In the previous video, we introduced how derivatives can be applied to real-world situations, because we learned that if you take the first derivative of position, it's going to give you velocity. Now here's a question for you. What would happen if I instead took the first derivative of velocity? What is this going to give me? Well, if you take the derivative of your velocity function with respect to time, this is going to give you a function for acceleration.
Now, you may have heard acceleration used in the real world when, say, describing a car on the highway or some object that is moving. And that's what we're going to be talking about in this video, the concept of the math behind acceleration. So without further ado, let's get right into things. Now, acceleration is the change in velocity over time. And in math terms, we describe it like this:
It's going to be the first derivative of velocity, like we see right here or the second derivative of position. And I think this makes sense because if you start with a position function, the first derivative is going to give you velocity, and the second derivative is going to give you acceleration. So if you take 2 derivatives of your position function, it will get you straight to the acceleration function. Now, to understand how we can solve some problems when we come across them in this course, let's take a look at one example here of an acceleration problem you might see. So here, we're asked, given the velocity function, which we have right here, we're told this is given in meters per second, and we need to find the following information based on the time interval.
Now before I get into solving these examples, I want you to pay close attention to something that's happening in the language in this example. Notice that we are specifically given a velocity function, and this is important because previously, we've seen position functions given to us. But here, we're specifically given a velocity function. So you need to make sure you know what type of function you're dealing with before you get into just solving the example. So since we now are familiar with the fact that this is a velocity function, let's see how we can solve this first example.
This asks us to find the change in velocity. And finding the change in velocity is actually similar to how we found the change in position in previous videos. All you want to do is take the final velocity at the final time and then subtract off the velocity at the initial time. So doing this, we would have v3 minus v0. It's always going to be a final minus initial.
Now, what I'm first going to do is figure out what v3 is, and I can do that by taking 3 and replacing it with every place I see t in the top function. So we're going to have 332 minus 3×32 plus 2×3. Now, I trust this is math that you can do on your own. You can just cube this and then subtract off this, and then you can add this portion, which is going to give you 6. But what about v0?
Well, what I need to do is figure out what v0 is, but something actually very convenient is going to happen here. Notice how we have t in every single term. So because everything is being multiplied by t, which in this case is going to be 0, the entire thing is just going to collapse to 0. So what we end up having is that v3 is 6 and v0 is 0. So we have 6 minus 0, which is going to be 6, and then we're told this is in meters per second.
So 6 meters per second is the change in velocity and the solution to this first example a. But what about part b? Part b asks us to find the average acceleration. Now, what exactly is average acceleration? Well, we talked about in the previous video how average velocity is a change in position over time if you're looking at 2 points.
So it's like finding the slope on a secant line or just the average between two values, and that's exactly the same idea as average acceleration. So when average velocity was a change in position over time, average acceleration is a change in velocity over time. So the way I can do this is by looking at the time interval here, and I can say that our average acceleration will be a change in velocity divided by a change in time. Now the change in velocity, we already calculated up here to be 6 meters per second. So all I need to do is divide this by the change in time, which is going to be the final time of 3 minus the initial time of 0.
Now doing this math here, we'll have 6 divided by 3, which ends up being 2. So what we get here is 2 meters per second squared as our result. Now I want you to pay attention to the units that we just got here. Notice we got meters per squared seconds. This is always what happens when dealing with a type of acceleration.
I mean, it might not be these exact units, but it's always going to be a distance divided by a squared time. So you might also see miles per square hours or meters per square seconds or something along those lines, but it's always a distance per squared time. Now, for our last portion here, we're asked to find the instantaneous acceleration at the end of the time interval. So how do we go about doing this? Well, notice we're dealing with an instantaneous versus an average acceleration.
An instantaneous and average is the same idea that we see whenever we're dealing with these motion problems. So an average was looking at an average between two points, where an instantaneous is looking at the acceleration for one specific point. So if I want to find the instantaneous acceleration, I need a function for acceleration. And we know that we can do this by finding the first derivative of velocity. Now I can see here that we have our velocity function.
So to find the derivative for this, that's going to give me the acceleration. And I can do this by just taking the derivative of each of these terms. So I can use the power rule here to get that t3 becomes 3t2. I can then use the power rule on this negative 3t2 to get negative 6t. And then using these same rules on the 2t, that t is just going to come off, giving us plus 2.
So this right here would be the acceleration function that we're looking for, and I can use this acceleration function right here to solve the situation that we have down here. So let's go ahead and do that. Now I'm looking specifically for the acceleration at the end of the time interval, and I can see the end of the time interval up here is 3 seconds. So I'm looking for a3. So if I want to find that result, well, I just need to take 3 and plug it in for every place that I see t.
So I have 3×t2 minus 6×3 plus 2. Now 3×t2, that's 27, then it's going to be minus 18 plus 2, and this whole thing should come out to 7. So what you're going to get as your acceleration at 3 seconds, your instantaneous acceleration, is 7 meters per second squared. This whole thing comes out to 7, so that right there is going to be the result. So this would be the solution to part c, and that is how you can solve these situations where you're dealing with acceleration in motion problems.
So, hopefully, you found this video helpful, and let's try getting a bit more practice with this concept.
