Worked example: distance and displacement from position-time graphs | AP Physics 1 | Khan Academy

In other videos, we've already talked about the difference between distance and displacement, and we also saw what it meant to plot position versus time. What we're going to do in this video is use all of those skills. We're going to look at position versus time graphs and use them in order to figure out displacement and distance traveled.

So this first question says: A 3.2 kg iguana runs back and forth along the ground. The following graph shows the horizontal position of the iguana in meters over time. The first question is: What is the displacement of the iguana between 0 seconds and 5 seconds? So be very careful; it's not saying between 0 seconds and 6 seconds; it's saying between 0 seconds and 5 seconds. So pause the video and see if you can figure that out.

Well, displacement is just your change in position, and sign matters; we care about direction. So it's going to be your final position minus your initial position. Well, what is your final position? Well, we're not talking about the final—the last time that we have on the graph here. Our final is going to be at 5 seconds, so our final position is at a positive 6 m.

So our final position is positive 6 m, and from that, you're going to subtract our initial position. Well, our initial position was at -2 m. So this is going to be equal to 6 - (-2), which is positive 8 m. So one way to think about it is that this iguana shifts 8 m in the positive direction. If we think positive direction is, say, to the right, it would be 8 m to the right.

We could draw a number line here, so if we—it's sometimes confusing because we're talking about the horizontal direction, but we're plotting position vertically here. But we could take the same number line and make it horizontal, and you'd have -6, -4, -2, 0, 2, 4, 6, 8. So what's happening here is the iguana is starting at -2, and then over the next 3 seconds, it goes to positive 6. It goes over here, and so it shifts to the right by 8, and that's what we saw right over there; 6 - (-2) is 8.

Now, what about the distance traveled by the iguana over that same time period? Pause the video and see if you can figure it out. Well, the distance is the total length traveled, the total path length. But once again, it went from -2 all the way to positive 6 over the first 3 seconds, and then it just stays there. So if we think about distance, we're actually going to get the same result. Even though we don't care about direction here, we just care about the magnitude, we still get the same thing: it traveled 8 m.

So it travels 8 m, so we're going to get the exact same result. Now, what would be interesting is to think about what would be different if instead of 5 seconds, if this said 6 seconds, if this was between 0 and 6 seconds? Well then, for displacement, we would say, “Hey look, we went plus 8,” but then we're going to go another -6.

So, if you did it for all the way to the 6th second, then your displacement is going to be plus 2 m. You have the 8, and then you subtract the 6. Another way to think about it is you would finish at zero; you started at -2. So, 0 - (-2) is 2.

If you wanted to think about the distance between 0 seconds and 6 seconds, well you would have this 8 m, and then you would go— even though you're going back to zero, so you're going back to zero right over here, 6 m. You wouldn't subtract it because the total path you travel is 8 meters to the right and then 6 meters to the left. So you would add them.

So if you said over the first 6 seconds instead of the first 5 seconds, this would be 14 m. Let me make that clear again: displacement and distance. In both cases, you have plus 8 over that first leg, I guess you could say. When we thought about displacement, we subtracted because we're now moving to the left. We moved to the left by 6, so in the case of displacement, you subtract the 6 and you have a net displacement of plus 2.

But distance—the total path traveled—you have the 8 to the right and then the 6 to the left, which gives you a total path traveled of 14. Let’s do one more example here: A 2.7 kg armadillo rolls in a straight line in the desert. The following graph shows the horizontal position of the armadillo in meters over time.

So let’s think about the same thing over the first 24 seconds. Let’s go all the way to the 24th second and think about what the displacement is and what the distance traveled is. So first, pause this video and see if you can figure out the displacement over the first 24 seconds.

Well, this is going to be our final position minus our starting position. Our final position—we’re at 0 m; our starting position—we were at time zero, we were at 6 m, or let me just write the numbers down, so it is -6. Let me say that one more time: at time 24, notice our vertical coordinate we are at a position of zero—that's where that came from.

When we started at time zero, our position was right over here. So our final minus our starting is -6, and you could also see that if you just look along this line. We shifted from positive 6 to zero, which would be a shift of six to the left, or a displacement of -6.

Let me draw this on a horizontal line just to make this a little bit more clear. So if you have zero, four, oh, zero, three, three, six, nine, twelve, fifteen, so on and so forth, we are starting at six. Let me do that in purple color. We’re starting at six; we do a bunch of stuff in between, but then we end up after 24 seconds at zero.

So our shift—we went six to the left, or we have a displacement of -6. Zero - 6 is 6. So now, let’s try to figure out distance. Pause the video and figure out the distance that this armadillo travels over this 24 seconds. So this is interesting. It starts off at a position of six. So let me do it right here: it starts off at position of six.

It stays there for the first eight seconds, then from the 8th second to the 16th second, its position increases by nine to get to 15. So it does this; it goes to 15. So this is going to be plus—this is going to be plus 9. Then, on the 16th second, it goes from 15 back to zero. So it goes from 15 back to zero. If we were thinking about displacement, we would write a minus 5 here, and then we would net these out to get to a -6, but we’re thinking about distance.

So we want to think about the absolute value of the various parts of the path traveled, so all of these are going to be positive. We just say, “Hey, what is the total journey?” So this is going to be plus 15. We just care about the lengths of these arrows, not the direction.

So 9 + 15 is 24. So this is interesting: even though the armadillo traveled a total of 24 meters, its entire path was 24 meters long. Its net shift, its displacement, is 6 m to the left. And this is another thing to emphasize: this negative number is implying direction; it's saying if we're looking at this number line, it's saying to the left.

Even if it was a positive 6, because we're talking about displacement, it would imply positive would mean to the right. Distance doesn’t tell you about direction; it just tells you the absolute magnitude of the total distance traveled or the length of the path.