Answer
$\theta = 75.1^{\circ}$
Work Step by Step
We can find the linear speed at $t = 15.0~s$:
$v = (0.500~m/s^2)(15.0~s)$
$v = 7.50~m/s$
We can find the radial acceleration:
$a_r = \frac{v^2}{r}$
$a_r = \frac{(7.50~m/s)^2}{30.0~m}$
$a_r = 1.875~m/s^2$
We can find the angle $\theta$ that the net acceleration vector makes with the car's velocity vector:
$tan~\theta = \frac{a_r}{a_t}$
$\theta = tan^{-1}~(\frac{a_r}{a_t})$
$\theta = tan^{-1}~(\frac{1.875~m/s^2}{0.500~m/s^2})$
$\theta = 75.1^{\circ}$
