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Motion problems: when a particle is speeding up

The position of a particle moving along the x-axis is given by s(t)=t³-6t²+9t. Sal analyzes it to find the times when the particle is "speeding up.". Created by Sal Khan.

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  • stelly blue style avatar for user Julian Delgadillo Marin
    If acceleration = s(t) is the second derivative of s(t), whats are we finding out in the third derivative or even fourth derivative of s(t) ?
    (3 votes)
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    • blobby green style avatar for user Creeksider
      The third derivative is the rate of change in acceleration. For example, if you keep the accelerator pedal in a car pressed to the floor, the car will eventually reach maximum speed and stop accelerating (or minimum speed, crumpled against a tree). Anyway, the third derivative is often (but not universally) called jerk, and the rate of change in jerk is (again, not universally) called jounce. Jerk and jounce can be important in some systems, such as controlling the flight of drones.
      (14 votes)
  • piceratops sapling style avatar for user DEEP
    Shouldn't we be speeding up between second and third second?
    (4 votes)
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    • leaf grey style avatar for user James L.
      Think of it this way: Between second and third the particle is still moving in the left direction (because velocity is negative) but you are accelerating in the right direction (because acceleration is positive, i.e. the slope of velocity is positive). So think of it as the particle is slowing down in the left direction and therefore you are not speeding up, you are actually slowing down. It is only when the particle switches direction to the right that it is speeding up (i.e. velocity turns positive at t > 3). Hope this helps!
      (7 votes)
  • marcimus pink style avatar for user moooriah101
    At , why did he plug in 2 and 4 into the original equation to find the minimum?? Shouldn't he have just picked one number??
    (7 votes)
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    • spunky sam green style avatar for user Muhammad Nawal
      The derivative is a second degree polynomial thus it is a parabola .......... when sal found its two roots at 1 and 3 ........ it is understood that the vertex of parabola will exactly be between them because the symmetry of parabola.... i.e. x=2 ....and for y value he plugged x=2 .....3(2)^2-12(2)+9.......3(4)-12(2)+9....
      (2 votes)
  • blobby green style avatar for user keshavnemeli
    Is the acceleration decreasing or increasing in the interval 1<t<2? is the acceleration=0 at t=1 and at t=2 because the slopes at these points are zero. if between t=1 n t=2 if it is becoming more and more negative how can it become zero at t=2(slope =0) because it is increasing in the negative direction?
    (4 votes)
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    • leafers ultimate style avatar for user KrisSKing
      Your question "Is the acceleration decreasing or increasing" is asking about the change in acceleration. That concept is not needed to answer the question of when is the particle "speeding up", but it is a good question. On the interval 1 < t < 2 the acceleration is negative, but it is increasing. You can see this by imagining the tangent line to the velocity function in that interval. That tangent line is slanted down, so the slope of that line is negative, so the acceleration is negative. But as you move the tangent line to the right, its slope becomes less and less negative, the slope is increasing. Acceleration is the derivative of velocity. Sal didn't do this, but you can take the derivative of the velocity function and get the acceleration function:

      v'(t) = a(t) = 6t - 12

      From looking at the acceleration function you can also figure out the acceleration is negative but increasing from t = 0 to t = 2. From t = 0 to 2, the acceleration is going to be negative, at t = 2 the acceleration is zero, and at t > 2 the acceleration is positive. The function for acceleration is a linear function with a slope of positive 6, so the function is always increasing. Another way to see the acceleration is always increasing is to take the derivative of the acceleration function:

      a'(t) = 6

      So the acceleration is always increasing at a rate of 6/1.
      (7 votes)
  • piceratops ultimate style avatar for user Aubert Roy
    Since v(t) = ds / dt and a(t) = dv / dt, could you write it as a(t) = d(ds / dt) / dt?
    (4 votes)
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  • male robot johnny style avatar for user Hans Liu
    How did Sal automatically know the vertex of the parabola was at (2, -1)?
    (1 vote)
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  • starky seedling style avatar for user Alma Ionescu
    So the largest exponent on the original equation points out how many times is the particle speeding up? And the coefficient on the first term shows the magnitude of the speeding?
    (2 votes)
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    • purple pi purple style avatar for user doctorfoxphd
      Well, the short answer is "no and no".
      The original equation describing movement on a line was
      s(t) = t³ - 6t² +9t
      In this case, there were two times when the particle was speeding up
      That would be where the velocity and acceleration are both positive OR both negative in sign.
      t(1, 2) Time between 1 and 2 seconds
      t(3, ∞) Time greater than 3 seconds.
      We do not count anything that happens before the clock starts ticking at time = 0

      The coefficient on the first term doesn't show the magnitude of speeding. The coefficient on the first term is one. That doesn't match any way I can interpret your words, "magnitude of the speeding".

      If you look at the graph of the acceleration (the rate of change for the velocity), it has a slope of 6. The only equation that has a coefficient of 6 on the first term is the second derivative of s(t), which is the equation for the acceleration, and of course its slope would equal that coefficient.
      a(t) = s"(t) = 6t - 12

      However, if you look at the slope of the velocity, at every point it is changing: after all, it is a parabolic function. Sal walked us through how the slope is negative and flattens out to zero and then becomes a little positive, and then very steeply positive. So if you mean from that graph that it is speeding up according to the coefficient of that equation, that also doesn't work. The equation for the velocity is v(t) = s"(t) = 3t²- 12t + 9

      The main shortcut that I know is to quickly do the single and double derivatives and examine the behaviors of those curves. The powerful one for this purpose is the curve of the velocity or s'(t) as Sal showed us.
      (3 votes)
  • leafers seed style avatar for user Miquela Goodson
    At , why is the particle speeding up between 1<t<2? Shouldn't it be slowing down since the particle's velocity reaches 0 at t=2?
    (1 vote)
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  • piceratops seed style avatar for user Solomon
    How can we say that velocity is increasing only when it goes to right direction and decreases when it goes to left?
    (2 votes)
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    • spunky sam blue style avatar for user Vishnu Gopalakrishnan
      We don't. We say that the velocity is increasing when there is positive acceleration and when there is positive acceleration the velocity will either climb to higher negative numbers ( which is to say from -7 to -2 for example) if negative and will climb to higher positive numbers if positive. If your question was why did Sal define velocity increasing in the rightward direction, it is because in this video we picked the convention that if velocity is positive and increasing then it is travelling rightward.
      (1 vote)
  • leafers ultimate style avatar for user Sam Claxton
    In this video Sal says that for the particle to be "speeding up" (or in other words the magnitude of the velocity must be increasing). He states that the particle is "speeding up" when 1 < t < 2. However surely the object is speeding up at t = 1 as the velocity = 0 but the acceleration is negative and so its velocity will increase. Similar logic should also apply when t = 3 Therefore shouldn't the answer be 1 ≤ t < 2, t ≥ 3? Thanks for your replies in advance!
    (2 votes)
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    • female robot grace style avatar for user Peregrine Void
      I’m not sure I got it right but here is my reasoning:
      At t = t₀ (in our problem, 1 or 3), acceleration is non-zero, but velocity is zero. It means that our point is not moving − yet. It is about to move! At t = t₀ + δt, velocity becomes non-zero and we can at last say that the point is speeding up.
      (1 vote)

Video transcript

Let's say that we have some particle that's moving along the number line. So let me draw a number line right over here. So that's our number line right over there. And let's say it starts right over here at 0. And then as time passes, this little point is going to move around. Maybe it moves to the right, slows down, speeds up. Maybe it moves to the left, slows down, speeds up. It might do all sorts of things. And to describe this motion, its position as a function of time, we have a function s of t. This particle's position as a function of time we're given is t to the third power minus 6t t squared plus 9t. And we're going to restrict the domain to positive time. So we're going to assume that time is greater than or equal to 0. Now the question that we want to answer in this video is, when is this particle speeding up? So when are we speeding up? And I think that bears a little clarification. What does it mean to speed up? Well, there's two scenarios. If the particle is already moving in the rightward direction-- and the way we would know it's moving in the rightward direction is if its velocity is greater than 0. If it's moving in the rightward direction and it's also accelerating in the rightward direction-- so if its acceleration is also greater than 0-- then this is a situation where we are speeding up. Now another scenario where we would be speeding up is if we're moving in the leftward direction. In that case, our velocity is going to be negative. So if our velocity is negative and we want to go faster in the negative direction, then our acceleration should also be negative. That would make our velocity getting more and more and more negative with time. So then our acceleration needs to also be negative if we still want to be speeding up. If you have any other combination here, if your velocity is negative but your acceleration is positive, that means that your velocity is becoming less negative, or you would be slowing down the leftward direction. And vice versa, if your velocity is positive and your acceleration is negative, that means you're going to the right but you are slowing down in the rightward direction. So let's think about these two scenarios. And since velocity matters here so much, we just have to remind ourselves that the velocity-- remember, a derivative is just the rate of change with respect to a variable. So if you have your position function, the derivative of position with respect to time, this is really just what is the instantaneous rate of change of position with respect to time? Well, what is the change of position with respect to time? Well that is just going to be equal to our velocity function. That's going to be equal to our velocity function, v of t. Or we could write s prime of t, which could be also written this way, as ds dt, is equal to our velocity as a function of time. So let's take the derivative of this. Our velocity as a function of time is going to be equal to 3t squared minus 12t plus 9. So let's see if we can graph this velocity function to start making sense of it. When is the velocity positive? When is it negative? And what's the acceleration doing in those intervals? And so to help me graph it, we could say the v-intercept, or the vertical intercept, when v of 0 is going to be equal to 9. So that'll help us graph it. That's where we intersect the vertical axis. But also, let's plot-- let's figure out where it intersects the t-axis. So let's set this equal to 0. So 3t squared minus 12t plus 9 is equal to 0. Let's see. To simplify this, I can divide both sides by 3. And I get t squared minus 4t plus 3 is equal to 0. Now this is very factorable. This is t. Let's see. What two numbers, when you take a product, get 3, and when you add them, you get negative 4? Well, that's going to be t minus 3 times t minus 1 is equal to 0. How can this expression be equal to 0? Well if either of these are equal to 0, if either t minus 3 is 0 or t minus 1 is 0, it's going to be equal to 0. So t could be equal to 3, or t could be equal to 1. If t is 3 or t is 1, either of these are equal to 0, or this entire expression up here is going to be equal to 0. And since our coefficient on the t squared term is positive, we know this is going to be an upward opening parabola. So let's see if we can plot velocity as a function of time. So that is my velocity axis. This right over here is my time axis. And let's say this is 1 times 1 second, or I'm assuming this is in seconds-- 2, 3, 4. Actually, let me spread them apart a little bit more just because 1 and 3 are significant-- 1, 2, and 3. And they're not going to be-- I'm going to squash to the vertical scale a little bit. But this right over here, let's say that is 9, a velocity of 9. And so when t equals 0, our velocity is 9. When t equals 1, then our velocity is going to be 0. We get that right over here. 3 minus 12 plus 9, that's 0. And when t is equal to 3 our velocity is 0 again. Our vertex is going to be right in between those, when t is equal to 2-- right in between these two 0's. And we could figure out what that velocity is if we like. It's going to be 3 times 4 minus 12 times 2 plus 9. So what is that? That's 12 minus 24 plus 9. So that is negative 12 plus 9. So that's going to be equal to negative 3. Did I do that-- 12, yep, negative 3. So you're going to be-- negative 3 might be-- that's 9, so that's positive. So it might be something like this. So the graph of our velocity as a function of time is going to look something like this. And we only care about positive time. It's going to look something like this. So let's think. Remember, this is velocity. This is our velocity as a function of time. Now let's think about when is the velocity less than 0 and the acceleration is less than 0? So let's think about this case right over here? When is this the case? Both of them are going to be less than 0. Well, velocity is the less than 0 over this entire interval, this entire magenta interval. But the acceleration isn't less than 0 that entire time. Remember, the acceleration is the rate of change of velocity. We can write here that acceleration as a function of time, this is equal to the rate which velocity changes with respect to time. Or we could write, acceleration is equal to v prime of t, which is the same thing as the second derivative of position with respect to time. And so the acceleration, you could really think of the slope of the tangent line of the velocity function. And so over here, the place where this is downward sloping, where this has a negative slope, and the curve itself is below the t-axis, that's only over this interval right over here. Between this 0 right over here and the vertex, we get to this point right over here. And then our slope flattens out. So this interval right over here is t is going to be greater than 1, and it is going to be less than 2. That meets these constraints. Now let's think about where our velocity is greater than 0 and our acceleration is greater than 0. Well our velocity is greater than 0 over here. But notice, our acceleration, the slope here is negative. We're downward sloping, so that doesn't apply. Here our velocity is greater than 0, and the slope of the velocity, the rate of change of velocity, the acceleration, is also greater than 0. So that's this interval right over here, where we're speeding up in the rightward direction. So that interval is t is greater than 3. So when are we speeding up? We're speeding up between the first and second seconds, and then we're speeding up after the third second.