Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises - Page 238: 90

Answer

$$s\left( t \right) = 2{t^2} - 3t + 2$$

Work Step by Step

$$\eqalign{ & a\left( t \right) = 4;{\text{ }}v\left( 0 \right) = - 3;{\text{ }}s\left( 0 \right) = 2 \cr & v\left( t \right) = \int {a\left( t \right)dt} \cr & v\left( t \right) = \int {4dt} \cr & v\left( t \right) = 4t + C \cr & {\text{Use the initial condition }}v\left( 0 \right) = - 3 \cr & - 3 = 4\left( 0 \right) + C \cr & C = - 3 \cr & {\text{Thus,}} \cr & v\left( t \right) = 4t - 3 \cr & s\left( t \right) = \int {v\left( t \right)} dt \cr & s\left( t \right) = \int {\left( {4t - 3} \right)} dt \cr & s\left( t \right) = 2{t^2} - 3t + C \cr & {\text{Use the initial condition }}s\left( 0 \right) = 2 \cr & 2 = 2{\left( 0 \right)^2} - 3\left( 0 \right) + C \cr & C = 2 \cr & {\text{Therefore,}} \cr & s\left( t \right) = 2{t^2} - 3t + 2 \cr} $$
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