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### Course: 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

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# Worked example: motion problems (with definite integrals)

What can you say about the velocity and position of a particle given its acceleration. Created by Sal Khan.

## Want to join the conversation?

- What if the acceleration increases? Is there also a name for the derivative of acceleration in respect to time?(24 votes)
- There is actually a funny story behind this that involves cereal.

I believe the colloquialism began in Great Britain. As I understand it, the derivative of acceleration is jerk, the derivative of jerk is snap, the derivative of snap is crackle, the derivative of crackle is pop. Jerk and Jounce I think are the technical names of the fourth and fifth derivatives of position. However, I've heard tell that within the academic community there is some laughter over the fact that those derivatives (while rarely used outside of theoretical physics) are named after the Rice Crispies elves :P.

http://en.wikipedia.org/wiki/Snap,_Crackle_and_Pop

see the section of physics. :D(38 votes)

- Now I wonder: What quantity will we get by taking the derivative of acceleration?
`a'(t) = ?`

(14 votes)- The derivative of acceleration is usually (and I am not making this up) called "jerk". It is called that because, if I understand correctly, a lack of uniform acceleration gives a vehicle a ride that "jerks". The term is also sometimes called the "surge".

The derivative of the jerk is called the jounce or the snap.(35 votes)

- What if I take the antiderivative of s(t)? Is there anything such as that?(11 votes)
- Yes, but it is a bit obscure. It is called the "absement".(20 votes)

- Is there such a thing called a triple derivative? How would you apply it?(6 votes)
- Yes, there is. It's the same as a double derivative, except you take the derivative 3 times. From the information from other answers. the derivative of acceleration is "jerk" and the derivative of "jerk" is "jounce". So if you took the triple derivative of position, you'd get the jerk. Triple derivative of velocity, jounce.(1 vote)

- what is the integration of a constant?(1 vote)
`∫(2)dx`

2•∫(x^0)dx

2•1/(0 + 1)•x^(0 + 1)

2•1/1•x^1

[2•x](14 votes)

- what does "c" represent?(2 votes)
- The C represents an unknown constant value.

Consider:

Differentiation and integration are inverse processes, kind of like square root operator and the square operator, each "undoes" what the other did. But remember that if we have sqrt(x²) that there could be two solution since we don't know if the original x was negative or not since the square operation turns a negative into a positive, so in essence, we lost information when x was squared; we lost if it was originally positive or negative valued number.

Well, it is kind of the same with differentiation and integration.

Differentiate the following:

f(x) = x²,

f(x) = x² + 5,

f(x) = x² - 1000,

f(x) = x² + 185673

The derivative of all of them is f'(x) = 2x, right? We lost the constant value - we lost information about the original function f(x) when we took the derivative.

So now, when you take the indefinite integral of f'(x) = 2x, you do get back the x², butcomes back about any constant value that the function may also have had originally, so we say x² + C to account for the missing constant. The constant C might be zero, or it might be 5, it could be any value that was lost during the process of differentiation.**nothing**(9 votes)

- I cannot understand this for some reason. I understand differentiation and integration but this concept baffles me. Watched this video multiple times but no clue, anyone got any tips?(3 votes)
- shouldn't the acceleration be negative since velocity is get more and more negative .(2 votes)
- It certainly should not. In order to explain why, we need to be clear about why the velocity is a negative number and what that means.

Velocity is a vector, which means that it has a magnitude (called speed) as well as a direction. In this particular case, there are only two relevant directions because we are working in a single-dimensional space. This basically means that we are only focussing on two directions like: Up and down or Right and left. The way that we differentiate between these two directions in a single-dimensional space is by the use of + and - signs. Plus usually represents up or right, while minus usually represents down or left. So the reason why the velocity is a negative number, is not because the object in motion is slowing down or anything like that - it simply means that the object is moving left or down. In addition, the absolute value of the velocity is going to be the speed (speed cannot be negative - you cannot drive negative 70 mph). So looking only at the magnitude of the number which represent velocity, you know that the object is speeding up if that absolute value increases as the variable t (time) increases.

In the video, Sal is showing an example of an object moving with constant accelleration. This means that we are looking at an object which keeps speeding up at a constant rate. This is why the velocity gets more and more negative - because the object is speeding up in a left or down direction.

If the acceleration had been (for example) -1, then the object would have deaccelerated at a constant rate and eventuelle come to a standstill and accellerate in the opposite direction (deacceleration is basically just acceleration in the opposite direction of motion). For example: If you throw an object into the air, then that object is constantly deaccelerated (or accelerated in the opposite direction; towards earth) by the force of gravity. Due to this force, the object will eventuelly cease its upwards motion and come to a halt in midair before dropping back down to earth (accelerated towards earth).(6 votes)

- why can't we use this to find the position and momentum of a particle. It was something about the heisenberg uncertainty principle but i don't understand why.(2 votes)
- We can determine the exact position OR momentum, but not both at the same time. The act of measuring the position and momentum creates the uncertainty.(3 votes)

- Are there any "golden rules" per say to remember how to solve for velocity and position from acceleration?(2 votes)
- Let
`s`

= position,`v`

= velocity,`a`

= acceleration.

Starting with`s`

:`v = ds/dt`

,`a = dv/dt`

Starting with`a`

:`v = ∫(a)dt`

,`s = ∫(x)dt`

(2 votes)

## Video transcript

Let's review a little bit of what we
learned in differential calculus. Let's say we have some function S that it
gives us as a function of time the position of a
particle in one dimension. If we were to take the derivative with
respect to time. So if we were to take the derivative with
respect to time of this function S. What are we going to get? Well we're going to get ds, dt or the rate in which position changes with
respect to time. And what's another word for that, position
changes with respect to time? Well, that's just velocity. So that we can write as velocity as a
function of time. Now, what if we were to take the
derivative of that with respect to time. So we could either view this as the second
derivative, we're taking the derivative not once, but twice
of our position function. Or you could say that we're taking the derivative with respect to time of our
velocity function. Well this is going to be, we can write
this as, we can write this as dv, dt, the rate at which velocity
is changing with respect to time. And what's another word for that? Well, that's also called acceleration. This is going to be our acceleration as a
function of time. So, you start with the position function,
take it, the position as a function of time. Take its derivative with respected time,
you get velocity. Take that derivative with respected time,
you get acceleration. Well, you could go the other way around. If you started with acceleration, if you
started with acceleration, and you were to take
the antiderivative. If you were to take the antiderivative of
it, the anti, anti, an antiderivative of it is going to be, actually let me just
write it this way. So an antiderivative, I'll just use the
interval symbol to show that I'm taking the
antiderivative. Is going to be the integral of the
anti-derivative of a of t. And this is going to give you some
expression with a plus c. And we could say, well, that's a general
form of our velocity function. This is going to be equal to our velocity
function. And to find the particular velocity
function, we would have to know what the velocity is at a
particular time. And then, we could solve for our c. Whether then, if we're able to do that and
we were to take the anti-derivative again. Then, now we're taking the anti-derivative
of our velocity function, which would give us some expression as a
function of t. And then, some other constant. And, if we could solve for that constant,
then we know, then we know what the position is going to be, the
position is a function of time. Just like this, it would have some, plus c
here if we know our position at a given time we could
solve for that c. So now that we've reviewed it a little
bit, but we've rewritten it in. I guess you could say, thinking of it not
just from the differential point of view from the
derivative point of view. But thinking of it from the
anti-derivative point of view. Lets see if we can solve an interesting
problem. Lets say that we know that the
acceleration of a particle is a function of time is equal to
one. So it's always accelerating at one unit
per, and you know, I'm not giving you time. Let, let's just say that we're thinking in
terms of meters and seconds. So this is one meter per second, one meter
per second-squared, right over here. That's our acceleration as a function of
time. And, let's say we don't know the velocity
expressions, but we know the velocity at a particular time and we
don't know the position expressions. But we know the position at a particular
time. So, let's say we know that the velocity,
at time three. Let's say three seconds is negative three
meters per second. And actually I wanna write the units here,
just to make it a little, a little bit. So this is meters per second squared, that's going to be our unit for
acceleration. This is our unit for velocity. And let's say that we know, let's say that
we know that the position at time two, at two seconds is equal to
negative ten meters. So, if we're thinking in one dimension, of
if this is moving along the number line, this is ten to the
left of the origin. So, given this information right over
here, and everything that I wrote up here. Can we figure out the actual expressions
for velocity as a function of time? So not just velocity at time three, but
velocity generally as a function of time. And position as a function of time. And I encourage you to pause this video
right now. And try to figure it out on your own. So let's just work through this. What is, we know that velocity, as a function of time, is going to be the
anti-derivative. The anti-derivitive of our acceleration is
a function of time. Our acceleration is just one. So this is going to be the anti-derivitive
of this right over here is going to be t and then we can't forget
our constant plus c. And now we can solve for c because we know
v of 3 is negative 3. So lets just write that down. So v of 3 is going to be equal to 3, 3
plus c. I just replaced where I saw the, the t or
every place where I have the t I replaced it with this 3
right over here. Actually let me make it a little bit
clearer. So v of 3, v of 3 is equal to 3 plus c,
and they tell us that that's equal to negative 3. So that is equal to, that is equal to
negative 3. So what's c going to be? So if we just look at this part of the equation or just this equation right
over here. If you subtract 3 from both sides, you
get, you get c is equal to negative 6. And so now we know the exact, we know the exact expression that defines velocity as
a function of time. V of t, v of t is equal to t, t plus negative 6 or, t minus 6. And we can verify that. The derivative of this with respect to
time is just one. And when time is equal to 3, time minus 6
is indeed negative 3. So we've been able to figure out velocity
a s a function of time. So now let's do a similar thing to figure
out position as a function of time. We know that position is gonna be an
anti-derivative of the velocity function, so let's write
that down. So, position, as a function of time, is
going to be equal to the anti-derivative of v of t,
dt. Which is equal to the anti-derivative of t
minus 6, dt which is equal to well the anti-derivative of t, is
t squared over 2. So, t squared over 2, we've seen that
before. The anti-derivative of negative 6 is
negative 6 t, and of course, we can't forget our constant, so,
plus, plus, c. So this is what s of t is equal to, s of t is equal to all of this business
right over here. And now we can try to solve, for our
constant. And we do that, using this information
right over here. At two seconds were at, our position is
negative two meters. So s of 2, or I could just write it this
way. Well, let me write it this way. S of 2, at 2 seconds, is going to be equal
to 2 squared over 2. That is, let's see, that's 4 over 2,
that's going to be 2. Minus 6 times 2. So, minus 12 + c is equal to negative 10,
is equal to negative 10. So, let's see, we get 2 minus 12 is negative, is negative 10 plus c equals
negative 10. So you add 10 to both sides you get c in
this case is equal to 0. So we figured out what our position
function is as well. The c right over here is just going to be
0. So our position as a function of time is
equal to t squared over 2 minus 6 t. And you can verify. When t is equal to 2, t squared over 2 is
2, minus 12 is negative 10. You take the derivative here, you get t
minus 6. And you can see and we already verified
that v of 3 is negative 3. And you take the derivative here, you get
a of t, just like that. Anyway, hopefully, you found this
enjoyable.