Motion Analysis - Online Tutor, Practice Problems & Exam Prep

Derivatives Applied To Velocity

We've talked about what derivatives are and how we can calculate them, and even how we can graph derivatives. Now, one of the most common questions that you're going to hear asked in just about any math course is, why would I ever need to know this information? How does any of this apply to the real world? Well, over the course of this video and the next few videos, that's going to be the question that we're answering here. So without further ado, let's take a look at how derivatives can be applied to some real-world situations, like, say, finding the motion of an object or an object's velocity.

So here, if we're trying to find the motion of an object over time, we can describe this using a position and velocity. Now the position tells us where our object is at a certain time we call t, and velocity would be the function that tells us how fast our object is moving at a certain time. So this is what the position and velocity functions look like, respectively. Now, if you want to understand how this type of function works, well, let's take a look at an example we have down here. In this example, we are told an object's position s(t) is given in meters, and we're asked to find the missing values based on the time interval that we have right here.

Now, this is the position function that we have, and what this is basically telling us is the position of our object as we move forward in time. So let's say, for example, we were looking at a car driving on the freeway. What we can see here based on this graph that we have is at one second of time, the car has only moved 1 meter, but at 2 seconds of time, the car has moved 4 meters. At 3 seconds of time, the car has moved 9 meters. And this should actually make sense given the function that we have.

Since we have t², we get this kind of parabola shape here where the farther we progress in time, the more severely our car has moved. So that's what this function here is telling us. But what we can do is use this position function that we have to find what the velocity function looks like, and the way we can do that is by recognizing the relationship between position and velocity. Velocity is just the change in position over time, which we can write as ds/dt. So another way of if we want to find this velocity here is recognizing that velocity is just the first derivative of position.

So I can find the velocity function by taking the derivative of my position function, and I can do that using the power rule on this function that we have here. We know the power rule looks something like this. So moving this 2 to the front and reducing the power by 1, we end up getting 2t as our velocity function. And 2t is just going to be a line with a slope of 2. So this right here would be our velocity function, and that is how you can find the relationship between position and velocity.

Now it turns out there's a lot more information that you can extract by being given this position function and the time interval. So what I'm going to do is write the time interval down here from 0 to 2 seconds. That's the time we have up here, and let's see if we can figure out other information that is missing in this problem. Now some of the other missing values that you might be asked to calculate are these values down here. We'll first take a look at this one, which is displacement.

Now displacement is just a change in position. That's what we call displacement. So if I want to calculate displacement, I can use my time interval. I can recognize it's going to be the final position minus the initial position. Now the end of the time interval is 2 seconds.

The start of the time interval is 0 seconds. So what I can do is plug in these values, 2 and 0, into our position function to get the final and initial positions. So the final position is going to be 2² since we have t² as our function, and the initial position is going to be 0². 2² is 4, 0² is 0, and 4 minus 0 is 4. Since we have a change in position and our problem here says that we're dealing with meters, then our change in position is going to be 4 meters.

And that's the solution to this portion of the example. Now another thing that we may be asked to find is average velocity, and something that you need to understand is the difference between average velocity and instantaneous velocity. Now when dealing with average velocity, that's going to be the average between 2 specific times or 2 specific points. So in this example here, if we're trying to find an average velocity, say that we're looking at this point right here and that point right there, 2 different points in time, we could just draw a secant line here and find the slope of this line. So the average velocity is just going to be the change in position from 2 different positions divided by the change in time.

So if I want to find the average velocity over this time interval, well, it's going to be the change in position we have over this time interval, which we already figured out was 4 meters. And that's going to be divided by the change in time, which is going to be 2 seconds minus 0 seconds. So we have 4 divided by 2, which gives us 2, and then that's going to be meters per second. So that right there is going to be the average velocity over this time interval that we have. Now another thing that we may be asked to do is to find instantaneous velocity.

Well, we already talked about how average velocity is just going to be an average between two points, so what would an instantaneous velocity be? Well, like an average velocity was a secant line, an instantaneous velocity would be a tangent line. So rather than looking at 2 points, we're trying to find the slope of a line at exactly one point. So this would be the tangent line, and we could find this using the equation that we already derived. So we can see that we have that our velocity is 2t.

So to find the velocity as, say, 2 seconds, the instantaneous velocity, we would just take this t here and replace it with 2. So it's going to be 2 times 2, which is 4, and that's going to be in meters per second. So that's the instantaneous velocity at 2 seconds. Now the last thing we're going to try calculating here is the speed. And speed and velocity are oftentimes used interchangeably when people talk about them, but they don't actually mean the same thing.

The speed always has to be positive, so it's always going to be the absolute value of whatever your velocity is. It's possible for us to get a negative velocity, but we cannot get a negative speed. So what we can do here is take the absolute value of our velocity, which we calculated to be 4 meters per second. Now in this case, the absolute value of 4 is still just going to be positive 4. So 4 meters per second is going to be the speed.

Now in this case, the speed and velocity were the same, but if we calculated this to be negative 4 meters per second, then they would not be the same. So that is how you can find speed, instantaneous velocity, average velocity, and displacement, and notice how we were able to find all of this by only being given the position function and a certain time interval. So this is how you can solve these types of problems where you're dealing with motions or situations in the real world. Hope you found this video helpful, and let's try getting some more practice.

Given the position equation $s\left(t\right)$, calculate the average velocity (in meters per second) based on the given interval, and the instantaneous velocity (in meters per second) at the end of the time interval.

$s\left(t\right)=9t-t^2$, $0\le t\le3$

• A. Average: $v_{avg}=6\frac{m}{s}$, Instantaneous: $v\left(3\right)=3\frac{m}{s}$
• B. Average: $v_{avg}=10\frac{m}{s}$ , Instantaneous: $v\left(3\right)=18\frac{m}{s}$
• C. Average: $v_{avg}=18\frac{m}{s}$ , Instantaneous: $v\left(3\right)=10\frac{m}{s}$
• D. Average: $v_{avg}=6\frac{m}{s}$ , Instantaneous: $v\left(3\right)=0$

Given the position equation $s\left(t\right)$ , calculate the average velocity (in meters per second) based on the given time interval, and the instantaneous velocity (in meters per second) at the end of the time interval.

$s\left(t\right)=\frac{30}{t+5}$, $-4\le t\le0$

• A. Average: $v_{avg}=-6\frac{m}{s}$, Instantaneous: $v\left(0\right)=0$
• B. Average: $v_{avg}=-6\frac{m}{s}$, Instantaneous: $v\left(0\right)=30\frac{m}{s}$
• C. Average: $v_{avg}=6\frac{m}{s}$ , Instantaneous: $v\left(0\right)=18\frac{m}{s}$
• D. Average: $v_{avg}=-6\frac{m}{s}$, Instantaneous: $v\left(0\right)=-1.2\frac{m}{s}$

Derivatives Applied To Velocity Example 1

We should now be familiar with how position and velocity relate to each other using derivatives. And one of the things that you're going to need to do in this course is you're going to need to know how to solve real-world application problems when it comes to velocity or position or just motion in general. We're going to take a look at one of the longer, more complicated types of problems that you're going to see in this course, where you have the real-world situation. So let's take a look. In this problem, we're told the height in meters of a projectile shot vertically upward from a point of 5 meters above ground level with an initial velocity of 20 meters per second looks like this.

So this would be our equation right here. And part a asks us, what is the vertical velocity at t equals 3 seconds? Now we do have a part b and a part c to this problem, but I'm going to take this one step at a time. So starting with part a, we're asked what is the vertical velocity at t equals 3 seconds? Now, the way that you solve these types of problems, if you have a projectile situation, where you are given the equation, this equation is the key to solving the whole problem.

Because this is basically telling you the vertical position. This is the height with respect to time. So if you recognize that this equation of 5t+20t1-4.9t2 is the equation representing the position of your projectile, you can just manipulate this equation in different ways to get all the answers or all the solutions you need in this problem. Now our first piece here asks us to find the vertical velocity at t equals 3 seconds. So if we have a position equation here, how could we find the velocity at one instance in time?

Well, we've talked about how to calculate instantaneous velocity, and we need to recognize that velocity is just the first derivative of position. Now in this case, since we're dealing with a height specifically, a vertical position, we have h of t rather than s of t, but it's really the same thing. So in this case, it's gonna be h prime of t, and that's going to tell you the vertical velocity that you're dealing with. So what I can do here is take the derivative up here. Now the derivative of 5, that's just gonna be 0 because the derivative of any constant is 0.

And we also have 20 t, and the derivative of 20 t is just going to be 20 that t will derive off, and then we're going to have minus 4.9 t squared. Now to find the derivative here, I can use the power rule. Now what I can do is take this 2 here, this squared, I can bring it down to the front front, multiply it by this first number, and then reduce the power by 1. Doing that is going to give me 4.9 times 2, which is 9.8, and we're going to have t to the 2 minus 1, which is just t to the 1 or t. So this right here would be our function.

And to make the velocity function look a little more nice, what I'm going to do is get rid of the 0 because we don't have to write it, and then I'm just gonna write it like this. We're gonna add that our velocity function is 20 minus 9.8 t. So this right here is the velocity function. Now we're specifically asked, what is the velocity at t equals 3 seconds? Well, since we just found our velocity function, I can just take 3 and plug it in.

So we're going to have 20 minus 9.8 times 3. So this is what we get. So all you need to do is plug in these numbers, and if you go ahead and do this on a calculator, you should get negative 9.4. And the units that we're going to have for this is gonna be in meters per second. I know this because I can see that our height here is going to be in meters, and you see the time we're dealing with is in seconds.

A velocity is always gonna be measured in distance per time, so we can see that we have meters per second. So that is going to be the instantaneous velocity at t equals 3 seconds, and that is part a of this problem. But now let's move to part b. Part b asks us, when does the projectile reach its maximum height? Now this is a little bit trickier, because how exactly do we find when the maximum height is going to occur based on this equation that we were given?

Well, to do this, you need to think about the situation. It says that what we do is we start at this position that is 5 meters above the ground. So we're starting 5 meters above the ground, and we have some kind of object. Let's just say it's a ball. And what's gonna happen is this ball is going to be shot directly vertically up in the air, and then eventually it's gonna start coming back down, straight back down.

And we need to figure out exactly when does it reach this maximum height. Well, let's ask ourselves something. What exactly is going to be happening at that position right there? Well, the ball is going or the object of some kind is going to be flying up and up and up, and eventually just for a split second, that object's gonna hold still in the air, then it's gonna reroute and go back down. So we know for that split second it's gonna be holding still as it gets right there.

And because of that, that means that our vertical velocity at that specific instant will be 0. So we now have a definitive place that we know the object is gonna be at its maximum height. So what I can do is take the vertical velocity here and set the whole thing equal to 0, and that's going to give me the time we're at our maximum height. So the way I'm going to do this is I'll take our equation here, 20 minus 9.8, I'll do this in a different color, of 20 minus 9.8 t. I'll set the whole thing equal to 0.

Now all I need to do from here is solve for this t right here. So we're going to take 20 and subtract it on both sides of the equation. That will get the twenties to cancel on the left side, meaning we'll have negative 9.8 tiedade

Derivatives Applied To Acceleration

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.

Given the position of an object $s\left(t\right)=12t-t^2$ (in meters), find the acceleration of the object at $t=5$ seconds.

• A. Instantaneous Acceleration $a=9.8\frac{m}{s^2}$
• B. Instantaneous Acceleration $a=2\frac{m}{s^2}$
• C. Instantaneous Acceleration $a=-2\frac{m}{s^2}$
• D. Instantaneous Acceleration $a=0\frac{m}{s^2}$

Given below is the graph of velocity with respect to time. At which time(s) would acceleration be 0?

• A. At $t=0$
• B. At $t=1$ & $t=3$
• C. At $t=0$ & $t=5$
• D. At $t=1$, $t=2$ & $t=3$

Derivatives Applied To Acceleration Example 2

Welcome back, everyone. So let's take a look at one of the more complicated problems we're going to see in this course when it comes to solving motion problems. In this example, we're told the height in meters of a projectile shot vertically upward is given by this equation down here. The initial vertical velocity is 15 meters per second. And what we're first asked to do is find the vertical velocity at t equals 3 seconds.

Okay. Now, in this problem, the first step that I'm going to take since I'm looking for vertical velocity is to recognize how velocity relates to our function, which is in vertical position. Velocity is the first derivative of position with respect to time. So if I take the derivative of my height equation, that's going to give me the vertical velocity. The derivative of 15 t is just going to be 15.

The t is going to come off. And then the derivative of -4.9 2 t, well, we can use the power rule here. What we're going to get is 2×4.9t. Now I can clean this up a little bit by recognizing this is the same as 15 minus 9.8 t. So this right here would be our height equation, the first derivative of our height, which is going to be our velocity.

So we have the vertical velocity function right here, and we're asked to find this at a time of 3 seconds. So that means you need to take 3 and plug it in for t. So we're going to get 15-9.8×3, and this is math that you should be able to do. You could put this in on a calculator or check by hand, and you should get a result of -14.4 meters per second. So that's going to be the vertical velocity at t equals 3 seconds.

This is the solution to part a. Now next, we need to move to part b. This asks what is the vertical acceleration at a time of 3 seconds? We know that velocity is the first derivative of the position, and acceleration is going to be the second derivative of our position. So you could just go ahead and take the derivative twice of this function, or you could recognize that we just found what the derivative of our height was to get velocity and then just take the first derivative of our velocity function.

So that's what I'm going to do. Doing this, I'll take the derivative of this function right here. The derivative of 15 is just going to give us 0, and the derivative of 9.8 t is just going to give us 9.8. That t is just going to come off. So the second derivative of height or the first derivative of velocity will give us our acceleration, which is 0 minus 9.8 or -9.8 meters per second squared.

Now something that I'll tell you, a fun little fact here, this is actually the acceleration due to gravity on Earth. So gravitational acceleration is the constant -9.8 meters per second squared. Now we're asked for the acceleration at t equals 3 seconds, but notice we got a constant as our answer. So this right here will be the solution for our acceleration at a time of 3 seconds or any time that we were to check this. And the reason why we got a constant as our answer is that no matter what time you're looking at, gravity is always going to be pulling on this object that's been shot vertically upward.

So it's going to be the same acceleration no matter what time you're looking at. So that would be our solution. So this would be part b. We've gone ahead and solved this first part. We've solved the second part, but now we need to move to the third part, which asks us to find what it says is we have the maximum height of our projectile 11.5 meters.

We're asked to find when the projectile will initially reach half of its maximum height. How exactly could we do this? The maximum height that our projectile reaches is 11.5 meters. And if we want to know what half of this is, just divide it by 2. That's going to give you 5.75 meters.

But we're not asked what half the height of the projectile is. We're asked when we're going to reach half of our height. And how could we figure this out? Well, what I can do is recognize that this is the height equation right here. If I'm looking for a time that we're going to reach half of our max height, well, I can use the height equation to find this.

Because I can see that our height here that we're interested in is 5.75 meters. So if I take 5.75 meters and I set it equal to this function that we have on the right side here, we're going to get that this is equal to 15t-4.92t. And we're going to ignore the units just for now, so I'm going to focus on the math right here. So what I can do is subtract this 5.75 on both sides of the equation, and that's going to get the 5.75 to cancel on the left side. So what this is going to give me is the equation -4.92t, then that's going to be plus 15 t, and then we're going to have minus this 5.75 here.

So we have minus 5.75. That's all going to be equal to what we have nothing on this side now, so that's going to be 0. Notice how I just set up a little quadratic equation right here. Now what I'm going to do to make things a little more simple is I'm going to multiply both sides of the equation by -1. This is just going to allow me to cancel some of the extra negative sides I have on this left side here.

So we're going to get this negative sign to cancel and that negative sign to cancel. But, of course, this middle term here is going to become negative when we do that. So this whole thing is going to end up being 4.9t squared minus 15 t plus 5.75, and that's all going to be equal to 0. Now at this point, what I'm trying to do is solve for t. Because remember, we set this whole function.

We set the projectile function up here for the height equal to the height at 5.75 meters. So if I solve for t, that's going to tell me at what time we're going to reach that height. To do this, well, we need to solve this as a quadratic, so I can use the quadratic formula. Now I'm not going to go through an in-depth process in this video because I trust that we should know what the quadratic formula is and how to use it, but this is going to be your a. That's going to be your b term, -15, and this would be your c.

And the quadratic formula is that t, in this case, is going to be equal to negative b plus or minus the square root of b squared minus 4 times a times c, and all of this is going to be over 2a. We've talked about the quadratic formula before. So you can take these values for a, b, and c, and plug them into this formula, and that will give you your result for time. If you do this, the times that you should calculate, which I'll write up here, should be a time of 0.45 seconds approximately. So we're going to get about 0.45 seconds actually is what we're going to get for our first time, and then the other time that we're going to get is approximately equal to 2.61 seconds.

So these are the two times that we arrive at when we use our quadratic formula. But why exactly did we get 2 times? How does this make sense? Well, the reason for this is that if you imagine a situation where we have our projectile, I'll write it down here, and we have some object like a bullet that is vertically shot upward, it's going to go up, and then it's going to come back down. This would be the maximum height.

Around here would be the half height, and notice how we're going to hit this height twice. It's initially going to reach this point. It's going to go up and then come back down and reach this point again. So this would be the first time we reach the height. That would be the second time we reach the height.

But since we're looking for the time we initially reach half of our maximum height, we can go ahead and get rid of this final answer here, the larger number. We could only keep the smaller number, which would represent the time we initially reach half of our maximum height. So this would be the time right here, and this is the solution to the problem. And this is how you can solve more complicated situations when dealing with motion. Hope you found this video helpful.