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/s 2 . (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.

Guys. Let's check out this practice problem together. We've got a police car that is passed by a speeder and then basically the cop car is gonna accelerate, the police car is gonna accelerate and then overtake the speeder. So we've got a catch and overtake type situation here, so let's just go ahead and stick to the steps. So we're gonna draw the diagram and we're gonna list out the variables. So So what's going on here? The speeder passes by the police car and then eventually this police car because it's accelerating is gonna overtake the speeder. So that means that right here actually at the beginning is when they're both at the same points. So the speeder kind of blows past the police car like this going at constant velocity. So we know that this constant velocity here is 36. So this is my v naught a. We also know that the acceleration is equal to 0 for a. Right? Because it's constant speed. Now the police car is gonna be right here. Right? So this is sort of like the passing points, but eventually the police car is gonna accelerate and then overtake the speeder. So that's gonna look like this and they're both eventually going to be at the same point once again. But there's a, basically, there's a difference here because now we know the initial velocity for this police car is to 0, but it is constantly accelerating. I'm doing that acceleration is equal to 2. So that's it. That's the first step. That's everything we know. This is gonna be part a where we're calculating how long it takes for this to happen. So what is basically the time? What's the time that it takes for the police car to get back to this point and then overtake the speeder? So this is the first step, drawing out the diagram and listing out the known variables. The second thing we do is just now we write the objects full position equations. So let's get to it. So we've got the second part here. We got the position equation for a is just gonna follow this format over here, which has got all the terms. So my position equation for a, so this is the speeder, which I'll call a, and then this is the police car, which I'll call b. So my a is gonna be my initial position. Well, if they both start at the same exact position which is right here, then we can call it whatever we want, but we'll just call it 0. So let's just say that the X naught for both of them is just equal to 0 and we can kind of choose that and make these problems a little bit simpler. So that means that this term just goes away. So we've got 0 plus, now we've got the initial velocity for a, so the initial velocity knows 36 times t and plus, and then we've got one half of the acceleration times time squared. Well, we know that the acceleration for a is actually 0, so that means this whole term just goes away, which is to get 0, and this is just 30 six t. Let's do the same thing for b. So the initial position is 0. Now so we know that this term goes away. Now the we've got the initial velocity times time. Well, the initial velocity for the cop car is 0 because it starts at rest. So this is also gonna be 0 because this term goes away. And now we have one half times the acceleration times times squared. So we've got one half a, a is 2 times t squared. So we just end up with because we have one half and 2 that cancel, we just end up with t squared. So, actually, you'll see that a lot of times, some of these terms will just cancel out and make your equation a lot simpler. So we've got 36 t for 1, t squared for the other. So now that we have these two equations, the third step is we just set these position equations equal to each other. Remember, we're trying to figure out the time that it takes for these two positions to be at the same for these 2, vehicles or objects to be at the same position at the same time. So So now we have to just set their position equations equal. So we've got 36t4a equalst squared4b. And so now we just solve for t, which is the only remaining variable. Right? So this is the time that we're looking for and it's the only variable left. So if we go ahead and solve for this, what you'll notice is that I have t's on both sides of the equation, so I can just cancel one of the terms out. I can just cancel one power of t off from both sides. And so I'm only just left with one number. So t is equal to 36 seconds. That's our answer, guys. It takes 36 seconds for the police car to overtake and catch up to the speeder. So this t is equal to 36. So now, let's just quickly do part b over here. So now, we wanna calculate the speed of the police car at the overtaking points. So really what variable is that in our equations? Well, we know that basically both of these things are gonna be at the same points. One of the ways to kind of visualize this, you probably should have done this earlier, is if you look at a position time graph for both of these objects here, they would actually both start at the same points. The speeder's traveling a constant velocity here like this, and the police car is actually gonna start from the same position, but they're gonna accelerate. So that their graph is actually gonna be a little bit curvy and we know that it's speeding up that's gonna be going and getting more vertical. So basically, what we're looking for here is we're looking for the time that it takes for them both to get back to the same points. Okay. So we're gonna solve for t, sorry, we're gonna solve for the speed now of this police car. So that is actually gonna be the variable variable v b. So what is the what is this velocity here? So one thing we could do is just look at the, variables that we know for the police car. Remember, we're gonna need 3 out of the 5 to use our motion equations. We actually do know what the 3 out of 5 are. We know v naughts, we know a, and we have the time. So now we can use Basically, we can just pick the ignored variable in this equation which is gonna be delta x for b. We don't know what this variable is. And so we're just gonna come to our equations here, figure out which one we can use. Well, the first one here doesn't have delta x and it just relates the acceleration of time and the change in the velocity. So this is what we're gonna use here. So we're just gonna use my VB, this is the first equation, is equal to V naught b, which we already know is 0 plus the acceleration, which is 2 times the time, which is 36. And so we just end up with a final velocity of 72 meters per second. So this is actually double what it was for the speeder, and that makes sense because the cop car has to be going faster in order to be able to overtake them. Right? The car's velocity the cop car's velocity has to be steeper in order for there to be an overtaking point. Alright, guys. That's it for this one. Let me know if you have any questions.

Table of contents

Skip topic navigation

0. Math Review31m
- Math Review
  31m
1. Intro to Physics Units1h 29m
2. 1D Motion / Kinematics3h 56m
3. Vectors2h 43m
4. 2D Kinematics1h 42m
5. Projectile Motion3h 6m
6. Intro to Forces (Dynamics)3h 22m
7. Friction, Inclines, Systems2h 44m
8. Centripetal Forces & Gravitation7h 26m
9. Work & Energy1h 59m
10. Conservation of Energy2h 54m
11. Momentum & Impulse3h 40m
12. Rotational Kinematics2h 59m
13. Rotational Inertia & Energy7h 4m
14. Torque & Rotational Dynamics2h 5m
15. Rotational Equilibrium3h 39m
16. Angular Momentum3h 6m
17. Periodic Motion2h 9m
18. Waves & Sound3h 40m
19. Fluid Mechanics4h 27m
20. Heat and Temperature3h 7m
21. Kinetic Theory of Ideal Gases1h 50m
22. The First Law of Thermodynamics1h 26m
23. The Second Law of Thermodynamics3h 11m
24. Electric Force & Field; Gauss' Law3h 42m
25. Electric Potential1h 51m
26. Capacitors & Dielectrics2h 2m
27. Resistors & DC Circuits3h 8m
28. Magnetic Fields and Forces2h 23m
29. Sources of Magnetic Field2h 30m
30. Induction and Inductance3h 38m
31. Alternating Current2h 37m
32. Electromagnetic Waves2h 14m
33. Geometric Optics2h 57m
34. Wave Optics1h 15m
35. Special Relativity2h 10m

2. 1D Motion / Kinematics

Catch/Overtake Problems

Physics

2. 1D Motion / Kinematics

Catch/Overtake Problems

Previous problemNext problem

Struggling with Physics?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

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) t = 12 s (b) v = 72 m/s

(a) t = 12 s (b) v = 48 m/s

(a) t = 36 s (b) v = 48 m/s

(a) t = 36 s (b) v = 72 m/s

Previous problemNext problem

Verified step by step guidance

Identify the known values: The speeder's velocity is constant at 36 m/s, and the police car starts from rest with an acceleration of 2.00 m/s².

Set up the equations of motion for both vehicles. For the speeder, use the equation for constant velocity: \( x_s = v_s \cdot t \), where \( x_s \) is the distance traveled by the speeder, \( v_s = 36 \text{ m/s} \), and \( t \) is the time.

For the police car, use the equation for motion with constant acceleration: \( x_p = \frac{1}{2} a t^2 \), where \( x_p \) is the distance traveled by the police car, \( a = 2.00 \text{ m/s}^2 \), and \( t \) is the time.

Set the distances equal to each other to find the time when the police car overtakes the speeder: \( v_s \cdot t = \frac{1}{2} a t^2 \). Solve this equation for \( t \).

Once you have the time \( t \), calculate the speed of the police car at that moment using the formula \( v = a \cdot t \), where \( v \) is the final velocity of the police car.