 So we learned that if we have a position time graph given, we can find the velocity time graph by looking at the derivative or the slope of the position time graph. But what do we do if you have to go the other way around? If we are given a velocity time graph and we have to find the position time graph. Well, what's the opposite operation for derivations integration? So those of you that know calculus, the position as a function of time is the integration of the velocity as a function of time. Now those that didn't take calculus yet, this corresponds to the area between the x-axis, which is the time axis in our VT graph and the graph itself. So in this example, from zero to one second, the area, if you calculate it, was one meter per second times one second. So we had plus one meter of displacement. That means we know from zero to one seconds, we advanced by one meter. Now the problem is that doesn't tell us anywhere where to start. Those of you that took calculus know if you do an integration. There's always an integration constant plus C that sometimes you forget because it's just plus C in math, but it actually has a meaning in physics. In this case, it is the initial position. So in order to be able to draw the position time graph from the velocity time graph, you need to know the initial position. So let's assume my initial position was at one meter. So that means within one second, I went plus one meter. So at one second, I was at two meters. And I go on from one to two. Well, this is another one meter, one second times one meter per second plus another one meter. So at two, I'm at three. I go to three plus one meter again. So at three, I'm at four. And at four plus one meter again, I'm at five. Then for the next two seconds, absolutely nothing happens. I have no area between my x-axis and the graph. So plus zero meters for the next two seconds. So nothing happens. So five, six, nothing happens. And now it gets interesting. My velocity time graph is below the axis. So what do we do? From six to seven, what is my area? My area is minus two meters per second times one meter. So minus, sorry, times one second. So times, excuse me, minus two meters. So from six to seven, I went down by two meters. So starting at five, going down by two means I'm gonna end up at three from six to seven. So I'm here. Starting at five meters. And then the next square here is another minus two meters. And again, one second times minus two meters per second gives me minus two meters. So I'm going back to one meters per second. Now let's just try to make the best fit graph out of this. So it looks like six. This was our initial position time graph. And if you look at the previous videos, this is the graph that I used to create that BT graph.