Answer
The graph is shown below:
Work Step by Step
İf the angular acceleration is constant then the angle time function can be expressed as:
$\theta \left( t\right) =\theta _{0}+w_{0}t+\dfrac {\alpha t^{2}}{2} .............(1)$
Using equation (1), we get:
$\dfrac {\alpha t^{2}}{2}+w_{0}t+\left( \theta _{0}-\theta \left( t\right) \right) =0\left( 2\right) $
If we solve equation (2), we get
$t_{1,2}=\dfrac {-w\pm \sqrt {\left( w^{2}-4\times \dfrac {a}{2}\times \left( \theta _{0}-\theta \left( t\right) \right) \right) }}{a}..........\left( 3\right) $
So the graph will be a parabola. Lets mark two $t$ values for $\theta (t) =0$.
Given the numbers $w_{0}=4.7\dfrac {rad}{s};\alpha =-0.25\dfrac {rad}{s^{2}}; \theta_{0}=0$
Using (3), we get $t_{1,2}=\dfrac {-w_{0}\pm \sqrt {\left( w^{2}_{0}-4\times \dfrac {a}{2}\times \left( \theta _{0}-\theta \left( t\right) \right) \right) }}{a}=\dfrac {-4.7\pm \sqrt {\left( 4.7^{2}-4\times \left( \dfrac {-0.25}{2}\right) x\left( 0-0\right) \right) }}{-0.25}\approx 0s, 37.65s $
Now lets find the maximum value of $\theta$ using (3). Since the inside square root cant be negative and minimum can be zero, we get:
$w^{2}_{0}-4\times \dfrac {\alpha }{2}\left( \theta _{0}-\theta \right) \geq 0\Rightarrow \dfrac {\omega ^{2}_{0}}{2\alpha }\geq \theta _{0}-\theta \Rightarrow \theta \leq \theta _{0}-\dfrac {\omega ^{2}_{0}}{2\alpha }\Rightarrow \theta \leq 0-\dfrac {4.7^{2}}{2\times -\left( 0.25\right) }=44.18rad\Rightarrow \theta _{\max }=44.18rad$
İf we look at $α$, we see it is negative so the parabola will be concave up.
