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### Course: AP®︎/College Calculus AB>Unit 8

Lesson 2: Connecting position, velocity, and acceleration functions using integrals

# Motion problems with integrals: displacement vs. distance

The definite integral of a velocity function gives us the displacement. To find the actual distance traveled, we need to use the speed function, which is the absolute value of the velocity.

## Want to join the conversation?

• Is, is this what people learn at 10th grade in America?
(7 votes)
• In America, 10th graders usually learn Algebra 2, but talented 10th graders might learn pre-calculus or even calculus. Usually, calculus is a college level course in America.

Calculus is like algebra, but with the concept of a limit. This concept then leads to the concept of a derivative (think of the slope of a curve at a single point) and the concept of an integral (think of the area under a curve but above the x-axis). Furthermore, taking an integral is essentially the inverse of taking a derivative!

Have a blessed, wonderful day!
(17 votes)
• At he starts working out the distance portion using the mentioned |v(t)|, but I am struggling to apply that to this problem: integral |-16t+48|dt from t=2 to 5. How can I use the absolute value formula without using integration by parts? Is integration by parts the only way to go on these un-graphed problems that don't have a range that starts at the origin?

Once worked out using integration by parts, it is supposed to come out to 40, but thats only after you split it into two different integral functions with bounds that I don't exactly understand how to solve for to establish the two integrals. What am I missing?
(3 votes)
• The only way to integrate absolute value functions like this is by splitting the integral as you describe. If there is a formula or other such thing, it would be derived by splitting the integral.

Now, the key to splitting the integral is to do it in a way that lets you ignore the absolute value function. You do this by picking the bounds of integration so that the function doesn't change sign in the middle of the integration region.

So for this example, -16t+48 is positive for t<3 and negative for t>3. So we can have one integral from 2 to 3, and on this interval, |-16t+48|=-16t+48. On the other interval, from 3 to 5, |-16t+48|=16t-48. Now these are just simple linear functions, easy to integrate.

(Also, 'integration by parts' is a specific technique that you'll encounter later on, and is very different from this splitting of the interval here. I would advise not mixing the terminology.)
(7 votes)
• In regard to the difference between displacement and distance traveled, could it be said that displacement is a state function while distance traveled is a path function?
(3 votes)
• Howdy eskry,

That is absolutely correct, and once you study thermodynamics in physics the difference between state and path functions will become very important.

Happy learning! :-)
(3 votes)
• I was trying to find the distance traveled without a graph; so I integrated the absolute value of 5-t and plugged in 10 then subtracted the value of plugging in 0 and I got 0 which isn't what is said in this video. What'd I do wrong?
(3 votes)
• Without seeing your work, I can't know what you did wrong, but it is possible that you integrated the absolute value wrong. You cannot just integrate it like a linear function, you must split the absolute value into two parts.

Since it is absolute value, our answers must always be positive. for t > 5, 5-t is negative, so for the interval [5, 10] the absolute value function will be equal to -(5-t)

for t < 5, 5 - t will be positive, so for the interval [0, 5], the absolute value function will be equal to 5 - t

this leaves you with the definite integral from 0 to 5 of (5 - t), and the definite integral from 5 to 10 of - (5 - t) = (t - 5)

adding the results of these two integrals gives you the correct answer of 25
(1 vote)
• Sal defined displacement as "change in position". But wouldn't that mean that the derivative of displacement is the rate of change of the change in position with respect to time? I can't even understand what that would mean neither geometrically nor algebraically. Wouldn't the rate of change of the change of position with respect to time be acceleration? I'm confused.
(2 votes)
• We don't actually use displacement as a function, because displacement requires a time interval, whereas a function gives instants in time. The derivative of the vector-valued position function x(t) is the "rate of change of position", also known as velocity v(t).

I would also be careful using the words "change" and "rate of change" interchangeably; "change" refers to a difference in one quantity, whereas "rate of change" refers to a ratio of changes in different quantities. Acceleration is the "rate of change of the rate of change of position", not the "rate of change of the change of position".
(2 votes)
• Velocity is change in position/change in time, or in other words displacement/change in time. The derivative of position (with respects to time) is displacement/change in time, and so it is velocity.

So why do we say that the derivative of displacement is velocity? Isn't velocity the derivative of position, not change in position? And so how can velocity be the slope of the tangent line to a displacement vs. time graph?

EDIT:
So i'm reading that since displacement is measured as position minus some initial position, and that initial position is constant, the two derivatives are the same.

What really is position? Whenever we indicate position with a number, are we really indicating displacement but with initial position at zero?
(1 vote)
• Position is a vector. It tells us in which direction and at what distance an object is located relative a given reference point.

(Since we're dealing with vectors I decided to let them be two-dimensional.
In one dimension there are only two directions in which a vector can point, which we can represent with ordinary variables being either positive or negative. The calculations are the same, just omit all the 𝑦-coordinates.)

If (ℎ, 𝑘) are the coordinates of the reference point and (𝑥, 𝑦) are the coordinates of the object, then the object's position is
𝒑 = (𝑥, 𝑦) − (ℎ, 𝑘) = (𝑥 − ℎ, 𝑦 −𝑘)

As the object moves over time we get
𝒑(𝑡) = (𝑥(𝑡) − ℎ, 𝑦(𝑡) − 𝑘)
⇒ 𝒑'(𝑡) = (𝑥'(𝑡), 𝑦'(𝑡))

𝒑'(𝑡) is the velocity of the object, which is not the same as its displacement.

Instead, displacement is also a position vector, but tells us the object's position relative to the object's starting point, (𝑥(𝑡₀), 𝑦(𝑡₀)) at some given time 𝑡 = 𝑡₀ :

𝒅(𝑡) = (𝑥(𝑡) − 𝑥(𝑡₀), 𝑦(𝑡) − 𝑦(𝑡₀))

You can verify for yourself that with 𝒑(𝑡₀) = (𝑥(𝑡₀) − ℎ, 𝑦(𝑡₀) − 𝑘) we have
𝒅(𝑡) = 𝒑(𝑡) − 𝒑(𝑡₀)

Either way, we get
𝒅 '(𝑡) = 𝒑'(𝑡)
(3 votes)
• At , he says the integral of the rate function (v(t)) is equal to the change in quantity (aka distance traveled). I understand how to use this, but I'm struggling to understand the concept. I know the integral of v(t) is the area under the curve of v(t), but how does that equal the distance traveled?
(1 vote)
• Well, you know that velocity is the derivative of position/distance, since it defines a rate (think meters travelled, distance, changing to m/s, a rate at which an object travels). Velocity also gives the slope of a distance vs. time graph, since you take how many units are travelled over a specific time parameter. Since an integral is the opposite of a derivative, velocity is the antiderivative of position.

To answer your question, looking at the graph of velocity, it is "m/s" vs. seconds. By finding the area, you would essentially multiply the two units of measure, leaving you with just meters or whatever unit is used, leaving you with the "total distance" (finding the total distance would actually require taking the integral of the absolute value of velocity).

Hope this helps!!
(2 votes)
• When doing problems that involve taking the derivative or anti-derivative of something, should I have all the derivatives of the all the trigonometric functions memorized?
I have sine, cosine, and tangent memorized, but not the inverses of those nor the reciprocals (secant, cotangent, and cosecant).
I do have the derivatives of them written down in a math scratchpad of mine, but should I definitely have those derivatives of memorized to?
(1 vote)
• If you can derive the derivative/antiderivative fairly quickly, then there is no need to memorize it. It might be useful to memorize the inverse trig derivatives, because I’ve seen a lot of integral problems that simplify to some form of arctan.
Hope this helps!
(1 vote)
• In the next exercise I ran into a problem that was rather confusing:

"A particle moves in a straight line with velocity v(t)=-t^4+t^3 where t is time in seconds. At t=0, the particle's distance from the starting point was 4 meters in the positive direction. What is the particle's position at t=9 seconds?"

The answer was the definite integral from 0 to 9 of v(t)dt + 4. (I wish I could type mathematical symbols in questions). Why is it that the starting distance is added to the integral instead of the starting displacement (which wasn't given?). Also, if it's distance, shouldn't the direction not matter?
Thanks in advance.
(1 vote)
• x = -t^5/5 + t^4/4 + C from integration.

Given t=0 we are told displacement is 4. In other words x=4 when t=0.

Substituting x=4, t=0

yields 4=C.

To find the total distance travelled you would need to use integrational calculus.

You said the starting displacement was not given however it was given: "4 meters in the positive direction". X-Displacement is the how left or right an object is from it origin. In this case the particle was 4 metres to the right of the origin.
(1 vote)
• How does finding the area under curves relate to distance and displacement?
(1 vote)
• Remember the area of a rectangle formula, width times height? You can think about the area of a curve the same way, width times average height. When we take the area of a curve with the y value in m/s and the x value in s, the resulting unit is (m/s)*s, which is just m.

Another way to think about it is by taking advantage of the fact that an integral is an antiderivative. When we differentiate displacement, we get velocity, so when we integrate velocity, we get displacement.
(1 vote)

## Video transcript

- [Instructor] What we're going to do in this video is start thinking about the position of an object traveling in one dimension. And to get our bearings there, I'm going to introduce a few ideas. So the first idea is that of displacement. So you might use that word in everyday language, and it literally means your change in position, your change in position. Now a related idea that sometimes get confused with displacement is a notion of distance traveled. You might say well doesn't that, isn't that just the same thing as change in position? And you will see shortly, no, it isn't always the same thing. The distance traveled, this is the total length of path, total length of path. So what are we talking about? Well let's say, and we're going to introduce a little bit of calculus now, let's say that we have a particle's velocity function. And so let's say our velocity as a function of time is equal to five minus t. Now this is a one-dimensional velocity function. Let's say it's just telling us our velocity in the horizontal direction. And oftentimes when something's one dimension, people forget well that too can be a vector quantity. In fact this velocity is a vector quantity because you could think of it if it's positive it's moving to the right, and if it's negative it's moving to the left. So it has a direction. And so sometimes you will see a vector quantity like this have a little arrow on it, or you will see it bolded, or you will see it bolded like that. I like to write an arrow in, although that's not always the convention used in different classes. Now let's plot what this velocity function actually looks like, and I did that ahead of time. So you can see here, at time equals zero, let's say time is in seconds, and our velocity's in meters per second. So this is meters per second right over here, and this is seconds on this axis. At exactly time zero, this object is traveling at five meters per second. We can say to the right, it has a velocity of positive five meters per second. But then it keeps decelerating at a constant rate, so five seconds into it, right at five seconds, the particle has no velocity, and then it starts having negative velocity, which you could interpret as moving to the left. So let's think about a few things. First let's think about what is the displacement over the first five seconds, over first five seconds? Well we've seen already multiple times, if you wanna find the change in quantity, you can take the integral of the rate function of it. And so velocity is actually the rate of displacement is one way to think about it. So displacement over the first five seconds, we can take the integral from zero to five, zero to five, of our velocity function, of our velocity function. Just like that. And we can even calculate this really fast. That would just be this area right over here, which we can just use a little bit of geometry. This is a five by five triangle, so five times five is 25, times 1/2, remember area of a triangle's 1/2 base times height. So this is going to be 12.5, and let's see this is going to be meters per second times seconds, so 12.5 meters. So that's the change in position for that particle over the first five seconds. Wherever it started, it's now going to be 12.5 meters to the right of it, assuming that positive is to the right. Now what about over, over the first 10 seconds? Now this gets interesting, and I encourage you to pause your video and think about it. What would be the displacement over the first 10 seconds? Well we would just do the same thing, the integral from zero to 10 of our velocity function, our one-dimensional velocity function, dt. And so that would be the area from here all the way to right over there. So this entire area. But you might appreciate, when you're taking a definite integral, if we are below the t-axis and above the function like this, this is gonna be negative area. And in fact this area and this area are going to exactly cancel out, and you're going to get zero meters. Now you might be saying how can that be? After 10 seconds how do we, what why is our displacement only zero meters? This particle's been moving the entire time. Well remember what's going on. The first five seconds, it's moving to the right, it's decelerating the whole time, and then right at five seconds, it has gone 12.5 meters to the right. But then it starts, it's velocity starts becoming negative, and the particle starts moving to the left. And so over the next five seconds, it actually moves 12.5 meters to the left, and then these two things net out. So the particle has gone over 10 seconds 12.5 meters to the right and then 12.5 meters to the left, and so its change in position is zero meters. It has not changed. Now you might start, you might start to be appreciating what the difference between displacement and distance traveled is. So distance, if you're talking about your total length of path, you don't care as much about direction. And so instead of thinking about velocity, what we would do is think about speed. And speed is, you could view in this case, especially in this one-dimensional case, this is equal to the absolute value of velocity. Later on when we do multiple dimensions, it would be the magnitude of the velocity function, which is what the absolute function, which is what the absolute value function does, in one dimension. So what would this look like if we plotted it? Well the absolute value of the velocity function would just look like that. And so if you want the distance, you would find, the distance traveled I should say, you would find the integral over the appropriate change in time of the speed function right over here, which we have graphed. So notice, if we want the distance traveled, so I'll just say I'll write it out, distance traveled over first five seconds, first five seconds, what would it be? Well it would be the integral from zero to five of the absolute value of our velocity function, which is you can just view it as our speed function right over here, dt. And so it would be this area, which we already know to be 12.5 meters. So for the first five seconds, your distance and displacement are consistent. Well that's because you have in this case the velocity function is positive, so the absolute value of it is still going to be positive. But if you think about over the first 10 seconds, your distance, 10 seconds, what is it going to be? Pause the video and try to think about it. Well that's gonna be the integral from zero to 10 of the absolute value of our velocity function, which is going to be equal to what? It's going to be this area plus this area right over here. So plus this area right over here. And so this is going to five times five times 1/2 plus five times five times 1/2, which is going to be 25 meters. The particle has gone 12.5 meters to the right, and then it goes back 12.5 meters to the left. Your displacement, your net change in position is a zero, but the total length of path traveled is 25 meters.