 A typical type of kinematics problem is the so-called catch-up problem. It can have different variations of it, but in the end, it's always two objects moving on different paths, and the question is, where and when do they meet? So let's look at an example where a car, A, leaves Montreal heading towards Toronto with a speed of 50 kilometers an hour, and car B leaves Montreal one hour after car A, driving at a speed of 100 kilometers an hour towards Toronto as well. And the question is, where and when will car B catch up with car A? Now, there are two ways of solving it. One is doing an ST graph, and one is with the equation. We're going to solve both of them, and both have the same steps that we're always going to be looking at. The first thing that you should look at is when is time zero? Which one do we consider time as zero? You can actually choose yourself. Do we consider time as zero when A leaves Montreal, or do we consider time as zero when B leaves Montreal? The advantage of using B is that then you can do it with graphs and equations. If you were to choose A, the equations for constant exploration will not work because B will not have a constant exploration for the entire time. B will have zero velocity, then a huge peak of exploration, then suddenly 100 kilometers an hour, so you would have to split up the movement somehow by using a graph. So if you want to be using equations, there's an advantage to make sure that both objects are having a constant exploration, and this is where you consider time as t equals zero. Now, once you've established when time is zero, so for us time is zero, when car B leaves Montreal, so let's add this here at time equals zero, B leaves Montreal, and then we have to look, okay, when B leaves Montreal, at this moment, where is A? So B leaves Montreal one hour after A, so that means A was already driving 50 kilometers an hour for one hour, that means in our position time graph after when B is starting at zero, our car A already traveled 50 kilometers, 50 kilometers an hour times one hour, that means our car A was already here. Now let's draw the entire graph of A, what it's doing. In each hour, A will be driving 50 kilometers further, so at one hour A is at 100 kilometers, at two hours A will be at 150, three hours will be at 200 kilometers. So this is a rough sketch of what A is doing. Now let's have a look what B is doing, B is driving at 100 kilometers an hour, and we say starting at equal zero in Montreal at kilometers zero, after one hour, where is B? B was driving for one hour with 100 kilometers an hour, so in this case it will be at 100 kilometers, at two hours C will be at 200 kilometers, at three hours at 300 kilometers and so on. So here is what the movement will look like more or less. When and where do they meet? Well I think it's pretty obvious, that's when the two lines cross. So here is where they meet 100 kilometers outside of Montreal and one hour after B left Montreal, so two hours after A left Montreal. So we found the solution from the ground. Now how could you have done the same thing with the equations? With the equations we simply have to write down both equations of car A and car B. So car A, the blue one, is the position as a function of time, is its initial position, so in this case 50 kilometers plus its initial velocity, so 50 kilometers per hour times time, and we considered we are ignoring accelerations, traffic jams and all the other things. Car B, as there's a function of time, is where it was initially, zero kilometers plus V initial times times 100 kilometers per hour times time, also we were ignoring acceleration, traffic jams, etc. etc. And now what we have to do to find when and where they meet is simply do S A equals S B as a function of time and you're going to be solving. You have to be at the same position and you find at what time are they at the same position, so I can write my S A 50 kilometers plus 50 kilometers per hour times time equals 100 kilometers per hour times time, and I'm solving for time, so I have 50 kilometers is 100 minus 50, 50 kilometers per hour times time, so I get that time is one hour, and then I plug it in either in A or B, in this case I would definitely use B, and one hour times 100 kilometer gives me position is 100 kilometers, and I get exactly the same answer. Now this was a pretty straightforward simple example where we had no accelerations or we were ignoring them because over a long period of time we can totally do that. What however if you have something where a car is accelerating, well the curve that would simply look a bit different, for example if I had a car C that accelerates it could have a curve like this, and this would be the point where I'm meeting or catching up with car A, and in the equation then you would have a term with the t square and your equation here to solve will become a quadratic equation to solve, where you probably will have two times, you're called your calculator will always give you two possibilities because the calculator will not know or not distinguish between positive and negative time, so your calculator most likely will give you a negative time and a positive time, of course it's the positive time, that's the one where really the meeting happens.