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(II) An unmarked police car traveling a constant 95 $\mathrm{km} / \mathrm{h}$ ispassed by a speeder traveling 135 $\mathrm{km} / \mathrm{h}$ . Precisely 1.00 $\mathrm{s}$after the speeder passes, the police officer steps on theaccelerator; if the police car's acceleration is $2.00 \mathrm{m} / \mathrm{s}^{2},$ howmuch time passes before the police car overtakes thespeeder (assumed moving at constant speed)?

$13.0 \mathrm{s}$

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

University of Michigan - Ann Arbor

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Lectures

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So we have a police car that is traveling at a velocity of 95 kilometers an hour and we also have a speeding car that's gonna pass the police car going at 135 kilometers per hour. And before I go on, I'm just gonna convert these both two meters per second. So this one, I'll do one kilometer. It's 1000 meters and ah, now with two hours So one hour is 3600 seconds and check the kilometers. Cancel ours. Cancel were left meters per second and this is gonna give us velocity around 26.389 meters per second. This decimal keeps going on. But I'm just gonna round for now and we do the same thing here and you get 37 0.5 meters per second. And we know that initially we have one second that passes before the police car starts accelerating. And we know that the police car can accelerate two meters per second squared. And so first, I want to figure out how far the car gets in front of the police car before he starts accelerating. So I know that are Delta X. How far we travel, I was going to be be zero cheap plus 1/2 a T squared. And for this first period, they're both going at a constant velocity. So we don't need to worry about this term. Okay, for the police car, it's displacement. It's gonna be its velocity 26.389 times one second. And that's just 26.389 meters. And we do the same for the speeding car and we get its displacement. And this time is 37.5 meters per second time's one second. Because those 37.5 meters And now what I'm gonna do is I'm gonna set the police car to a position of zero. So that means the speeding car initial position. It's gonna be 37.5 minus 26.389 which is round 11.1. And this keeps going on because I haven't rounded this number yet. And now we can figure out the final position of the police car in the speeder after the police car catches up and then we'll set them equal because they're the same eso our ex final. I'm gonna write. This not dealt the Xbox final is their ex initial. Plus this velocity term, plus our acceleration term and for the police car, this is gonna be But we're starting at zero. And our initial velocity is this number 26.389 same as always. And our acceleration is two meters per second squared. That's given the problem. And for the speeding car we have, our final position is 11.111 post 37.5 t and then it has no acceleration. So this last time is just zero, and now these two are the same, so we can set them equal and we get 26.389 She plus is just gonna be t squared co vision of one. It's 11.1 plus 37.5 tea, and I'll bring all the terms to one side, and then we can sell the quadratic. So we have cheese squared, uh, minus 11.11 t nice. 11.111 zero and this gives us t value is gonna be 12.4 seconds. And that's how long it takes for the police car to catch up to the speeder after it passes and one seconds passed before the police car starts accelerating

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