Hey, guys. You're going to have to solve motion problems in which you have 2 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 that you're commonly going to see in those problems. Now, the way that you solve these problems is actually really, really repetitive. You're going to do the same thing over and over again. So I'm going to give you a list of steps in this video that's going to help you get the right answer all the time. Let's check it out.

So guys, the main idea we're going to use here is that when one object catches up to or overtakes another, what that means is that they're at the same position at the same time. For example, let's say I've got a race between these two runners. The starting line is over here. Runner a has a head start, which means that they have an initial, position of \( X_a(0) \) and they're running along with a velocity of \( V_a \). Now runner b is a little bit farther behind \( X_b(0) \), but they're a little bit more determined to finish the race and win. So they're actually running a little bit faster, which means that \( V_b > V_a \). So eventually, what happens is if runner b is going faster, but they're behind then 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 point. It's the place where they're at the same position at the same time. So if we have the position for a, so I'm going to call that \( X_a \) and the position for b \( X_b \), then these two positions are equal to each other. And the other thing is that a time it takes for a to get there, \( T_A \), and the time it takes for b to get there, \( T_B \), those things are also going to be equal as well. The best way to see this is actually on a position-time diagram or a position-time graph. So I've got one right over here. I'm going to use red for a and then blue for b. So red, a starts at \( X_a(0) \) with a head start. So they're going to start a little bit higher on the graph. So this is my \( X_a(0) \). B is a little bit behind so this is my \( X_b(0) \). But what happens is if the velocity of b is higher, remember, the velocity is the slope of the position graph, that means their slope is going to be a little bit steeper. So this is what the position-time graph would look like. And this overtaking point 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 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 a 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 going to need numbers and equations to solve them. And so that's why I'm going to give you a list of steps here that's going to 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 going to use a very particular list of steps to solve them. Let's check it out. So we've got these 2 cars. They're driving along the same road. Car a is at this initial position with an initial speed, and then car b is a little bit farther ahead, 280 meters far ahead and traveling at a constant 36. So the first step here is we're going to draw the diagram and list the known variables. This is just going to help us for any motion problem no matter what. So let's go ahead and do that. So we've got car a is going to look like this. So this is my \( A \). So I've got initial velocity. Let's see. My \( V_A \) is equal to 50 and I know this initial position \( X_a(0) \) is equal to 0. So something looks like this. Now, car b is a little bit farther ahead. So they're a little bit you know, over here somewhere. But eventually, what happens is that car a is going to catch up to car b. So that means that at some later points, these two things are going to have the same exact value over here. So this is car b and I know that the velocity for b is actually 36. It's going to be slower than the than the 50, but it starts starts a little bit farther ahead. So my initial position is 280 meters. So that's it. That's it for the first step. The next thing I want to do is I want to write the objects full position equations. So remember that the overtaking points, the black line that I've 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, I've 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 and this actually just comes from our UAM equations. We take a look at, equation number 3, our \(\Delta x\) equation. This has, this is the one that we've been working with so far, but we're going to use actually this version of this position equation because this is going to give us the final position, relative to the initial position. So remember, these two things are just kind of like different versions of each other, but we're going to use this one, the bottom one over here because because we're actually looking for the final positions of both of these objects. Alright. So let's get to it. So that was the first step which is the diagram. The second step right here is going to be me writing the equations. So I've got my \( X_a \) and remember, this is going to be the initial position plus velocity and time plus acceleration and, time. So we've got the initial position over here, which is just 0 plus. Now I've got the velocity. The velocity is a constant 50 meters 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 for both of them is actually just 0. So, that means that there is no acceleration for this particular term here. It's going to be \( \frac{1}{2} \) of 0 times \( T^2 \), and so we're just going to cancel out that term and cancel out this term. So if we do the same thing for b, b is going to have an initial position of 280 that's going to that's going to be 280 here plus the velocity is 36 \( t \). And then this is going to be also no acceleration so \( \frac{1}{2} \) of 0 times \( t^2 \) 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, then 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. So we just take those equations and we just literally set the right sides equal to each other. So that means that my \( X_a \) which is 50 \( t \) is going to be equal to the right side of this equation which is just 280 +36 times \( t \). So this is the 3rd 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 car b? So now that we've solved for and now that we've set these equations equal to each other, all you do is just solve for the time. So, basically, we're going to do is we're going to move this 36 over to the other side. So we have \( 50 t - 36 t \) is equal to 280. So now we can group these terms together \( 14 t = 280 \). And so therefore, \( t = \frac{280}{14} \) which is just 20 seconds. So this is the time. This is how long it takes for car a to catch up to car b. 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 4th step is we're going to solve for time and also any additional variables and so this is some of the additional variables that you might be asked for. Sometimes you might be asked for what's the speed of 1 or the speed of the other or the position of 1. But now that we have the time, we 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 \( X_a \), my \( X_b \) and remember I have the equations for those, those equations are right here. And now now that I have the time that it takes those two things to meet each other, now, I can just figure out what the positions of those things are. So my \( X_a \) is just going to be 50 \( t \). So this is just going to be 50 times 20 and this is just a 1000 meters. And if I do the same exact thing for a, I'm sorry for b, my \( X_b \), this is going to be 280 + 36 \( t \), so it's going to be 280 + 36 times 20 and if you plug this in, you're also just going to get a 1000 meters. 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 questions.