Answer
(a) $a=\pi/4 \space rad$
$\space \space \space \space
\space b=2 \space rad/s$
$\space \space \space \space
\space c=-0.139 \space rad/s^3$
(b) $t=0\space s$
(c) $\theta=19.5 \space rad$
$\space \space \space \space
\space \omega = 9.36 \space rad/s$
Work Step by Step
(a)Derive $\omega$ and $\alpha$ from $\theta$ and plug in the given values:
$\theta(t)=a+bt-ct^3$
$\theta(0)=a+b(0s)-c(0s)^3=\frac{\pi}{4}rad$
$a=\frac{\pi}{4}rad$
$\omega(t)=\frac{d\theta}{dt}$
$\omega(t)=b-3ct^2$
$\omega(0)=b-3c(0s)^2=2 \space rad/s$
$b=2 \space rad/s$
$\alpha(t)=\frac{d\omega}{dt}$
$\alpha(t)=-6ct$
$\alpha(1.5s)=-6c(1.5s)=1.25 \space rad/s^2$
$c=\frac{1.25 \space rad/s^2}{-6(1.5s)}$
$c=-0.139 \space rad/s^3$
(b)$\theta(t)=\frac{\pi}{4}+2t+0.139t^3=\frac{\pi}{4}$
$2t+0.139t^3=0$
Factor t then solve for t:
$t(2+0.139t^2)=0$
$t=0 \space$
or
$t=\sqrt{\frac{-2}{0.139}}=imaginary \space number$
Since t has to be a real number:
$t=0$
(c)$\alpha(t)=-6ct=3.5 \space rad/s^2$
$t=\frac{3.5 \space rad/s^2}{-6(-0.139 \space rad/s^3)}$
$t=4.2 \space s$
$\theta(t)=a+bt-ct^3$
$\theta(4.2s)=\frac{\pi}{4}rad+(2 \space rad/s)(4.2s)-(-0.139 \space rad/s^3)(4.2s)^3$
$\theta(4.2s)=19.5 \space rad$
$\omega(t)=b-3ct^2$
$\omega(4.2s)=2 \space rad/s-3(-0.139 \space rad/s^3)(4.2s)^2$
$\omega(4.2s)=9.36 \space rad/s$
