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The position of a particle as a function of time is given by$x=(-5 \mathrm{m} / \mathrm{s}) t+\left(3 \mathrm{m} / \mathrm{s}^{2}\right) t^{2}$ (a) Plot $x$ versus $t$ for $t=0$ to $t=2 \mathrm{s}$ .(b) Find the average velocity of the particle from $t=0$ to $t=1 \mathrm{s}$ .(c) Find the average speed from $t=0$ to $t=1 \mathrm{s}$ .

(a) GRAPH(b) $-2 \mathrm{m} / \mathrm{s}$(c) $6 \mathrm{~m} / \mathrm{s}^{2}$

Physics 101 Mechanics

Chapter 2

One-Dimensional Kinematics

Motion Along a Straight Line

Cornell University

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Hope College

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So in problem 22 we have to We have a function to describe the movement of a particle which is Let's see, X X equals two minus five, two less tree T square. And the first item we have to go. We have to plot the graph or off the movement of this particle. When she goes from T equals zero, open to T equals two seconds. Okay, this is a lot off Ex vs see? Let's see. So first we need to calculate what is the position with a particle 20 0 And when t goes to until you go zero particle is in the orange off the movement. So when t equals two, the position is going to be X equals minus five times two plus three times to a square square. This is it close to minus 10 plus 12. So the final position of the particle, it's just through meters. Okay, so two meters is around here and the lot off the entire movement from T cause zero up into teak was two seconds. It's going to be what we know. Dysfunction is the second degree function because off the G square in here and We know that in all second degree functions, the constant follows. The T square gives us the comparative depart Herbal. Since the constant, it's positive plus tree. We have honorable christened. And this is the graph for the first item. The movement of the particle. Okay, the second item. We have to calculate the average velocity off the particle average velocity when he she goes from times Deco zero upon two times t equals one second. We know that the average velocity is defined as the difference in the position. The rider by the difference in time, the different same time is just one second the difference in position. We have to find out. Where is the final position of the particle? The same as the first? I think so. The position of the particle is going to be minus five times one plus three times one. This is a close to minus two meters. Okay, so difference in position. Find her position. Miners Danish. Your position. It is going to be minus two meters per second. This is the average velocity of the particle. And if final biting, we have to calculate the average speed of the particle. Bro this doesn't even to calculate. We just I need to know that by definition, a function that describes the movement of a particle accelerated general function is going to be ex off t. It goes X zero. That is the initial position plus the initial velocity more to play by the time plus the acceleration divided by two multiplied by the Times Square. This is the if you go. If we compare function that we have, that is six. Sorry, It's not a function that is minus five T plus three D square. We can see that Just comparing this through. In here we have the acceleration divided by two equals tree. So the average speed is going to be six meters per second squared. This is the final answer for this question. Thank you for watching. No. What does she want? Me. Me, Me. Me, sir. Seem correct.

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