A police car at rest is passed by a speeder traveling at a con...

Question

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.

Accepted Answer

Hey, 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 cars and accelerate and overtake the speeder. So we've got to 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 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 point. So that speeder kind of blows past the police car like this, going a constant velocity. So we know that this constant velocity here is 36. So this is my V, not a. We also know the acceleration is equal to zero 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 point, but eventually the police guards and 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 basically there's a difference here because now we know that the initial velocity for the speeder is equal to zero. But it is constantly accelerating. We know the acceleration is equal to two. So that's it. That's the first step. That's everything we know. This is gonna be part A. 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 on the diagram and listing of 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. We just 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 started the same exact position, which is right here, then we can call it whatever we want. But we'll just call it zero. So let's just say that the X not for both of them just equal to zero. We can kind of choose that would make these problems a little bit simpler. So that means this term just goes away. So we've got zero plus now we've got the initial velocity for a So the additional velocity knows 36 times teeth and plus, and then we've got one half of the acceleration times times squared. Well, we know that the acceleration for A is actually zero. So that means this whole term just goes away, which is to get zero. And this is just 30 16. Let's do the same thing for B. So the initial position zero now, So you 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 zero, because it starts at rest. So this is also gonna be zero, because this term goes away. And now we have one half times acceleration Times Times Square. So we've got one half a is to times t squared. So we just end up with because we have one half and two that cancel. We just end up with T Square. 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 got 30 60 for won 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 the same for these two vehicles or objects to be in the same position at the same time. So now we have to just set their position equations equal. So you've got 30 60 for a equals t squared for B. And so now we just solved 40 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 well, you'll notice that I have tease on both sides of the equation, so I can just cancel one of the terms out. Just cancel one power of tee 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 he go to 36. So now let's just quickly do part B over here. So now we want to calculate the speed of the police car at the overtaking points. So, really, what variable is that in our equations? What? We know that basically both of these things are gonna be at the same point. One of the ways to kind of visualized this 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 point that speeders 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 that graph is actually gonna be a little bit curvy. And we know if 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 point. Okay, so we're going to solve for T. Sorry, we're going to solve for the speed now of this police car. So that is actually gonna be the variable variable vb. 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 three out of the five to use our motion equations. We actually do know what the three out of five are. We know V knots. We know a and we have the time. So now we can use basically, we could 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 while the first one here doesn't have Delta X, and it just relates the acceleration of time and the change in velocity. So this is we're gonna use here, So we're just gonna use my V B. This is the first equation is equal to V not be which we already know. Zero plus the acceleration, which is two times the time, which is 36. And so we just end up with a final velocity of 72 m 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 cars velocity the cop cars. 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.

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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