Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises: 50

Answer

(a) $v(t)=4t^{3}-6t^{2}+2t-1$ $a(t)=12t^{2}-12t+2$ (b) a(1) = 2 s (c) See graph.
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Work Step by Step

$s(t)=t^{4}-2t^{3}+t^{2}-t$ (a) Velocity is the derivative of position, so to get the velocity function in terms of t, we just need to take the derivative of s(t): $s'(t) = v(t)=(4)t^{3}-2(3)t^{2}+2t-1$ $v(t)=4t^{3}-6t^{2}+2t-1.$ Acceleration is the derivative of velocity: $v'(t) =a(t)=4(3)t^{2}-6(2)t+2-0$ $a(t)=12t^{2}-12t+2.$ (b) To find the acceleration at t = 1, just plug it into a(t): $a(1)=12(1)^{2}-12(1)+2 = 2 s$
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