The velocity ( mi ⁡ \operatorname{mi} / h r hr ) of a drone flying in the air is given by v ( t ) = 12 + 4 t 2 v\left(t\right)=12+4t^2 for 0 ≤ t ≤ 4 0\le t\le4 hours. Let s ( 0 ) = 0 s\left(0\right)=0 . Determine s ( t ) s\left(t\right) for 0 ≤ t ≤ 4 0\le t\le4 .

s ( t ) = 4 t 3 3 + 12 t s\left(t\right)=\frac{4t^3}{3}+12t

s ( t ) = 4 t 2 3 + 12 t s\left(t\right)=\frac{4t^2}{3}+12t

s ( t ) = 8 t s\left(t\right)=8t

s ( t ) = 8 t 3 3 + 4 t s\left(t\right)=\frac{8t^3}{3}+4t^{}

At t = 0 t=0 , a car approaching a stop sign decelerates from a speed of 50 mi ⁡ \operatorname{mi} / h r hr according to the acceleration function a ( t ) = 4 t + 3 a\left(t\right)=4t+3 , where t ≥ 0 t\ge0 and is measured in hours. How far does the car travel between t = 0 t=0 and t = 0.1 t=0.1 h r hr ?

5.02 mi ⁡ 5.02\operatorname{mi}

0.02 mi ⁡ 0.02\operatorname{mi}

50.32 mi ⁡ 50.32\operatorname{mi}

0.32 mi ⁡ 0.32\operatorname{mi}

A particle moves along the x x -axis and its acceleration is given by a ( t ) = cos ⁡ π t a\left(t\right)=\cos\pi t . Find v ( t ) v\left(t\right) if v ( 0 ) = 0 v\left(0\right)=0

v ( t ) = 1 π sin ⁡ ( π t ) v\left(t\right)=\frac{1}{\pi}\sin\left(\pi t\right)

v ( t ) = − π sin ⁡ ( π t ) v\left(t\right)=-\pi\sin\left(\pi t\right)

v ( t ) = − 1 π sin ⁡ ( π t ) v\left(t\right)=-\frac{1}{\pi}\sin\left(\pi t\right)

v ( t ) = 1 π cos ⁡ ( π t ) − 1 v\left(t\right)=\frac{1}{\pi}\cos\left(\pi t\right)-1

A particle moves along the x x x -axis and its acceleration is given by a ( t ) = cos ⁡ π t a\left(t\right)=\cos\pi t a ( t ) = cos π t . Find s ( t ) s\left(t\right) if s ( 0 ) = 1 s\left(0\right)=1

s ( t ) = π 2 − cos ⁡ ( π t ) + 1 π 2 s\left(t\right)=\frac{\pi^2-\cos\left(\pi t\right)+1}{\pi^2}

s ( t ) = π 2 − cos ⁡ ( π t ) π 2 s\left(t\right)=\frac{\pi^2-\cos\left(\pi t\right)}{\pi^2}

s ( t ) = π 2 + sin ⁡ ( π t ) + 1 π 2 s\left(t\right)=\frac{\pi^2+\sin\left(\pi t\right)+1}{\pi^2}

s ( t ) = π 2 + sin ⁡ ( π t ) π 2 s\left(t\right)=\frac{\pi^2+\sin\left(\pi t\right)}{\pi^2}

A rock is thrown from a height of 2 ft ⁡ 2\operatorname{ft} with an initial speed of 25 ft ⁡ 25\operatorname{ft} / s s . Acceleration resulting from gravity is − 32 ft ⁡ -32\operatorname{ft} / s 2 s^2 . Find v ( t ) v\left(t\right)

v ( t ) = 25 − 32 t v\left(t\right)=25-32t

v ( t ) = − 32 t v\left(t\right)=-32t

v ( t ) = 25 − 16 t 2 v\left(t\right)=25-16t^2

v ( t ) = 16 t 2 v\left(t\right)=16t^2

A rock is thrown from a height of 2 ft ⁡ 2\operatorname{ft} 2 ft with an initial speed of 25 ft ⁡ 25\operatorname{ft} 25 ft / s s s . Acceleration resulting from gravity is − 32 ft ⁡ -32\operatorname{ft} − 32 ft / s 2 s^2 s 2 . Find s ( t ) s\left(t\right)

s ( t ) = − 16 t 2 + 25 t + 2 s\left(t\right)=-16t^2+25t+2

s ( t ) = − 16 t 2 + 25 t s\left(t\right)=-16t^2+25t

s ( t ) = − 64 t 2 + 25 t + 2 s\left(t\right)=-64t^2+25t+2

s ( t ) = 16 t 2 s\left(t\right)=16t^2

10. Physics Applications of Integrals

Kinematics

10. Physics Applications of Integrals

Kinematics: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

Understanding motion involves the relationships between position, velocity, and acceleration. To find displacement, integrate the velocity function over a time interval, while total distance requires the absolute value of the velocity function to avoid negative values. For position, integrate the velocity function and add the initial position. The equations used include $s_{0}^{+}$ for position and $v_{0}^{+}$ for velocity.

concept

Using The Velocity Function

Video duration:

10m

Video transcript

So you may recall in earlier videos where we talked about solving motion problems. And recall that these motion problems involved us trying to find things like velocity, position, or even displacement. Now it turns out you're going to see some more complicated motion problems in this course, where rather than using derivatives to solve them, we're going to need to use integrals. This might sound a bit scary at first, but don't sweat it, because we're going to take a look at some examples in this video that you'll likely come across in this course. And we're going to see how we can actually implement this knowledge that we already have on motion and integration to solve these problems in hopefully the most clear way that we can. So let's just go ahead and jump right in to see what these examples look like.

So let's say we had a problem like this that said, suppose that an object moves in a line such that its velocity is given by this function right here, v(t). Now let's say we have an example that asks us to find the displacement of the object on the interval from zero to five. So we have this interval right here, and we're asked to find displacement. How could we go about doing that? Well, what I want you to do is recall what we already know about velocity, position, and displacement.

We've learned earlier on how velocity is going to be the derivative of position with respect to time. And we know that displacement is just a change in position. So this is something that we know already. But if I want to go ahead and find the displacement with the velocity function, well, if it's a derivative that gets me from position to velocity, it makes sense that it would be an integral that gets me from velocity to position, since we know that that integral is, in essence, just the reverse process of taking a derivative. So what I can do is go down here and say, if I take my velocity function and I integrate it with respect to time over the time interval given, that's going to give me the change in position, which would be my displacement.

And that's what we're trying to find. So going to this example here, what I can do is take this integral here, and I can bound it from zero to five. And I can see that my velocity function is 2t - 4. Now from here, we should know how to solve an integral like this, so let's just go ahead and do it. I'll first find the antiderivative of 2t, that's going to be t^2 minus the antiderivative of four, which is 4t. And then this whole thing is going to be bounded from zero to five. Now notice that zero is my bottom bound here. That's just going to set everything here equal to zero, so I only need to plug in my high bound here, which is five. So doing that, we're going to have this whole thing is 5^2 - 4 \times 5, and this whole thing should just come out to five.

So five would be the solution to this problem, and that's how we can calculate our displacement. Now graphically, this displacement would represent the area underneath the curve of our velocity function, because recall that that's really what's happening whenever we integrate something. We're finding the area underneath that curve. So if I were to look at my velocity curve right here, what I'm looking for is this area and plus that area. Now all of this area, keep in mind, that the area that dips beneath the curve here, this is all going to be negative, and the area above the curve here is going to be positive.

So when you add both of these up, you end up getting positive five. And it makes sense because there's a bit more area that comes above the curve than goes below, so that's why we didn't get a negative number. But it's possible that for displacement, you are going to see situations where the negative value, the value of the dips below the curve, is actually bigger, and that's when you get negative numbers for displacement. Now, let's try another problem. So, let's say instead we have an example like down here.

The example that we see here says find the total distance of the object on the interval from zero to five. Now how could we calculate distance, and what's really the main difference between distance and displacement? Well, the main difference with distance is that distance cannot be a negative value. So any negative values we could get, we can't actually incorporate. So it's gonna be the same process where we have to integrate the velocity function, but what I need to do is take the absolute value of my velocity function.

I'm doing this because I want to ensure that I get no negative values when I do this. So the total distance is going to be the same integral from zero to five, and it's going to be the integral of our velocity function, but notice we have to include these absolute value bars. So it's gonna be our velocity function, which is 2t - 4, and this is the integral we could use to solve this problem. Now how exactly do I find this integral when it comes to dealing with absolute value? Well, keep in mind, none of these values can be negative.

And I can see since there's a four being subtracted, we definitely could get negative values right now. So here's what I'm going to do. What I'm first going to do is set up an integral of this function, 2t - 4. I'm going to set the bounds up so that way these bounds here are only going to account for the negative values. Now looking at this, if I plug a zero in right here, that would clearly give us a negative value, negative four.

If I plug one in here, that would also give us a negative value. If I plug two in here, that would give us zero. So notice that we reach our zero value when we get to two. So what that means is this first integral is gonna go from zero all the way up to two. Now I know that this is only going to give us negative numbers.

So every single output that I get from this integral is going to be negative. So what that means is I need to make the entire integral negative. So if I make this integral negative here, that's gonna cancel out all the negative numbers I get from this integral, meaning that it's actually going to turn into a positive result. So take this negative portion right here, and then I'm going to add it to the rest of the integral all the way up to five. So we finished at two here, so I'll make my lower bound two and my upper bound five.

And then I don't need to make this integral negative since this portion of the integral right here is going to be all the positive values. Basically, what I'm doing here is I'm taking this portion of the graph that I see right here, and I'm, in essence, just reflecting it up. Because since I'm taking this and multiplying it by a negative value, which we can see right there, it's going ahead and focusing on all the area that just shows up right here, which means that I'm only going to get positive numbers here when I do this integral, because remember integrating is that area under the curve. And I'm also only going to get positive values right here. So that means I'm only gonna get positive answers when I go ahead and do the integral like this.

And keep in mind, we're doing this integral that goes from zero right here all the way over to five. So this is the area that I should calculate. So you can go ahead and do each of these integrals separately. I trust that you know this process for dealing with the definite integrals. And if you work this all out and you multiply this result by a negative number, you should get 13 as your total answer.

So 13 would be this total area under the curve of this function right here, and that would be our solution for the distance. So we've seen how we can handle displacement, and we've also seen how we can handle the distance. But now let's see how we can deal with just the position. Because it's possible that rather than wanting a specific value for your displacement or your distance, you'll just want some kind of function that represents the position at any possible time t. How can we deal with something like this?

Well, we should know that it has something to do with integrating the velocity function, because we've seen that the position and velocity are related with a derivative. So it makes sense that we can relate velocity back to position with an integral. We've seen that integrating from just this time interval would give you the displacement, so how exactly could we get a position at a certain time? Well, to get a function for position, what you first need to do is take your position initially, which is gonna be s(t_{\text{initial}}). Oftentimes, this is just gonna be zero in here, by the way, and then add it to the integral of velocity, which would basically be your displacement.

So it's gonna be from t_{\text{initial}}, but rather than going up to t_{\text{final}}, we're actually just gonna go up to t. And the reason why is because we want our function of position with respect to time. So what I need to actually do here is throw in a dummy variable for my velocity. So rather than v(t), it's going to be v(x), and that's gonna be integrated with respect to x. This dummy variable here just ensures that I'll get a function with respect to time, which is what I'm looking for for my position.

So this right here is the equation that you need to use. So what you're just doing is you're adding your initial position at the start right there, and then you're adding on this integral right here, and that's going to give you the solution for the position function. So going over here, we're told find the position function given that s(0) = 3. Now I can see here that zero is our initial time. So what we're going to have at the start here is s(0), then we're going to have that plus, and then notice our initial time, well, that's zero, so we go from zero all the way up to some time t.

And what we're gonna be doing is integrating our velocity function. Now we've seen that the velocity function we were given throughout this entire problem is 2t - 4. But notice that we have to switch this t to an x when it comes to finding our position function, so that means that rather than 2t - 4, we're going to have 2x - 4 integrated with respect to x. So we've gone ahead and plugged in our velocity function. Now from here, what I can do is I can plug in the numbers that I know.

So I know that s(0) = 3. And at this point, this is just the integral we should know how to deal with. So I can first find the antiderivative of 2x. That's gonna give me x^2. Then I can subtract off the antiderivative of four with respect to x.

It's gonna give me 4x. Then this whole thing is gonna be bounded from zero to t. Now what you wanna do from here is take this t and plug it in for x. Now I don't need to worry about putting in the zero because that's just gonna set everything to zero right here. So I just need to worry about the t.

So doing this, we're going to have three, and I'll go ahead and remove these parentheses since we're applying this bound here. So we're gonna have t^2, that's gonna be minus four times t. Now I can rearrange this to look a bit more proper when it comes to polynomials, so we're going to have t^2 - 4t + 3, just moving this three to the other side here, and that's going to give me my position function s(t). So this right here is the solution to the problem, and that's how we can find our position function. Now once you have this position function, you're also likely gonna see problems that ask you to find a future position.

And you can find a future position by just taking whatever the final time is that you're given. So you might be given, say, what's the position of the object in three seconds, or five seconds, or a hundred seconds. Just take whatever that final time is, plug it into your position function right there, and then that's going to allow you to find the position at that specific time. You'll see some problems like this. We'll also try practicing with some of these problems later.

But for now, this is how you can deal with displacement, distance, and position when you're given a velocity function. So hope you found this video helpful, and let's move on.

Problem

The velocity ( $mi$ / $h r$ ) of a drone flying in the air is given by $v (t) = 12 + 4 t^{2}$ for $0 is less than or equal to t is less than or equal to 4$ hours. Let $s of 0 equals 0$ .
Determine $s (t)$ for $0 is less than or equal to t is less than or equal to 4$ .

$s (t) = \frac{4 t^{2}}{3} + 12 t$

$s of t equals 8 t$

$s (t) = \frac{4 t^{3}}{3} + 12 t$

$s (t) = \frac{8 t^{3}}{3} + 4 t^{}$

Problem

The velocity ( $\operatorname{mi}$ / $hr$ ) of a drone flying in the air is given by $v\left(t\right)=12+4t^2$ for $0\le t\le4$ hours. Let $s\left(0\right)=0$ .
How far does the drone travel during the first hour?

13.3 mi

133.3 mi

8.00 mi

6.67 mi

Problem

The velocity ( $\operatorname{mi}$ / $h r$ ) of a drone flying in the air is given by $v\left(t\right)=12+4t^2$ for $0\le t\le4$ hours. Let $s\left(0\right)=0$ .
How far has the drone traveled by the time it has reached $48 mi$ / $h r$ ?

24 mi

34.7 mi

48 mi

72 mi

Problem

Suppose that a particle travels along the $x$ -axis and its velocity is given by $v of t equals 2 cosine t$ for $0 is less than or equal to t is less than or equal to 2 pi$ .
Find the particle's displacement on $open bracket 0 comma 5 pi over 4 close bracket$

$- \sqrt{2}$

$- \frac{\sqrt{2}}{2}$

$- 2 - \sqrt{2}$

$2 + \sqrt{2}$

Problem

Suppose that a particle travels along the $x$ -axis and its velocity is given by $v\left(t\right)=2\cos t$ for $0\le t\le2\pi$ .
Find total distance traveled by the particle on $\left\lbrack0,\frac{5\pi}{4}\right\rbrack$ .

1.41

2.00

5.41

0.027

Problem

A particle travels along the 𝑥-axis and its velocity is the given graph of $v (t)$ .
Find displacement on $open bracket 0 comma 5 close bracket$

2.5

4.5

-2

6.5

Problem

A particle travels along the 𝑥-axis and its velocity is the given graph of $v\left(t\right)$ .
Find total distance on $\left\lbrack0,5\right\rbrack$

2.5

4.5

6.5

concept

Using The Acceleration Function

Video duration:

Video transcript

In recent videos, we've been spending some time talking about how to solve motion problems with integrals. Now we saw how to do this with velocity in the previous video, but it turns out that you can also do this with acceleration. Now I want you to recall how position, velocity, and acceleration are all related to each other. And we know we can use derivatives to get from position to velocity, or from velocity to acceleration. We learned recently that an integral would allow us to go backwards from velocity to position.

Now it turns out if we're dealing with an acceleration in our problem, we can use the same idea of integrating acceleration to go backwards from acceleration to velocity. So let's check out an example of what this would look like to see how we could solve these types of problems. Here, we have an example that says the acceleration of a particle moving along the x-axis is given by this function right here. Notice we have an acceleration function with respect to time, and notice that we are given this acceleration in meters per second squared. That means we're going to be dealing with meters and seconds when we solve these problems.

We're given our acceleration function, and example a asks us to find the velocity function given this initial velocity condition right here. So how do we solve this problem? Well, I should recognize that if we're using the same logic that we've learned about for these types of motion problems, to go from acceleration, which we're given, to velocity, which we're looking for, well, I can just integrate my acceleration function. So the equation that I would want to use is something like this. Notice here how to find our velocity function as a function of time, we're integrating the acceleration function.

There's a couple of things we need to change here. First off, we need to put x as our dummy variable here, since we're ultimately trying to get a function of time. Notice we also have to add on the initial velocities since we're looking for a function right here. This is actually very similar to the approach we used in the previous video when trying to find the position function. Because recall that when we wanted to find position as a function of time, we needed to add the initial position and then also put on the integral of velocity.

Likewise, here we're adding the initial velocity and then integrating acceleration. So to solve example a here, what I would do is just set up this integral. So my velocity as a function of time is going to be equal to the initial velocity, which I can see is v0. That means we're starting at a time of zero right here. Then I'm going to go ahead and add on the integral from zero, the initial time, all the way up to some time, t.

And what I'll do from here is integrate the acceleration as a function of x or some dummy variable. So that means I need to take the t in my acceleration function and replace it with x, meaning we're going to have 6x-12 integrated with respect to x. Now from here, I can go ahead and plug in my initial velocity, which I can see is negative nine, and then I can also integrate this function. Now to integrate this, I can use the fundamental theorem of calculus where I first do the power rule for finding this antiderivative of the function inside. So the antiderivative of 6x, that's going to be 3x2, and the antiderivative of 12 is going to be 12x.

This whole thing is going to be bounded from zero all the way up to the high bound of t. Now notice here that zero is the bottom bound of this integral. So that means the zero is just going to set everything in this function equal to zero. So I only need to focus on the upper bound of t. So that means you need to take t and plug it in for x right here.

So doing this, we're going to have negative nine, that's gonna be plus 3t2 doing that right there, minus 12t right here. Now as a final step, I'll go ahead and rearrange this entire equation to move this negative nine to the other side here, just so it looks more like a polynomial that we're used to seeing. So this is going to be 3t2-12t-9. And then keep in mind, this is a velocity in meters and seconds we're dealing with, so the units will be meters per second. So this right here is my velocity function with respect to time, and that's gonna be the solution to example a.

So as you can see, it's a pretty straightforward process based on what we already know about position and velocity. We can do the same logic using acceleration. So since we knew that we could integrate our acceleration function to get velocity, we could just follow the same general process that we also followed when dealing with position and velocity. But now let's actually take this a step further, or I should say a step backwards, by seeing if we can now find our position function given the velocity function. So notice that we started with acceleration in this problem, then we just found velocity, now we're looking for position.

So how can I go about doing that? Well, we know the equation that we need to use to find position if given velocity, and we were able to find the velocity right here. So we can just use this equation again in order to find position. So st is going to be equal to the initial position, which I can see would be s

Problem

At $t equals 0$ , a car approaching a stop sign decelerates from a speed of 50 $mi$ / $h r$ according to the acceleration function $a (t) = 4 t + 3$ , where $t is greater than or equal to 0$ and is measured in hours. How far does the car travel between $t equals 0$ and $t equals 0.1$ $h r$ ?

$0.02 mi$

$50.32 mi$

$5.02 mi$

$0.32 mi$

Problem

A particle moves along the $x$ -axis and its acceleration is given by $a (t) = \cos π t$ .
Find $v (t)$ if $v of 0 equals 0$

$v (t) = \frac{1}{π} \sin (π t)$

$v (t) = - π \sin (π t)$

$v (t) = - \frac{1}{π} \sin (π t)$

$v (t) = \frac{1}{π} \cos (π t) - 1$

Problem

A particle moves along the $x$ -axis and its acceleration is given by $a\left(t\right)=\cos\pi t$ .
Find $s (t)$ if $s of 0 equals 1$

$s (t) = \frac{π^{2} - \cos (π t)}{π^{2}}$

$s (t) = \frac{π^{2} + \sin (π t) + 1}{π^{2}}$

$s (t) = \frac{π^{2} + \sin (π t)}{π^{2}}$

$s (t) = \frac{π^{2} - \cos (π t) + 1}{π^{2}}$

Problem

A rock is thrown from a height of $2 ft$ with an initial speed of $25 ft$ / $s$ . Acceleration resulting from gravity is $negative 32 ft$ / $s^{2}$ .
Find $v (t)$

$v (t) = - 32 t$

$v (t) = 25 - 32 t$

$v (t) = 25 - 16 t^{2}$

$v of t equals 16 t squared$

Problem

A rock is thrown from a height of $2\operatorname{ft}$ with an initial speed of $25\operatorname{ft}$ / $s$ . Acceleration resulting from gravity is $-32\operatorname{ft}$ / $s^2$ .
Find $s (t)$

$s (t) = - 16 t^{2} + 25 t + 2$

$s (t) = - 16 t^{2} + 25 t$

$s (t) = - 64 t^{2} + 25 t + 2$

$s of t equals 16 t squared$

example

Using The Acceleration Function Example 1

Video duration:

Video transcript

A particle moves so that its acceleration is given in meters per second squared, and is defined by this function right here a(t). We're told that t, the time, is greater than or equal to zero and is measured in seconds. Now this first part here asks us to find the velocity function v if the initial velocity v is six meters per second. So how can we go about solving this problem? Well, the first part of this problem asks us to find velocity, and notice that we're starting with acceleration.

One of the things that I know is that if I'm trying to get from acceleration to velocity, I can use an integral of the acceleration function, and that should get me from point a to point b, where I'm trying to get from acceleration to my velocity function. But recall there's an equation we can use when it comes to solving these types of problems. The equation looks like this. V(t) is equal to the initial velocity, which in this case is v(0), we can see right there, plus the integral of the acceleration function from our initial time to some time t, and then the acceleration function is going to now be a function with respect to x. We use x here as a dummy variable to just go ahead and say, hey, we want our upper bound here to be t, and we need to integrate our acceleration function to do that.

So to find our velocity function, let's go ahead and plug in what we know. I know that v(0) is given to us as six meters per second squared. Then I'm going to add this to the integral from zero to t of my acceleration function as a function of x. So I'm just going to take that t in there and replace it with x, giving us $ 2x - 1 $. So we'll integrate this with respect to x.

v ( t ) = 6 + ∫ 0 t x 2 - x dx

Now this is a process that I should know how to do when it comes to solving this integral right here. So we're going to have six plus, and I'm going to integrate this right here. Well, the integral of 2x, that's going to be $ x^2 $. They're going to have minus the integral of one, which is just x using this rule for integrals, just taking the antiderivative of each term right here, and this is all going to be bounded from zero to t. Now notice what would happen if I took zero and plugged it in for x.

Notice every place that I see x would just become zero, meaning everything within these brackets would just go away. So because of that, I only need to focus on my upper bound of t. So you need to take t and plug it in for x. So I'm going to do this, and simultaneously, I'm going to take this six and move it to the other side here. So plugging t in for x and then moving the six, we're going to have $ t^2 - t + 6 $.

So notice I just went ahead and moved the six to the other side, and I plugged in t, and this gave me my velocity function $ v(t) $. This right here would be the solution for part a. That's how we can find a velocity function. So as you can see, this process is similar to what we already know. We just needed to go ahead and find our velocity function based on what was given here in this first part.

But notice what part b asked us to do. Part b asked us to find the displacement of the particle during the first five seconds. So how do we go about doing this? Well, what we're now trying to do is find a displacement. And if I want to find a displacement, recall we have an equation that allows us to do that.

The displacement is going to be, in essence, equivalent to the integral of our velocity function. So what we need to do is take our velocity function, v(t), and we need to integrate it with respect to time. And the interval is going to be the time that we're looking at. Now this asks us for the first five seconds, and I can see that our time is greater than or equal to zero seconds. So what that means is the start of our interval would be zero seconds, and we would go for the first five seconds of our interval.

So that's going to be how we can find displacement. So if I want to find displacement in this case here, what I'm going to do is integrate from zero to five my velocity function. Now you can see we just calculated our velocity function to be $ t^2 - t + 6 $, and this whole thing is being integrated with respect to t. So we now have our displacement set up. And to find this, well, I could use the same strategy where I take the antiderivative of each portion right here, and then I go ahead and plug in my bounds.

s = ∫ 0 5 t 3 3 - t 2 2 + 6

So doing this, well, the antiderivative of $ t^2 $ using the power rule, that's going to be $ t^3/3 $. The antiderivative of $ t $, it's gonna be $ t^2/2 $. Yes. The antiderivative of six is going to be $ 6t $, and then we're going to go ahead and bound this all from our bounds right here, which are going to go from zero to five. So at this point, I can plug in my bounds.

Now notice once again, my lower bound is zero. If I take zero and I plug it in for t, each one of these terms is going to become zero. So everything within these brackets is just going to go away. So I only need to focus on my upper bound, which is five. So doing this, I'll plug in five for t.

5 3 3 - 5 2 2 + 30

So what this is gonna come out to is $ (125/3) - (25/2) + 30 $. You can go ahead and check these on a calculator you should get an approximate value of 76.83 meters. So this right here would be our displacement. Now recall that when we started this problem, the acceleration that we were dealing with was in meters per second squared. So because of that, we could say that our velocity that we calculated was in meters per second.

So that was when we found our velocity function, that was gonna be our output, and so that means that our displacement here would have units of meters. So this is how we can calculate our velocity and our displacement given the acceleration function we started with. So now we've solved parts a and b of this problem, and lastly, we'll move to part c. Part c asks us to find the total distance traveled by the particle during the first five seconds. Now how do we find distance, and how exactly is this different from displacement, which we calculated in the previous step?

Well, if we want to find distance, which I'm going to go ahead and say distance is $ d $, it's going to take an integral of our velocity function once again. However, the difference between this and displacement is that for distance, we take the absolute value of our velocity function. This is going to be the integral of the absolute value of our velocity function, and again, this is the first five seconds, so we're going to go from zero to five. So basically, when it comes to our distance, we're not allowed to get any negative numbers.

So any output that shows up below the x-axis or that shows up negative at all is going to be not allowed when it comes to calculating our distance. So what that means is we're going to use our same velocity function, $ t^2 - t + 6 $, and we're going to integrate that. So our distance is going to be the integral from zero to five of the absolute value of $ t^2 - t + 6 $. And we're integrating that with respect to time. Now how exactly do I figure out what the absolute value is right here?

Well, think about this. All the outputs we have here cannot be negative. So what would happen if I took zero and plugged it in right there? Well, we would get zero squared minus zero, plus six meaning we would get a positive output. What if I put in one right there?

Well, we would get one squared minus one, which would just be zero, plus six, meaning we would get a positive output again. What if I put in two there? We get two squared, which is four, minus two, which is two, plus six, which is and here's something interesting that happens when you go ahead and try and plug in numbers. It turns out there's no way of getting a negative output in here if you're just putting in real numbers. In fact, if you were to actually graph this curve of $ t^2 - t + 6 $, what we would do is we would get some curve that looks something kind of like this.

You get this kind of parabola that is above here, and what this parabola does is notice it would never dip below the $ t $-axis. Right? So the horizontal axis here, whether it be the x-axis or the $ t $-axis, that means that none of our outputs are going to be negative. So because of that, we can say that we're not going to get a negative result depending on what we plug in to this function. So what does that mean?

Well, that means that we can just integrate our original function, and it's totally fine because we're not going to get any negative outputs when we do that. So we can just say we're going to go from zero to five, and we're just integrating $ t^2 - t + 6 $ all with respect to $ t $. And to do this, well, I know how to do this integral here. We just solved it up here. It's the same integral that we had before.

So we should end up getting is $ t^3/3 $, and that's going to be minus $ t^2/2 $ plus $ 6t $. All of this bounded from zero to five. And we know that all we need to do is plug in our high bound of five, and we should get their distance at approximately 76.83 meters, the same result that we got for our displacement. So this is how we can calculate our distance, and that is the solution to the problem. So as you can see, when it comes to solving these more complicated types of motion problems where you have multiple steps, it's really just about using the same process of recognizing how the acceleration, velocity, and position are all related to each other.

It's just rather than finding the position function, we were either calculating our displacement, which is a change in position, or we were calculating our distance, which is the same thing as displacement, a change in position, just only focusing on the positive values, which is what we did. So that is how you can solve these types of problems. Hope you found this video helpful, and let's move on.

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Here’s what students ask on this topic:

How do you find displacement using integrals in kinematics?

To find displacement in kinematics, you integrate the velocity function over a specified time interval. Displacement is essentially the change in position, and since velocity is the derivative of position with respect to time, integrating velocity gives you the position change. Mathematically, if $v (t)$ is the velocity function, the displacement $d$ over the interval $[a, b]$ is given by the integral: $d_{ab} = \int_{a, b} v (t) dt$ . This integral calculates the net area under the velocity-time curve, representing the total displacement.

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What is the difference between displacement and total distance in kinematics?

In kinematics, displacement and total distance are related but distinct concepts. Displacement is the net change in position of an object and can be positive, negative, or zero. It is calculated by integrating the velocity function over a time interval. Total distance, on the other hand, is the sum of all distances traveled, regardless of direction, and is always a non-negative value. To find total distance, you integrate the absolute value of the velocity function over the interval. This ensures that any negative values, which indicate a change in direction, are accounted for as positive contributions to the total distance.

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How do you find the position function from a velocity function in kinematics?

To find the position function from a velocity function in kinematics, you integrate the velocity function with respect to time and add the initial position. If $v (t)$ is the velocity function, the position function $s (t)$ is given by: $s (t) = s (t) + \int_{t, t} v (x) dx$ , where $s (t)$ is the initial position. This integration process provides a function that describes the position of the object at any time $t$ .

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How can you find the velocity function from an acceleration function in kinematics?

To find the velocity function from an acceleration function in kinematics, you integrate the acceleration function with respect to time and add the initial velocity. If $a (t)$ is the acceleration function, the velocity function $v (t)$ is given by: $v (t) = v (t) + \int_{t, t} a (x) dx$ , where $v (t)$ is the initial velocity. This integration provides a function that describes the velocity of the object at any time $t$ .

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What is the role of initial conditions in solving kinematics problems using integrals?

Initial conditions play a crucial role in solving kinematics problems using integrals. They provide the necessary constants needed to determine the specific solution to a problem. When integrating to find velocity from acceleration or position from velocity, the initial conditions, such as initial velocity or initial position, are added to the integral result. This ensures that the solution accurately reflects the motion of the object from the specified starting point. Without these initial conditions, the solution would be incomplete, as integration introduces an arbitrary constant that must be resolved using known values at a specific time.

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Kinematics: Videos & Practice Problems

Using The Velocity Function

Video transcript

The velocity (mi⁡\operatorname{mi}/hrhr) of a drone flying in the air is given by v(t)=12+4t2v\left(t\right)=12+4t^2 for 0≤t≤40\le t\le4 hours. Let s(0)=0s\left(0\right)=0. Determine s(t)s\left(t\right) for 0≤t≤40\le t\le4.

The velocity (﻿mi⁡\operatorname{mi}mi﻿/﻿hrhrhr﻿) of a drone flying in the air is given by ﻿v(t)=12+4t2v\left(t\right)=12+4t^2v(t)=12+4t2﻿ for ﻿0≤t≤40\le t\le40≤t≤4﻿ hours. Let ﻿s(0)=0s\left(0\right)=0s(0)=0﻿.How far does the drone travel during the first hour?

Suppose that a particle travels along the xx-axis and its velocity is given by v(t)=2cos⁡tv\left(t\right)=2\cos t for 0≤t≤2π0\le t\le2\pi.Find the particle's displacement on [0,5π4]\left\lbrack0,\frac{5\pi}{4}\right\rbrack

A particle travels along the 𝑥-axis and its velocity is the given graph of v(t)v\left(t\right). Find displacement on [0,5]\left\lbrack0,5\right\rbrack

A particle travels along the 𝑥-axis and its velocity is the given graph of ﻿v(t)v\left(t\right)v(t)﻿. Find total distance on ﻿[0,5]\left\lbrack0,5\right\rbrack[0,5]﻿

Using The Acceleration Function

Video transcript

At t=0t=0, a car approaching a stop sign decelerates from a speed of 50 mi⁡\operatorname{mi}/hrhr according to the acceleration function a(t)=4t+3a\left(t\right)=4t+3, where t≥0t\ge0 and is measured in hours. How far does the car travel between t=0t=0 and t=0.1t=0.1 hrhr?

A particle moves along the xx-axis and its acceleration is given by a(t)=cos⁡πta\left(t\right)=\cos\pi t. Find v(t)v\left(t\right) if v(0)=0v\left(0\right)=0

A particle moves along the ﻿xxx﻿-axis and its acceleration is given by ﻿a(t)=cos⁡πta\left(t\right)=\cos\pi ta(t)=cosπt﻿. Find s(t)s\left(t\right) if s(0)=1s\left(0\right)=1

A rock is thrown from a height of 2ft⁡2\operatorname{ft} with an initial speed of 25ft⁡25\operatorname{ft}/ss. Acceleration resulting from gravity is −32ft⁡-32\operatorname{ft}/s2s^2. Find v(t)v\left(t\right)

A rock is thrown from a height of ﻿2ft⁡2\operatorname{ft}2ft﻿ with an initial speed of ﻿25ft⁡25\operatorname{ft}25ft﻿/﻿sss﻿. Acceleration resulting from gravity is ﻿−32ft⁡-32\operatorname{ft}−32ft﻿/﻿s2s^2s2﻿. Find s(t)s\left(t\right)

Using The Acceleration Function Example 1

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How do you find displacement using integrals in kinematics?

What is the difference between displacement and total distance in kinematics?

How do you find the position function from a velocity function in kinematics?

How can you find the velocity function from an acceleration function in kinematics?

What is the role of initial conditions in solving kinematics problems using integrals?

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The velocity ( $mi$ / $h r$ ) of a drone flying in the air is given by $v (t) = 12 + 4 t^{2}$ for $0 is less than or equal to t is less than or equal to 4$ hours. Let $s of 0 equals 0$ .
Determine $s (t)$ for $0 is less than or equal to t is less than or equal to 4$ .

The velocity ( $\operatorname{mi}$ / $hr$ ) of a drone flying in the air is given by $v\left(t\right)=12+4t^2$ for $0\le t\le4$ hours. Let $s\left(0\right)=0$ .
How far does the drone travel during the first hour?

Suppose that a particle travels along the $x$ -axis and its velocity is given by $v of t equals 2 cosine t$ for $0 is less than or equal to t is less than or equal to 2 pi$ .
Find the particle's displacement on $open bracket 0 comma 5 pi over 4 close bracket$

A particle travels along the 𝑥-axis and its velocity is the given graph of $v (t)$ .
Find displacement on $open bracket 0 comma 5 close bracket$

A particle travels along the 𝑥-axis and its velocity is the given graph of $v\left(t\right)$ .
Find total distance on $\left\lbrack0,5\right\rbrack$

At $t equals 0$ , a car approaching a stop sign decelerates from a speed of 50 $mi$ / $h r$ according to the acceleration function $a (t) = 4 t + 3$ , where $t is greater than or equal to 0$ and is measured in hours. How far does the car travel between $t equals 0$ and $t equals 0.1$ $h r$ ?

A particle moves along the $x$ -axis and its acceleration is given by $a (t) = \cos π t$ .
Find $v (t)$ if $v of 0 equals 0$

A particle moves along the $x$ -axis and its acceleration is given by $a\left(t\right)=\cos\pi t$ .
Find $s (t)$ if $s of 0 equals 1$

A rock is thrown from a height of $2 ft$ with an initial speed of $25 ft$ / $s$ . Acceleration resulting from gravity is $negative 32 ft$ / $s^{2}$ .
Find $v (t)$

A rock is thrown from a height of $2\operatorname{ft}$ with an initial speed of $25\operatorname{ft}$ / $s$ . Acceleration resulting from gravity is $-32\operatorname{ft}$ / $s^2$ .
Find $s (t)$