Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.4 Derivatives of Trigonometric Functions - 2.4 Exercises - Page 151: 35

Answer

(a) $x(t)$ = $8sint$ $v(t)$ = $x'(t)$ = $8cost$ $a(t)$ = $v'(t)$ = $-8sint$ (b) the mass at time $t$ = $\frac{2\pi}{3}$ position is $x(\frac{2\pi}{3})$ = $8sint$ = $8sin(\frac{2\pi}{3})$ = $4\sqrt {3}$ velocity is $v(\frac{2\pi}{3})$ = $8cos(\frac{2\pi}{3})$ = $-4$ acceleration $a(\frac{2\pi}{3})$ = $v'(t)$ = $-8sin(\frac{2\pi}{3})$ = $-4\sqrt {3}$ since $v$ $\lt$ $0$ so the particle is moving to the left

Work Step by Step

(a) $x(t)$ = $8sint$ $v(t)$ = $x'(t)$ = $8cost$ $a(t)$ = $v'(t)$ = $-8sint$ (b) the mass at time $t$ = $\frac{2\pi}{3}$ position is $x(\frac{2\pi}{3})$ = $8sint$ = $8sin(\frac{2\pi}{3})$ = $4\sqrt {3}$ velocity is $v(\frac{2\pi}{3})$ = $8cos(\frac{2\pi}{3})$ = $-4$ acceleration $a(\frac{2\pi}{3})$ = $v'(t)$ = $-8sin(\frac{2\pi}{3})$ = $-4\sqrt {3}$ since $v$ $\lt$ $0$ so the particle is moving to the left
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