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The equation of motion of a particle is $ = t^3 - 3t, $ where is in meters and is in seconds. Find(a) the velocity and acceleration as functions of $ t, $(b) the acceleration after $ 2 s, $ and(c) the acceleration when the velocity is 0.

(a) $$a(t)=v^{\prime}(t)=6 t$$(b) $$a(2)=6(2)=12 \mathrm{m} / \mathrm{s}^{2}$$(c) $$v(t)=3 t^{2}-3=0 \text { when } t^{2}=1, \text { that is, } t=1[t \geq 0] \text { and } a(1)=6 \mathrm{m} / \mathrm{s}^{2}$$

01:07

Frank L.

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 1

Derivatives of Polynomials and Exponential Functions

Derivatives

Differentiation

Missouri State University

University of Nottingham

Boston College

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Hey, it's clear. So in Yuma, read here. So we're going to note that velocity is the derivative of the displacement. He s over DT, which is equal to de over de ti Pretty cute. Minus three T, which is equal to three. T square minus three and we need acceleration. And that's the derivative of velocity. Do you be over DT? That is therefore equal to 60. For part B, we have acceleration, which is equal to 60 to find it. After two seconds, we just have to plug in two for tea that becomes equal to 12 meters per second square. We're seeing we have our velocity equation, which is three t square minus three. We're gonna find out at what time? The velocities zero. So we're gonna plug in zero for B A and Saul for tea. We get T is equal to one. Now we just plug it into acceleration. We get six meters per second square

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