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Solving "Catch Up"or "Overtake"Problems

Patrick Ford
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Hey, guys, you're gonna have to solve motion problems in which you have two moving objects and one of them is trying to catch up to another one. I call these problems catch up or overtake problems because these are words your commonly going to see in those problems. Now, the way that you solve these problems is actually really, really repetitive. You're gonna do the same thing over and over again. So I'm gonna give you a list of section this video that's gonna help you get the right answer all the time. Let's check it out. So, guys, the main idea we're gonna use here is that when one object catches up to or overtakes another, what that means is that there at the same position at the same time, for example, let's say I've got a race between these two runners. Starting line is over here. Runner A has a head start, which means that they have in initial position of x, not a. And they're running along with a velocity of V A. Now runner B is a little bit farther behind x not be, but there are a little bit more determined to finish the race and win. So they're actually running a little bit faster, which means that VB is greater than V A. So eventually what happens is if Runner B is going faster there behind than Runner B is going to catch up and overtake Runner A And this point right here, the place where we have these two dots. This is the overtaking points. It's the place where the same position at the same time. So if we have the position for a so I'm gonna call the X A and the position for B XB than these two positions are equal to each other. And the other thing is that the time it takes for a to get their t A and the time it takes for B to get their TB those things are also going to be equal as well. The best way to see this is actually on a position, time diagram or position time graph. So I've got one right over here. I'm gonna use red for a and then blew for B. So read a starts at a little bit of a with, you know, with a little bit of that head start. So they're going to start a little bit higher on the graph. So this is my ex, not a B is a little bit behind. So this is my ex. Not be. But what happens is if the velocity of B is higher, Remember, the velocity is the slope of the position graph. That means they're slope is gonna be a little bit steeper. So this is what the position time graph would look like. And this overtaking points is really just the place where the lines are going to cross. So this point right here the overtaking point is really just the places where the lines intersect on a position time graph. So these two things are the same thing. So notice how also, that these these lines basically the red and the blue lines here both have the same position. They're both right here at the same time. Now, again, this is kind of just ah, way to visualize what's going on with the overtaking. Now. A lot of times you won't have these position time graphs, so you're gonna need numbers and equations to solve them. And so that's why I'm gonna give you a list of steps here that's gonna help you get the right answer every time. Let's just go ahead and take a look at this problem right here. Now, what I want to point out first about these steps is that these are actually pretty different from how we've solved motion problems with acceleration. And that's because there's a very particular kind of problem. So we're gonna use a very particular list of steps to solve them. Let's check it out. So we've got these two cars, they're driving along the same road car A is that this initial position with initial speed and then Carby is a little bit farther ahead, 280 m, far ahead and traveling at a constant 36. So the first step here is gonna draw the diagram and this unknown variables. This is just gonna help us for any motion problem, no matter what. So let's go ahead and do that. So we've got car A is gonna look like this. So this is my A. So I've got an initial velocity. Let's see, my va is equal to 50 and I know this initial position X, not a is equal to zero, so something looks like this. Now, Carby is a little bit farther ahead, so they're a little bit, you know, over here somewhere. But eventually, what happens is that car is gonna catch up to Carby. So that means that at some later point these two things are gonna have the same exact value over here. So this is Carby, and I know that the velocity for B is actually 36. It's gonna be slower than the 50 but it starts a little bit farther ahead. So my initial position is 280 m. So that's it? That's it for the first step. The next thing I wanna dio is I want to write the objects full position equations. So remember that the overtaking points the black line that have got over here is the place where the position of A and the position of B are actually going to be the same. So I have to write equations for both of these variables and the equations of actually listed them in this table right here. This is the equation for a and this is the equation for B. You might be wondering where we get those equations from its actually just comes from our, um, equations. We take a look at equation number three are Delta X equation. This has, um this is the one that we've been working with so far. But we're gonna use actually this version of this position equation, because this is gonna give us the final position relative to the initial position. So remember these two things were just kind of, like, different versions of each other, but we're gonna use this one the bottom one over here because, um, because we're actually looking for the final positions of both of these objects. All right, so let's get to it. So that was the first step. Which the diagram? The second step right here is gonna be me writing the equations. So I got my ex a And remember, this is gonna be the initial position, plus velocity and time, plus acceleration and time. So we've got the initial position over here, which is just zero plus. Now I've got the velocity. The velocity is a constant 50 m per second. So I've got 50 times T then plus any acceleration. Remember that both of these cars are driving at Constant 50 and constant 36. So that means that the acceleration of both things actually just zero. So that means that there is no acceleration for this particular term here. It's gonna be one half of zero times t squared. And so we're just gonna cancel out that term and cancel out this term. So if we do the same thing for B B is gonna have an initial position of 280 that's gonna that's gonna be 280 here. Plus the velocity is 30 60 and then this is gonna be also no acceleration. So one half of zero times t squared doesn't matter. That term goes away. So that means these two equations here are our position equations for A and B. So that's the second step. And if you remember, that in the third step is actually pretty straightforward, because now that we have both of these position equations, all we do is just we set them equal to each other. Remember that these two position equations are going to be equal to each other. Eso we just take those equations and we just literally set the right sides equal to each other. So that means that my ex A, which is 50 t, is going to be equal to the right side of this equation, which is just 280 plus 36 times t. So this is the third step over here. And so now, now that we've set those equations equal to each other, notice how the one thing that's missing is just the time. That's the only variable that's left. And that's exactly what we're solving for in part A. When does car a catch up to Karbi? So now that we've solved for now that we've set these equations equal to each other, all you do is just solved for the time. So basically, we're gonna do is we're gonna move this 36 over to the other side. So we have 30 50 minus 30 16 50 t minus 30 is equal to 280. So now we can group these terms together. 50 T minus 30 60 is 14 t equals 280 so therefore, t equals 2 80/14 which is just seconds. So this is the time. This is how long it takes for car A to catch up to Carby. So this is my part? A So let's move on to part B. Now, Part B is asking us for at what position in meters do the cars meet? So remember the fourth step. We're going to stall for time and also any additional variables. And so this is something additional variables that you might be asked for. Sometimes you might be asked for what's the speed of one or the speed of the other or the position of one? But now that we have the time, well, we will be able to use much more equations here. So let's figure this out. So we're looking for the position of a or we're looking for the position of B. So this is my ex a my ex B. And remember, I have the equations for those. Those equations are right here and now. Now that I have the time that it takes for those two things to meet each other now, I could just figure out what the positions of those things are. So my ex A it's just gonna be 50 t so this is just gonna be 50 times 20 and this is just m. And if I do the same exact thing for a I'm sorry for be my ex B, this is gonna be 280 plus 36 t. So it's gonna be 280 plus times 20. And if you plug this in, your also just gonna get 1000 m. So it's no coincidence that we get the same number, because remember, these two things will have the same position at the same time. Alright, guys, that's it for this one. Let me know if you have any question.