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(II) An inattentive driver is traveling 18.0 $\mathrm{m} / \mathrm{s}$ when henotices a red light ahead. His car is capable of deceleratingat a rate of 3.65 $\mathrm{m} / \mathrm{s}^{2}$ . If it takes him 0.200 s to get thebrakes on and he is 20.0 $\mathrm{m}$ from the intersection when hesees the light, will he be able to stop in time?

he will not be able to stop in time.

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

Rutgers, The State University of New Jersey

Hope College

University of Sheffield

Lectures

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(II) A driver is traveling…

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we have an inattentive driver who is initially going at a speed of 18 meters per second. And we know the car is capable of decelerating once breaking. And I'm going to call this negative because deceleration is slowing down 3.65 meters per second squared. And now we have to split the problem into two parts because, uh, it takes the driver a second to react. Well, it actually takes him 20.2 seconds. So the displacement for this first part is pretty simple. It's just the velocity times, however long it takes because the velocity is constant and this is gonna be 18 times 0.2. Miss comes out to be 3.6 meters. So that's the first displacement. And at one more piece of information we know is that the final velocity zero right, because the car comes to a stop. Hopefully And so now we can use B squared this equation and now let me set be equal to zero and we'll solve for Delta X here. So bring this to the other side. Now we know that Delta X is negative. Zero squared over to a Now you can pluck in this is gonna be. The zero is 80 meters per second and a is negative 3.65 So, Thea, the negatives cancel out nicely and we get 44 lips, 44.4 meters. And so now the total displacement was gonna add on when I got here 3.6 meters. And this is for th meters, which is much greater than the 20 meters that we had stop. So no, you're not stopping time.

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