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: 20


$ d=-5 \sin (\dfrac{ 4\pi}{5}) t $

Work Step by Step

Amplitude, $ A=-5$ (negative because the object is moving down initially.) Given: $ Period =2.5$ Now, period, $ P=\dfrac{2 \pi}{\omega} \implies \omega =\dfrac{2 \pi}{2.5}=\dfrac{ 4\pi}{5}$ So, $ d=A \sin \omega t=-5 \sin \dfrac{ 4\pi}{5} t $
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.