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A robot used in a pharmacy picks up a medicine bottle at$t=0 .$ It accelerates at 0.20 $\mathrm{m} / \mathrm{s}^{2}$ for 5.0 $\mathrm{s}$ , then travelswithout acceleration for 68 $\mathrm{s}$ and finally decelerates at$-0.40 \mathrm{m} / \mathrm{s}^{2}$ for 2.5 $\mathrm{s}$ to reach the counter where the pharmacist will take the medicine from the robot. From how faraway did the robot fetch the medicine?

$71.75 \mathrm{m} \approx 72 \mathrm{m}$

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

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 you want to find the total displacement of the robot. And to do that, I'm gonna break this problem into three parts since we have three time intervals and the 1st 1 is five seconds. And we know for this first interval that our acceleration is 0.20 meters per second squared. And we can also figure out that our initial velocity zero So our displacement for this interval it's just the zero t close 1/2 a T squared and playing in our numbers. We know this term just goes to zero. Acceleration is 00.2 and five seconds. This comes out 2.5 meters. Now, for the next interval, we know we're going on a constant velocity. Um, 68 seconds A equals zero. We don't know what this philosophy is, so I'm just gonna sew for it now. From the first part, we know that our velocity squared is our initial velocity plus two a Delta X and here we have B squared is again zero plus two times 20.2 times my answer from the first part. So 2.5 let me get V squared equals one so V equals one um, now, this V I'm gonna call be zero for the next part equals one meter per second. And so if we're looking for the displacement for this part, same equation except the acceleration term is gonna go to zero now. So this is just one time, 68? Um, that goes to zero. So we get 68 meters now for the third time interval, which lasts two and 1/2 seconds. We know our acceleration is negative now, so we're slowing down and still the same initial velocity from the last part of one meter per second and again, same equation. Now, finally, we have both terms that will contribute to our displacement. So one times 2.5, that's 1/2 times negative point for times 2.5 squared. And this is 1.25 meters. And so our total displacement I'll just say yes, it's gonna be 2.5 plus Ah, 68. So here we got 2.5 68 1.25 and that gives us a total of 71.75 meters

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