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Problem 65

Rectilinear Motion In Exercises $65-68$ , consider a particle moving along the $x$ -axis, where $x(t)$ is the position of the particle at time $t, x^{\prime}(t)$ is its velocity, and $x^{\prime \prime}(t)$ is its acceleration.

$x(t)=t^{3}-6 t^{2}+9 t-2, \quad 0 \leq t \leq 5$

(a) Find the velocity and acceleration of the particle.

(b) Find the open $t$ -intervals on which the particle is moving to the right.

(c) Find the velocity of the particle when the acceleration is $0 .$

Answer

please see explanation

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## Discussion

## Video Transcript

hearing question quickly. Father functions E c T equals pick. You go minus 60 squared off 19 lies too. So a is acceleration, which is a double prime and sees the velocity when it's hungry. She was zero. So I know I have to do my double prime. The person concerned with crying t it's been equal three t squared minus 12 t and then the double crime 60 minus 12 which can also be written as six. Factor at six. Get t minus two part eh, R B You have to grab your graphing calculator. So if you draft on your grasp calculator in the White Bulls and you put in the three X square Yeah, um, yeah, dizzy crime green here, my ex plus nine in your graphing calculator, you see, like that and the values are only interested in from want 0 to 5. So the open interval that we're looking for, where it positive or it's moving toe right, which would be positive from here. You hear our graph is negative. So on that interval, it is not moving to the right. So it's all moving from the right from some grow one union from three In the end, that would be. And then our c um, the velocity when intelligence. So Z double premises acceleration. So when acceleration is zero would be a t equals two so that I'm gonna put equals two into my Z prime function, huh? And I get Meg with three.

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