Motion problems with integrals: displacement vs. distance | AP Calculus AB | Khan Academy

What we're going to do in this video is start thinking about the position of an object traveling in one dimension. To get our bearings there, I'm going to introduce a few ideas.

So the first idea is that of displacement. 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 gets confused with displacement is the 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 5 minus t. Now, this is a one-dimensional velocity function. Let's say it's just telling us our velocity in the horizontal direction.

Oftentimes, when something's one dimensional, 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. 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 it, although that's not always the convention used in different classes.

Now, let's plot what this velocity function actually looks like. I did that ahead of time. So you can see here time equals zero. Let's say time is in seconds, and our velocity is in meters per second. So this is meters per second right over here, and this is seconds in this axis.

At exactly time zero, this object is traveling at five meters per second. We could say to the right it has a velocity of positive 5 meters per second. But then it keeps decelerating at a constant rate. So, 5 seconds into it, right at 5 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 the first five seconds, well, we've seen already multiple times if you want to 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 could take the integral from zero to 5 of our velocity function, just like that. We can even calculate this really fast. That would just be this area right over here, which we could just use a little bit of geometry. This is a five by five triangle, so five times five is twenty-five times one-half. Remember, the area of a triangle is one-half base times height, so this is going to be 12.5.

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 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 0 to 10 of our velocity function, our one-dimensional velocity function, dt.

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 going to be negative area. In fact, this area and this area are going to exactly cancel out, and you're going to get 0 meters.

Now you might be saying, "How can that be? After 10 seconds, how do we—why is our displacement only zero meters? This particle has 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—its velocity starts becoming negative, and the particle starts moving to the left. 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 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.

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. 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 could just view it as our speed function right over here, dt.

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 going to be the integral from 0 to 10 of the absolute value of our velocity function, which is going to be equal to what? Well, 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 be 5 times 5 times 1/2 plus 5 times 5 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 zero, but the total length of path traveled is 25 meters.