Catch/Overtake Problems - Online Tutor, Practice Problems & Exam Prep

In catch-up problems, two objects move towards the same position at the same time. To solve these, start by drawing a diagram and listing known variables. Write position equations for both objects, typically in the form $x=x₀+vt$. Set these equations equal to find the time when they meet. Finally, substitute this time back into either equation to find the position where they meet, ensuring both objects have the same position at that time.

Solving "Catch Up"or "Overtake"Problems

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.

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.

• A. (a) t = 12 s (b) v = 72 m/s
• B. (a) t = 12 s (b) v = 48 m/s
• C. (a) t = 36 s (b) v = 48 m/s
• D. (a) t = 36 s (b) v = 72 m/s

Superman & the Watermelon

Hey, guys. We've got an interesting little example problem here. We're going to be dropping a watermelon off the Empire State Building, and then Superman is going to fly by the instance that we release it, and we're going to calculate how fast the watermelon is going once it passes Superman again on the way down. Let's check it out. So, basically, we've got a bunch of things passing each other, but now we've got this thing happening in vertical motion because we're going to have things falling. But we're still just going to 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. This is step one. We've got the Empire State Building that looks like this. Well, it's pretty good. So, we're all the way up here at the top, and we're going to drop a watermelon, and it's going to fall. But the instant that we drop it, Superman is going to be coming by, and he is essentially going to be going in the same exact direction. And eventually, the watermelon is going to catch up to Superman, so this is going to be the overtaking point.

Now, I'm going to use red for the 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 can write out the position equations. That's going to be the second step. So we're dropping the watermelon from rest, which means that the initial velocity for the watermelon, which I'll call \(v_{0a}\), is equal to 0. But I know that the watermelon does accelerate, so it is actually accelerating here. But because this is vertical motion, I already know what that acceleration is going to be. I already know that the acceleration in the y-direction, so I'll call this \(\ a_{ay} \), is just equal to negative \( g \), which is going to 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 going to 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 about Superman? Superman is going to fly down, heading at a constant 35 meters per second. So that means at this point, Superman does have an initial velocity. So \( v_{0b} \) is equal to 35. But be cautious 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 \( a_b \) in the y-direction is actually 0 because it says gravitational force doesn't matter, doesn't apply when you're Superman. He can fly at whatever speed he wants to. We also know the last thing, that the distance of the Empire State Building is 320 meters. So what we can do is say that this is actually the 0 points, and we can say that the position that we're already starting off from, our \( y_0 \), is equal to 320, since we're actually heading downwards from it. So we can just say that \( y_0 \) is equal to 320, and it's actually going to be the same for both of them since they're starting at the same place. That's it. That's all we know about this problem.

So the second part here is we're just going to write out the full position equations for each. So the second part here is we write \( y_a \) or actually we're going to 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 going to be my initial position in the y, which is 320. 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've got 0 plus, and now we've got 1 half times the acceleration in the y direction, which we know is negative 9.8, times \( t^2 \). So if you simplify this by canceling out, so, you know by, doing the one half times negative 9.8, we're going to get negative 4.9 \( t^2 \) plus 320. So let's do the same thing for Superman. So \( y_b \) for Superman, 320, the same initial position plus. Now, we've got the 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 0 over here. So we rewrite out the same thing. We're going to get negative 35 \( t \) plus 320.

So that means these are the position equations over here, which bring us to the 3rd step. Now we're just going to set those position equations equal to each other. Because remember, we're trying to solve for time. That's always what we're trying to solve for in these catch and overtake problems. So I've got negative 4.9 \( t^2 \) plus 320 equals negative 35 \( t \) plus 320. So these things are actually going to have the same 320 on both sides of the equation. They're starting off in the same place. So one of the things we could already do is just cancel them both out. We're going to subtract 320 from both sides. It's basically just a wash. They basically just go both go away. And so we're only left with these last two terms, 4.9 \( t^2 \) and negative 35 \( t \). Now the other thing that we can do is we can also cancel out the negative signs that are for both of them because they're both negative. What we end up with is 4.9 \( t^2 \) equals 35 \( t \). Now remember, we're trying to solve for this \( t \) 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 4.9 \( t \) is equal to 35. And now the last thing I do is just divide by 4.9. So my \( t \) is equal to 35 over 4.9, and what I get is I get 7.1 seconds. So this is the time it takes for the watermelon to catch up back to Superman because the watermelon is accelerating. But I'm not done yet. Remember, this isn't my final answer because what I'm really looking for is 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 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 watermelon. So let's do just that over here. So if I want to know my \( v_a \) here, then that's a motion variable. I need to use one of my motion equations. But to do that, I need 3 out of my 5 variables. So do I have that? Well, I've got the initial velocity. I've got the acceleration in the y direction, and I also now have the time. So that means that I have 3 of my 5 variables, and I can figure out my final velocity. What this means is that my \( \Delta y_a \), basically, the distance that it takes for them to 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, where we just pick any equation that doesn't have the ignored variable, which is going to be the first one. The first one relates just the final velocity, initial velocities, and the accelerations of the times. So that means that my \( v_a \) so I'm just going to pick equation 1, which 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 0, right? It starts from rest, and so my \( v_a \) is just equal to my acceleration in the y-direction which is negative 9.8. It's always it for vertical motion times time, which we now know is 7.1. So if you work this out, what you're going to get is negative 69.6 meters 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 to catch up to Superman once it passes. Anyway, guys, so this is your answer. Negative 69.6 meters 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.

Here’s what students ask on this topic:

What are the steps to solve a catch-up or overtake problem in physics?

To solve a catch-up or overtake problem, follow these steps:

1. Draw a diagram and list known variables.
2. Write the position equations for both objects, typically in the form $x=x_0+vt$.
3. Set these equations equal to find the time when they meet.
4. Substitute this time back into either equation to find the position where they meet.

This ensures both objects have the same position at the same time.

How do you write the position equations for objects in a catch-up problem?

To write the position equations for objects in a catch-up problem, use the formula:

$x=x_0+vt$

Here, $x_0$ is the initial position, $v$ is the velocity, and $t$ is the time. Write separate equations for each object and include their respective initial positions and velocities.

How do you find the time when two objects meet in a catch-up problem?

To find the time when two objects meet in a catch-up problem, set their position equations equal to each other:

$x_1=x_2$

For example, if the equations are $x_1=x_0+v_{1}t$ and $x_2=x_0+v_{2}t$, set them equal and solve for $t$:

$x_0+v_{1}t=x_0+v_{2}t$

Rearrange to isolate $t$ and solve.

What is the significance of the overtaking point in a catch-up problem?

The overtaking point in a catch-up problem is the position where two objects meet at the same time. It signifies that both objects have the same position at that specific time. This point is crucial for solving the problem as it allows you to set the position equations of both objects equal to each other to find the time and position where they meet.

How do you find the position where two objects meet in a catch-up problem?

To find the position where two objects meet in a catch-up problem, first determine the time $t$ when they meet by setting their position equations equal. Then, substitute this time back into either object's position equation:

$x=x_0+vt$

This will give you the position $x$ where both objects meet.