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Worked example: Motion problems with derivatives

Finding (and interpreting) the velocity and acceleration given position as a function of time.

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  • primosaur ultimate style avatar for user Steve
    hmmm so if Speed is always the magnitude of the Velocity.....can it be said that Speed is always the absolute value of whatever the Velocity is?
    (9 votes)
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    • leafers ultimate style avatar for user Yan
      That is correct. However, a more rigorous way of saying it is the "modulus" instead of the "absolute value".
      Velocity is a vector, which means it has both a magnitude and a direction, while speed is a scaler.
      We call this modulus. The modulus of a vector is a positive number which is the measure of the length of the line segment representing that vector.
      (7 votes)
  • blobby green style avatar for user Jakub Klisky
    At , can you please explain in more detail how can we get the particle's direction based on the velocity?
    (6 votes)
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    • hopper cool style avatar for user Math Hopper
      Hi Jakub!

      Velocity is a vector, which means it takes into account not only magnitude but direction. We can see this represented in velocity as it is defined as a change in position with regards to the origin, over time.

      When the slope of a position over time graph is negative (the derivative is negative), we see that it is moving to the left (we usually define the right to be positive) in relation to the origin.

      Hope this helps ;)

      Please feel free to ask if anything is still unclear to you.
      (10 votes)
  • blobby green style avatar for user tofanjake
    How does distance play into all this? Like how would I find the distance travelled by the particle, using these same equations?
    (2 votes)
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    • duskpin ultimate style avatar for user Andrzej Olsen
      If you want to find the displacement, you can subtract the final x from the starting x.

      If you want to find the full length of the path, that's more challenging, and probably what you're asking for, so I'm going to show it. Let's do it from x = 0 to 3.

      To do that, just like normal, we have to split the path up into when x is decreasing and when it's increasing. We can do that by finding each time the velocity dips above or below zero. Let's do just that:

      v(t) = 3t^2 - 8t + 3 set equal to 0
      t^2 - (8/3)t + 1 = 0

      I'm gonna complete the square.

      t^2 - (8/3)t + 16/9 - 7/9 = 0
      (t - 4/3)^2 = 7/9
      t - 4/3 = ±√(7/9)
      t - 4/3 = (±√7)/3
      t = (4 ± √7)/3

      Now we know the t values where the velocity goes from increasing to decreasing or vice versa. if you put both t values in a calculator, you'll get 0.451 and 2.215, which are both in our range of 0 to 3. So, we have 3 areas to keep track of.

      0 to 0.451 | 0.451 to 2.215 | 2.215 to 3

      In each of these areas, we're guaranteed to be going in the same direction, so we don't have to worry anymore. Now we can just get the displacement in each of those and arrive at our answer.

      0 to 0.451: x(0.451) - x(0) = -1.369 + 2 = 0.631
      0.451 to 2.215: x(2.215) - x(0.451) = -4.113 + 1.369 = -2.744
      2.215 to 3: x(3) - x(2.215) = -2 + 4.113 = 0.744

      Since we just want to know the distance and not the direction, we can get rid of the negatives and add these distances up.

      Distance traveled = 0.631 + 2.744 + 0.744 = 4.119

      Ugh, why does everything I write end up being so long? Hope you stayed with me.
      (9 votes)
  • leaf orange style avatar for user Brendan
    That does not make any sense.
    If speed is increasing or decreasing isn't that just acceleration? 1. Right?
    So if the second derivative of position (aka acceleration) is positive doesn't that mean speed is increasing? 2. Right?
    And cant speed increase in a positive or negative direction (aka positive/right or negative/left velocity)? 3. Right?
    THUS, if velocity (1nd derivative) is negative and acceleration (2nd derivative) is positive. Doesn't that mean we are increase speed (aka accelerating) in a negative/left direction?

    Just the different vs same signs comment between acceleration and velocity just completely through me off.
    (1 vote)
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    • leaf green style avatar for user kubleeka
      Your first three points are correct, but your conclusion is not. If velocity is negative, that means the object is moving in the negative direction (say, left).

      If acceleration is also positive, that means the velocity is increasing. As a negative number increases, it gets closer to 0. So if we apply a constant, positive acceleration to an object moving in the negative direction, we would see it slow down, stop for an instant, then begin moving at ever-increasing speed in the positive direction.

      This is what happens when you toss an object into the air. Gravity pulls constantly downward on the object, so we see it rise for a while, come to a brief stop, then begin moving downward again.
      (10 votes)
  • blobby green style avatar for user Bryant Syme
    I'm surprised no one has asked: why is x moving down "left" and moving up "right"? Like, in relation to what? I guess if I tilt my head to the left x is moving in those directions. Am I missing something?
    (3 votes)
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    • cacteye blue style avatar for user Jerry Nilsson
      You are right that from a bystander's point of view the 𝑥-axis can be aligned in any direction, not necessarily left to right.

      Therefore, if I were given this question on a test I would not answer that the particle is moving to the left, but rather that it is moving in the negative direction of the 𝑥-axis.
      (1 vote)
  • male robot hal style avatar for user Mikiyas Alemu
    Wait a minute, I just realized something. please just hear me out.
    So we can calculate the distance traveled by a particle by finding the area between velocity time graph because distance is velocity times time right? So if we were to know the equation of the velocity function with time as an input and somehow make a function from the velocity function such that our new function's derivative is the velocity function. So from definition, the derivative of the distance function is the velocity so our new function got to be the distance function of the velocity function right? So that means the area of the velocity time graph up to a time is equal to the distance function value at that point??
    And if this true then it means we will be able find the area under EVERY DIFFERENTIABLE FUNCTION up to a point by just creating a new function whose derivative is our first function and calculating the value at that point?
    PLEASE answer this question I am too curious.
    (2 votes)
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    • leaf green style avatar for user kubleeka
      You're correct. Your observation is (half of) the fundamental theorem of calculus, that the area under a curve is described by the antiderivative of that function.

      More exactly, if f(x) is differentiable, then for any constant a, ∫_a^x f'(t)dt=f(x).
      (2 votes)
  • blobby green style avatar for user Amarachi Nwosu
    What if the velocity is 0 and the acceleration is a positive number both at t=2? Would the particle be speeding up, slowing down, or neither?
    (2 votes)
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  • blobby green style avatar for user ju lee
    when we trying to find out whether an object is speeding up or slowing down, can we just find the derivative of absolute value of velocity function? if the derivative is positive, then the object is speeding up, if the derivative is negative, then the object is slowing down. is my assumption correct?
    (2 votes)
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    • mr pink red style avatar for user V J
      Yes that is right. If derivative of the position function is > 0, velocity is increasing, and vice versa. Furthermore, to find if acceleration is increasing, you take the second derivative
      (0 votes)
  • leaf green style avatar for user 石乐志大师
    So what does the derivative of acceleration mean?
    (2 votes)
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  • purple pi purple style avatar for user shidoro
    It looks like for a function of n degree polynomial (assuming it's differentiable) if n>0 there will be at least n+1 derivatives, and for n<0 there will be infinite derivatives. Is this correct?

    Also, if a function is differentiable, does that imply all further derivatives will be differentiable until reaching a value of 0?

    How can we reason about the relationship between a function and all its nth derivatives?
    (1 vote)
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    • leaf green style avatar for user kubleeka
      First, remember that there are no polynomials of negative degree. Functions like x⁻² are rational functions, but are not polynomials.

      Both polynomials and rational functions have infinitely many derivatives. The fact that most of these derivatives are the 0 function doesn't change that.

      But you're correct that repeatedly differentiating a function eventually results in the zero function exactly when the function is a polynomial. All polynomials, and only polynomials, have this property.

      And no, a function being differentiable tells you nothing about whether its derivative is differentiable. Consider a non-differentiable function, call it f. The antiderivative of f is differentiable (since its derivative is f), but is not twice-differentiable (since f is not differentiable). And if you take the antiderivative of f a bunch of times (say, k times), the resulting function will be differentiable k times, but not k+1 times.
      (2 votes)

Video transcript

- [Instructor] A particle moves along the x-axis. The function x of t gives the particle's position at any time t is greater than or equal to zero, and they give us x of t right over here. What is the particle's velocity v of t at t is equal to two? So pause this video, see if you can figure that out. Well, the key thing to realize is that your velocity as a function of time is the derivative of position. And so this is going to be equal to, we just take the derivative with respect to t up here. So derivative of t to the third with respect to t is three t squared. If that's unfamiliar, I encourage you to review the power rule. The derivative of negative four t squared with respect to t is negative eight t. And derivative of three t with respect to t is plus three. Derivative of a constant doesn't change with respect to time, so that's just zero. And so here we have velocity as a function of time. And so if we want to know our velocity at time t equals two, we just substitute two wherever we see the t's. So it's gonna be three times four, three times two squared, so it's 12 minus eight times two, minus 16, plus three, which is equal to negative one. And you might say negative one by itself doesn't sound like a velocity. Well, if they gave us units, if they told us that x was in meters and that t was in seconds, well, then x would be, well, I already said would be in meters, and velocity would be negative one meters per second. You might also be saying, well, what does the negative means? Well, that means that we are moving to the left. Remember, we're moving along the x-axis. So if our velocity's negative, that means that x is decreasing or we're moving to the left. What is the particle's acceleration a of t at t equals three? So pause this video again, and see if you can do that. Well, here the realization is that acceleration is a function of time. It's just the derivative of velocity, which is the second derivative of our position, which is just going to be equal to the derivative of this right over here. And so I'm just going to get derivative of three t squared with respect to t is six t. Derivative of negative eight t with respect to t is minus eight. And derivative of a constant is zero. So it's just going to be six t minus eight. So our acceleration at time t equals three is going to be six times three, which is 18, minus eight, so minus eight, which is going to be equal to positive 10. All right, now they ask us what is the direction of the particle's motion at t equals two? Well, I already talked about this, but pause this video and see if you can answer that yourself. Well, we've already looked at the sign right over here. The fact that we have a negative sign on our velocity means we are moving towards the left. So I'll fill that in right over there. At t equals three, is the particle's speed increasing, decreasing, or neither? So pause this video, and try to answer that. All right, now we have to be very careful here. If it says is the particle's velocity increasing, decreasing, or neither, then we would just have to look at the acceleration. We see that the acceleration is positive, and so we know that the velocity is increasing. But here they're not saying velocity, they're saying speed. And just as a reminder, speed is the magnitude of velocity. So, for example, at time t equals two, our velocity is negative one. If the units were meters and second, it would be negative one meters per second. But our speed would just be one meter per second. Speed, you're not talking about the direction, so you would not have that sign there. And so in order to figure out if the speed is increasing or decreasing or neither, if the acceleration is positive and the velocity is positive, that means the magnitude of your velocity is increasing. So that means your speed is increasing. If your velocity is negative and your acceleration is also negative, that also means that your speed is increasing. But if your velocity and acceleration have different signs, well, that means that your speed is decreasing. The magnitude of your velocity would become less. So let's look at our velocity at time t equals three. Our velocity at time three, we just go back right over here, it's going to be three times nine, which is 27, three times three squared, minus 24 plus three, plus three. So this is going to be equal to six. So our velocity and acceleration are both, you could say, in the same direction. They are both positive. And so our velocity's only going to become more positive, or the magnitude of our velocity is only going to increase. So our speed is increasing. If our velocity was negative at time t equals three, then our speed would be decreasing because our acceleration and velocity would be going in different directions.