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

Worked example: motion problems (with definite integrals)

AP.CALC:
CHA‑4 (EU)
,
CHA‑4.C (LO)
,
CHA‑4.C.1 (EK)
What can you say about the velocity and position of a particle given its acceleration. Created by Sal Khan.

Want to join the conversation?

  • duskpin ultimate style avatar for user okdewit
    What if the acceleration increases? Is there also a name for the derivative of acceleration in respect to time?
    (23 votes)
    Default Khan Academy avatar avatar for user
    • purple pi purple style avatar for user Josiah Blaisdell
      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
      (29 votes)
  • ohnoes default style avatar for user Cyan Wind
    Now I wonder: What quantity will we get by taking the derivative of acceleration? a'(t) = ?
    (14 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Just Keith
      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)
  • piceratops seed style avatar for user Pedro Lopes
    What if I take the antiderivative of s(t)? Is there anything such as that?
    (11 votes)
    Default Khan Academy avatar avatar for user
  • leaf red style avatar for user Kyle Gatesman
    Is there such a thing called a triple derivative? How would you apply it?
    (6 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Matthew Chen
      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)
  • leaf yellow style avatar for user Mohd. Nomaan
    what is the integration of a constant?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user learner123
    what does "c" represent?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • leaf blue style avatar for user Stefen
      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, but nothing comes 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.
      (9 votes)
  • blobby green style avatar for user taranowiczmagda
    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)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user naveed
    shouldn't the acceleration be negative since velocity is get more and more negative .
    (2 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Jimmi Aastrøm
      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)
  • piceratops ultimate style avatar for user Emi Garcia
    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)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user gracemaher98
    Are there any "golden rules" per say to remember how to solve for velocity and position from acceleration?
    (2 votes)
    Default Khan Academy avatar avatar for user

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.