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A police car at rest, passed by a speeder traveling at a constant 130 $\mathrm{km} / \mathrm{h}$ , takes off in hot pursuit. The police officer catches up to the speeder in 750 $\mathrm{m}$ , maintaining a constant acceleration. (a) Qualitatively plot the position vs. time graph for both cars from the police car's start to the catch-up point. Calculate $(b)$ how long it took the police officer to overtake the speeder, (c) the required police car acceleration, and $(d)$ the speed of the police car at the overtaking point.

a) See graphb) 21 \mathrm{s}c) $a_{\mathrm{pol}}=3.5 m / s^{2}$d) 72 \mathrm{m} / \mathrm{s}

Physics 101 Mechanics

Chapter 2

Describing Motion: Kinematics in One Dimension

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Cornell University

Hope College

University of Sheffield

University of Winnipeg

Lectures

03:28

Newton's Laws of Motion are three physical laws that, laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. These three laws have been expressed in several ways, over nearly three centuries, and can be summarised as follows: In his 1687 "Philosophiæ Naturalis Principia Mathematica" ("Mathematical Principles of Natural Philosophy"), Isaac Newton set out three laws of motion. The first law defines the force F, the second law defines the mass m, and the third law defines the acceleration a. The first law states that if the net force acting upon a body is zero, its velocity will not change; the second law states that the acceleration of a body is proportional to the net force acting upon it, and the third law states that for every action there is an equal and opposite reaction.

04:16

In mathematics, a proof is a sequence of statements given to explain how a conclusion is derived from premises known or assumed to be true. The proof attempts to demonstrate that the conclusion is a logical consequence of the premises, and is one of the most important goals of mathematics.

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So in this problem we have a police car this catching a speeding car. Eso what that looks like. It's the police car here and then the other car is coming and it passes. And then after it passes, the police car accelerates from rest and this is going at 130 kilometers an hour. So first we just want a plots, the position versus time eso We know that they well, we're starting at the same point, right? When the car passes so that will be at position zero and then we know they also have to the police car catches up to it after 750 meters. So at 7 50 they're also gonna be at the same point. And now, first the speeding car, Well, it's going at a constant velocity and what that looks like on a position versus time graph is linear. So that's the graph for the speeding car on the police vehicle is starting from rest, and it has to accelerate. So first its velocity is going to be less than the speeding car, and then eventually it's gonna be greater than in order for it to catch up. And when we have a constant acceleration plotted on a position versus time graph that's gonna look quadratic. So the police graphs and start from here slowly accelerate and then you catch up. That's not great. You go and then I catches up with the car right there. And so if we were to extend this, this would look kind of quadratic like that s o. This shows that they're starting at the same point here and there, ending at the same point here. Now to figure out the time it takes for this to happen. Well, we know the velocity of this car, and I'm just gonna convert that two meters per second. So we have one kilometer is 1000 meters and one hour is 3600 seconds. This is gonna be, ah, 36.11 meters per second. So now we have everything in the right units and we know that Delta X for this car, the speeding car is gonna be its velocity. Times time. We're going on a constant velocity and the displacement we already are given in the problem is 750 meters. The velocity, we also know is 36.11 meters per second. Have that multiplied by T. So t is going to be 7 50 over 36.11 which is 20.77 seconds. About the next part. We want to know the acceleration of the police car. And now that we know how long it took, well, it's going to take the same time for the police car. So we know T is 20.77 We know our initial velocity zero because we're starting at rest and we're looking for the acceleration. And we also know our displacements is the same as well, 750 meters and the obvious Kinnah Matic equation we want to use here. This is all four of these variables. So tell to exit the initial times time plus 1/2 a T squared Delta X still 7 50 b zero is zero Uh, what we know the time, but it's just gonna go to zero. Then we add in 1/2 a times 20.77 squared. And if you do little arithmetic, you get a is 3.477 ish, which is around a 3.5 meters per second squared. And finally, we want to know the police cars velocity once it catches up to this car and the velocity of the police car here gonna be greater than the velocity of the speeding car does that has to in order to catch up so we can plug back into this. Actually, we don't use this question because we want to find, uh, final velocity. So we don't know that now we know our acceleration 3.477 ish each per second squared. And so now we want it. We want to find velocity final and the equation that uses that it's gonna be the final squared equals B initial squared plus to a Delta X. And again we know initial velocity zero. That's gonna make this pretty easy. Two times 3.477 ish and 7 50 And from this, we get our final velocity rounded to 70 72.2 meters per second. And that's the velocity when the police car catches up to the speeding car

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