Now you might start, you might start to be appreciating what the difference between displacement and distance traveled is. Over 10 seconds 12.5 meters to the right and thenġ2.5 meters to the left, and so its change in And so over the next five seconds, it actually moves 12.5 meters to the left, and then these two things net out. Starts becoming negative, and the particle starts
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. Now you might be saying how can that be? After 10 seconds how do we, what why is our displacement And in fact this area and this area are going to exactly cancel out, and you're going to get zero meters. 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 so that would be the area from here all the way to right over there. 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. Now what about over, over the first 10 seconds? Now this gets interesting, and I encourage you to pause Meters to the right of it, assuming that positive is to the right. Wherever it started, it's now going to be 12.5 So that's the change in position for that particle over To be meters per second times seconds, so 12.5 meters. So this is going to be 12.5, and let's see this is going This is a five by five triangle, so five times five is 25, times 1/2, remember area of a triangle'sġ/2 base times height. That would just be thisĪrea right over here, which we can just use a The first five seconds, we can take the integral from zero to five, zero to five, of our velocity function, of our velocity function. The rate of displacement is one way to think about it. 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 Having negative velocity, which you could interpret as moving to the left. But then it keeps deceleratingĪt a constant rate, so five seconds into it, right at five seconds, the particle has no velocity, and then it starts We can say to the right, it has a velocity of positive five meters per second. At exactly time zero, this object is travelingĪt five meters per second.
Second right over here, and this is seconds on this axis. Say time is in seconds, and our velocity's in meters per second. So you can see here, at time equals zero, let's Now let's plot what this velocity function actually looks like, and I did that ahead of time. I like to write an arrow in, although that's notĪlways the convention used in different classes. And so sometimes you will seeĪ vector quantity like this have a little arrow on it, or you will see it bolded, or you will see it bolded like that. Of it if it's positive it's moving to the right, and if it's negative In fact this velocity is a vector quantity because you could think Something's one dimension, people forget well that tooĬan be a vector quantity. Telling us our velocity in the horizontal direction. And so let's say our velocityĪs a function of time is equal to five minus t. We're going to introduce a little bit of calculus now, let's say that we have a So what are we talking about? Well let's say, and Is the total length of path, total length of path. Thing as change in position? And you will see shortly, no, it isn't always the same thing. You might say well doesn't that, isn't that just the same Sometimes get confused with displacement is a notion of distance traveled.
Your change in position, your change in position. Word in everyday language, and it literally means So the first idea is that of displacement. And to get our bearings there, I'm going to introduce a few ideas. Going to do in this video is start thinking about the position of an object traveling in one dimension.
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