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