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# Motion problems with integrals: displacement vs. distance

AP.CALC:
CHA‑4 (EU)
,
CHA‑4.C (LO)
,
CHA‑4.C.1 (EK)

## Video transcript

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 gets confused with displacement is the notion of distance traveled and 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 particles 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 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 and 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 time equals zero let's say time is in seconds and our velocities 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 and we could 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 meet 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 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 five zero to five of our velocity function of our velocity function and just like that and 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 area for triangles one-half base times height so this is going to be twelve point five and let's see this is going to be meters per second times seconds so twelve point five 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 twelve point five meters to the right of it assuming that positive is to the right now what about over over the first ten 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 ten seconds well we would just do the same thing the integral from zero to ten of our velocity function are one dimensional velocity function D T 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 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 and 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 what why is our displacement or only 0 meters this particles 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 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 it's 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 approach 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 down distance traveled over first five seconds four 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 and so it would be this area which we already know to be twelve point five 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 ten seconds your distance ten 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 zero to ten 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 five times five times one-half plus five times five times one half which is going to be twenty five meters the particle has gone 12.5 meters to the right and then it goes back twelve point five meters to the left your displacement your net change in position zero but the total length of path traveled is twenty five meters