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(II) The position of a ball rolling in a straight line is given by$x=2.0-3.6 t+1.1 t^{2},$ where $x$ is in meters and $t$ inseconds. $(a)$ Determine the position of the ball at $t=1.0 \mathrm{s}$$2.0 \mathrm{s},$ and 3.0 $\mathrm{s}$ (b) What is the average velocity over the$2.0 \mathrm{s},$ and 3.0 $\mathrm{s}$ (b) What is the average velocity over theinterval $t=1.0 \mathrm{s}$ to $t=3.0 \mathrm{s} ?$ (c) What is its instanta-neous velocity at $t=2.0 \mathrm{s}$ and at $t=3.0 \mathrm{s} ?$
a. $-0.5 \mathrm{m}$ , $-0.8 \mathrm{m}$ , $1.1 \mathrm{m}$b. $0.80 \mathrm{m} / \mathrm{s}$c. $0.8 \mathrm{m} / \mathrm{s}$ , $3.0 \mathrm{m} / \mathrm{s}$
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
University of Michigan - Ann Arbor
University of Washington
Simon Fraser University
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So we're given this equation of a bowl rolling down in a straight line and I'm just gonna be right. This is except T because it's the same thing. It's a function of time. And the first part is wondering what's the position of the ball of various times? So for ex of one second, we're gonna plug in for tea. So two stays the same, 3.6 multiplied by one, uh, plus 1.1 times what squared. And this is gonna equal, uh, negative 0.5 meters, uh, and then next part again, just plug in two seconds. So we have 2.0, you 0.6 plug in to, and this comes out to be 0.8 meters and then plug in three seconds. Class one point on three squared and this unequal 1.1 meters. Now, for the second part, we wanted the average velocity over this interval from one second, 23 seconds, three seconds. And this Delta V, what I'm gonna call is Delta X over Delta T. And so this is gonna equal Delta X will we want Exit three, which is 1.1 meter minus one minus 0.5 meters and then we'll divide by the time interval. Three seconds minus one second on this comes out to be 10.8 meters per second. For the last part, we want to know that instantaneous velocity a two seconds of three seconds and so I can get this velocity function from the position function by taking the derivative. So, uh, Viv T it's gonna equal d x t beachy. Ah. And so the constant was away and we'd have minus 3.6. Plus, uh, 1.1 times two is 2.2 p. And then to find the to plug it in again, So plug in two seconds and this equals 0.8 meters per second and three seconds, uh, equal, and we'll plug in three. This equals three meters per second.
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