A point moves along the x-axis with its position, x, at time t given by x(t) = t^3 - 3t^2 + 2t -...


A point moves along the {eq}x {/eq}-axis with its position, {eq}x {/eq}, at time {eq}t {/eq} given by {eq}x(t) = t^3 - 3t^2 + 2t - 1 {/eq}. Find the acceleration when {eq}t = 2 {/eq}?

Answer and Explanation:

Suppose that a point moves along the {eq}x {/eq}-axis according to {eq}x(t)=t^3-3t^2+2t-1 {/eq}. Its velocity at time {eq}t {/eq} is then

{eq}v(t)=x'(t)=3t^2-6t+2 {/eq}

and its acceleration is

{eq}\begin{align} a(t)=v'(t)=6t-6 \end{align} {/eq}

Therefore, its acceleration at {eq}t=2 {/eq} is {eq}a(2)=6(2)-6= \mathbf{6} {/eq}

The acceleration of the point when {eq}t=2 {/eq} is {eq}\mathbf{6} {/eq}.

Learn more about this topic:

Speed, Velocity & Acceleration

from TExES Physics/Mathematics 7-12 (243): Practice & Study Guide

Chapter 47 / Lesson 5

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