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2. 1D Motion / Kinematics

1

concept

9m

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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.

2

Problem

A police car at rest is passed by a speeder traveling at a constant 36 m/s. The police officer takes off in hot

pursuit, accelerating at a constant 2.00 m/s2. **(a)** How long does it take for the police officer to overtake the speeder? **(b)** Calculate the speed of the police car at the overtaking point.

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3

example

7m

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Hey, guys. We had an interesting little example. Problem here, we've got we're gonna be dropping a watermelon off the Empire State Building, and then Superman is gonna fly by the instance that we release it and we're gonna calculate how fast the watermelon is going. Once it passes Superman again on the way down, let's check it out. So basically, you've got a bunch of things passing each other. But now we've got this thing happening in vertical motion because you have things falling. But we're still just gonna follow the same exact steps that we use for catch and overtake problems. So first things first, let's just draw the diagram and list everything we know about the problem. So this is step one. So we've got the Empire State Building. It looks like this. Well, that's pretty good s So we got the Empire State Building that looks like this. So we're all the way up here at the top, and we're gonna drop a watermelon, and it's gonna fall. But the instant that we drop it, Superman is gonna be coming by and basically gonna be going in the same exact direction. And eventually the watermelon is going to catch up to Superman. So this is gonna be the overtaking points. All right? So basically, I'm gonna use red for the watermelon, so my red is gonna be watermelon, and my blue is going to be Superman. So now let's list what we know about each one of these objects so we could write out the position equations. That's gonna be the second step. So we're gonna draw the diagram list of known variables. So we're dropping the watermelon from rest. Which means that the initial velocity for the watermelon, which I'll call V not A is equal to zero. But I know the watermelon do does accelerate. So it is actually accelerating here, but because this is vertical motion, I already know what that acceleration is going to be already know that the acceleration in the Y direction so called this a a Y is equal to negative g, which is gonna be negative 9.8. And remember that when we do vertical motion problems, remember that the positive direction is always going to be up, and that's why it's negative 9.8. So then eventually, this thing is gonna get down here at some final position where these two things are equal to each other, the same position at the same time. And I'm trying to figure out what the final velocity is of the watermelon. So what is the speed of this watermelon when it passes? Superman? So what about Superman? Superman is gonna fly down, heading at a straight our constant 35 m per second. So that means at this point, Superman does have an initial velocity, so we not be is equal to 35. But be careful, because remember that the positive direction is upwards. So because this velocity is downwards, we actually have to put a negative sign here. You have to remember that. And so the acceleration for Superman. So the acceleration for B in the Y direction is actually zero because it says that gravity doesn't matter, doesn't apply. When you're Superman, he can fly whatever speed he wants. Thio, we also know the last thing that the distance of the Empire State Building is 320 m. So we can do is we can say that this is actually going to be the zero points on. We can say that the position that we're already starting off from our Y zero is equal to 320 since we're actually heading downwards from it. So that we can just say that why not is equal to 3 20. That's actually gonna be the same for both of them since just starting at the same place. That's it. That's all we know about this problem. So the second part is we're just gonna write out the full position equations for each. So the second part here is we're right, x A or actually, we're gonna use y A because we're talking about the vertical motion direction. So all these things still apply for vertical motion. So my y A is gonna be my initial position in the UAE, which is 3 20. That's what we said, right, that's over here, plus the initial velocity, times time. But we know that the watermelon gets released from rest, which means that this term actually just goes away. So we got zero plus and now we've got one half times the acceleration in the Y direction, which we know is negative 9.8 times t squared. So if you just simplify this by canceling out. So, you know, by doing the one half times negative 9.8, we're gonna get negative. 4.9 t squared, plus 320. So let's do the same thing for the Superman. So why be for y Superman? 3 20 The same initial position. Plus, now I've got initial velocity. Well, here. Now, we've got initial velocity of negative 35. So it's negative. 35 t plus. And now we've got any acceleration Times time. But remember that Superman doesn't have any acceleration, so that means this whole term goes away. So we've got zero over here. So we're right at the same thing. We're just getting to simplify. Some of these things were gonna get negative. 35 T plus 320. All right, so that means these air, the position equations over here, which brings us to the third step. Now, we're just going to set those position equations equal to each other. Because remember, we're trying to solve for the time. That's always we're trying to solve, solve for these catching overtake problems. So I've got negative 4.9 t squared, plus 320 equals negative. 35 t plus 320. So these things actually are gonna have the same 320 in both sides of the equation. They're starting off in the same place. So one of things, we could already dio it. Just cancel them both out. We're going to subtract 3 20 from both sides is basically just a wash. Their basically just go both both go away. And so we're only left with these last two terms 4.90 square and negative 35 t. Now, the other thing that we can do is we could also cancel out the negative signs that air for both of them because they're both gonna be negative. We end up with is we end up with 4.9 t squared equals 35 t. Remember, we're trying to solve for this tea here, but notice how I have a t on both sides of the equation, which means I can cancel out one term and one term, and I just end up with 4.90 is equal to 35 now the last thing I do is just divide by 4.9. So my T is equal to 35/4 0.9. And what I get is I get 7.1 seconds. So this is the time that it takes for the watermelon to catch up back to Superman because the watermelons accelerating. But I'm not done yet. Remember, this is my final answer. Because what I'm really looking for, I'm looking for the speed of the watermelon once it passes. So let's take a look at our diagram again. We know now that the time that it takes for both of these things to catch up to each other is 7.1 seconds. So now I just have to figure out the speed of the watermelons. So let's just do that over here. So if I want to know my V A here, then that's emotion. Variable. I need to use one of my motion equations. But to do that, I need three out of my five variables. So do I have that? Well, I've got the initial velocity. I've got the acceleration, the Y direction and I also now have time. So that means that I have three of my five variables and I can figure out for my final velocity. What this means is that my delta y a basically the distance that it takes for them over overtake each other. I don't know what that is, but it doesn't matter because I don't know and I don't care about it. So it's my ignored variable. So now that just leads us to that step in the motion equations, Um, where we just pick the equation that doesn't have the ignored variable, which is gonna be the first one First one relates just the final velocity, initial velocities and accelerations of the times. So that means that my va So I'm just gonna pick Equation One says that my final velocity for a is my initial velocity for a plus my acceleration in the Y direction times time. So I know that this is equal to zero, right starts from rest. And so my va is just equal to my acceleration. The y direction, which is negative. 9.8. It's always it for vertical motion, times time, which we know now is 7.1. So if you work this out, where you're gonna get is negative. 69 negative 69. m per second. Notice how the negative sign tells us that the watermelon is falling down and the number is higher than Superman, which actually makes sense. The watermelon has to be going faster in order to catch up to Superman once it passes. Anyway. Guys. So this is your answer. Negative 69.6 m per second. So notice how we can use all the same steps for vertical motion in terms of catching overtake problems. It's the same exact thing. You just go through all the steps and you get the right answer. Alright, guys, that's it for this one.

Additional resources for Catch/Overtake Problems

PRACTICE PROBLEMS AND ACTIVITIES (4)

- David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accele...
- David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accele...
- At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with...
- At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with...

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