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.
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.