Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.8 - Applications of Trigonometric Functions - Exercise Set - Page 637: 19


$ d=-3 \sin (\dfrac{ 4\pi}{3}) t $

Work Step by Step

Amplitude, $ A=-3$ (negative because the object is moving down initially.) Given: $ Period =1.5$ Now, period, $ P=\dfrac{2 \pi}{\omega} \implies \omega =\dfrac{2 \pi}{1.5}=\dfrac{ 4\pi}{3}$ So, $ d=A \sin \omega t=-3 \sin \dfrac{ 4\pi}{3} t $
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