Answer
$6 \sin \pi t $
Work Step by Step
Amplitude, $ A=|6|=6$
Now, $ f=\dfrac{\omega}{2 \pi} \implies \omega =2 \pi f $
So, $\omega =2 \pi (1/2)=\pi $
period, $ P=\dfrac{2 \pi}{\omega} $
So, $ d=A \sin \omega t=6 \sin \pi t $
Precalculus (6th Edition) Blitzer
by Blitzer, Robert F.
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 639: 59
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