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The motion of a particle is defined by the relation $x=6 t^{4}-2 t^{3}-$ $12 t^{2}+3 t+3,$ where $x$ and $t$ are expressed in meters and seconds respectively. Determine the time, the position, and the velocity when $a=0$.

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$t=0.667 \mathrm{s}$$x_{\frac{2}{3}}=0.259 \mathrm{m}$$v_{\frac{2}{3}}=-8.56 \mathrm{m} / \mathrm{s}$

Physics 103

Chapter 11

Kinematics of Particles

Kinetic Theory Of Gases

Simon Fraser University

University of Sheffield

McMaster University

Lectures

02:28

The motion of a particle i…

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The acceleration of a part…

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seen this problem were given a you relation that defines the emotion of a particle. And we want to determine the time, the position and the velocity when the acceleration is zero. So we're giving that X is equal to 60 to the fourth minus two t cubed minus 12 t squared plus 32 plus three mhm. So the first thing that we want to do is take the derivative with respect. Um, dx DT so we can find the lawsuit. So during that would give us mhm and then to find the acceleration we take the derivative of that should give us Yeah. So when a is equal to zero Yeah. So I need your t squared minus 12. T 24 is equal to zero, so and we can go ahead and factor this so it's easier to solve. So, b 12 times, sixties squared minus T minus two is equal to zero. Then we can, uh, see that there's two solutions using the quadratic formula. That would be t is two thirds No, and to use one half of a second. But we're going to go ahead and reject that one and go with this solution So at this time, we're just going to plug that in to our first equation. That's what we started with. So plugging this for T until equation one consulting give us a exposition of 0.259 leaders. Then we're also going to plug this into our equation for velocity. We'll call it equation, too. Okay, which will give us a velocity at that time negative 8.56 m per second. And all that negative sign tells us that they are going in the opposite direction as we've defined our access.

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