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# A mass on a spring vibrates horizontally on a smooth level surface (see the figure). Its equation of motion is $x(t) = 8 \sin t,$ where $t$ is in seconds and $x$ in centimeters.(a) Find the velocity and acceleration at time $t.$(b) Find the position, velocity, and acceleration of the mass at time $t = 2\pi/3.$ In what direction is it moving at that time?

## a) $V=8 \cos (t), A_{t}=-8 \sin (t)$b) position: $$4\sqrt{3} \mathrm{cm} \quad \text{moving to the left}$$ velocity: $$-4 \mathrm{cm} / \mathrm{sec}$$acceleration: $$-4 \sqrt{3} \mathrm{cm} / \mathrm{sec}^{2}$$

Derivatives

Differentiation

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##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

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### Video Transcript

he has clear. So when you right here. So we have. Except T is equal to eight sign. So this is given. Next We're gonna find the derivative. So we get eight co sign her derivative is equal to the velocity And to find acceleration, we just have to derive it one more time. We get negative eight sign. It's a party for part B. We have the velocity function. This is equal to V. This is equal. Today we have except t, which is the position function. So worried you were gonna plug in to pie over three pretty for all three equations when we get four square root of three, which is around six point 93 centimeters. Next we're gonna plug it into our velocity function. So we get negative four centimeters per second and finally our acceleration function which give this negative six points 93 centimeters per second square. Since the velocities negative, it's going to the left

#### Topics

Derivatives

Differentiation

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

Lectures

Join Bootcamp