the calculator out. We can use calculators for
this part of the AP exam. And I assume they intend us to
because this isn't something that's easy to
calculate by hand. So the acceleration at time 5.5,
we just have to say t is 5.5 and evaluate this function. So 1/2-- I'll just write
0.5-- times e to the t over 4. Well t is 5.5. 5.5 divided by 4. And then, times cosine of 5.5
divided by 4 gives us 0.38. Did I do that right? We have 0.5 times e to
the 5.5 divided by 4, times cosine-- oh, sorry. I made a mistake. That looked a little strange. It's not cosine of
5.5 divided by 4, it's cosine of e to
the 5.5 divided by 4. So let's look at that. So that's one
parentheses I close. And now that is the second
parentheses that I've closed. And I get negative 1.3-- well, I
just say roughly negative 1.36. So this is equal to,
or approximately equal to, negative 1.36 if I round it. And we don't care about so
much as the actual value. What we really care
about is its sign. So the acceleration at
time 5.5 is negative, which tells us that the
velocity is decreasing. Now, you might be tempted
to say, we're done. But remember, they're
not asking us, is the velocity of the particle
increasing or decreasing? They're asking us, is
the speed of the particle increasing or decreasing? And if you're saying,
hey, how did I know that, just remember
acceleration is just the rate of change of velocity. If acceleration
is negative, that means the rate of
change of velocity is negative It is going down. But anyway, how do we
address this speed issue? How do we think about it? Well, there's two scenarios. If our velocity is
positive at time 5.5, so if we have a
positive velocity, so let's say our velocity
is 5 meters per second-- although they don't
give us units here, so I won't use units. So let's say our velocity is 5. And then it's negative. So at some point our velocity is
going to be something smaller. Then that means that the speed
would also be decreasing. So if we have a
positive velocity, then the fact that
acceleration is negative means that both velocity and
speed would be decreasing. On the other hand, if we had
a negative velocity at time t equals 5.5, then the
fact that it's decreasing means that we're getting
even more negative. And if we're getting
even more negative, then that means the
speed is increasing. The magnitude is increasing
in the leftward direction. So what we really need to
do, beyond just evaluating the acceleration at
time 5.5, we also have to evaluate
the velocity to see if it's going in the left
or the rightward direction. So let's evaluate the velocity. The velocity at 5.5-- and we'll
just get our calculator out again. Velocity at 5.5-- this is
our velocity function-- is going to be equal
to 2 times the sine of-- let me write it this way. Just because I want to make
sure I get my parentheses right. 2 times the sine-- let me
write it this way-- 2 sine of e to the 5.5, that's our
time, divided by 4. So I did that part
right over here. And I'm going to
close the 2 as well. And then plus 1. So this is our velocity. So our velocity at
time 5.5 is negative. So negative 0.45. Negative 0.45 roughly. So this is negative. The velocity is negative. So we have this scenario where
the velocity is negative, which means we're going
in the left direction. And the fact that the
velocity is also decreasing means that over time-- at
least at this point in time-- as we go forward in time, it'll
become even more negative. And it'll become
even more negative if we wait a little bit longer. So that means that the magnitude
of the velocity is increasing. It's just going in the
leftward direction. So if the magnitude
of the velocity is increasing
although it's going in the leftward
direction, that means that the speed is increasing. So the velocity--
so this is one of those interesting scenarios--
the velocity is decreasing. But the speed, which is
what they're asking us in the question,
speed is increasing. And if you wanted to do
it really quick with all of this explanation I gave
you, you would say, hey, look. What's acceleration? Is it positive or negative? You would evaluate it. You'd say, hey, it's negative. So you know velocity
is decreasing. And then you would say,
hey, what is velocity? Is it positive or negative? You evaluate it. You say it's negative. So you have a negative
value that is decreasing. So it's becoming more negative. So that means its
magnitude is increasing, or speed is increasing.