Answer
$\theta=35\ rad$
Work Step by Step
We know that
$\omega_{\circ}=\frac{v_{\circ}}{t}$
We plug in the known values to obtain:
$\omega_{\circ}=\frac{17}{0.33}=52\frac{rad}{s}$
Similarly, angular acceleration is given as
$\alpha=\frac{a}{r}$
$\implies \alpha=\frac{1.12}{0.33}=3.4\frac{rad}{s^2}$
Now we can find the angular displacement as
$\theta=\omega_{\circ}t+\frac{1}{2}\alpha t^2$
We plug in the known values to obtain:
$\theta=(52)(0.65)+\frac{1}{2}(0.34)(0.65)^2$
$\theta=35\ rad$
