Answer
$a = 0.30~m/s^2~~$ and $~~\theta = 270^{\circ}$
Work Step by Step
It is given in the question that $r = 3.00~m$
The center of the circular path is at the position $(0, 3.00~m)$
We can find the angle the particle has moved through at $t = 10.0~s$:
$\phi = (\frac{10.0~s}{20.0~s})(360^{\circ}) = 180^{\circ}$
Then the particle is at the position $(0, 6.00~m)$
We can find the particle's speed:
$v = \frac{2\pi~r}{T}$
$v = \frac{(2\pi)~(3.00~m)}{20.0~s}$
$v = 0.942~m/s$
We can find the magnitude of the acceleration:
$a = \frac{v^2}{r}$
$a = \frac{(0.942~m/s)^2}{3.00~m}$
$a = 0.30~m/s^2$
Note that the acceleration vector is directed toward the center of the circular path, so at $t = 10.0~s$, the acceleration vector is directed in the negative direction of the y axis. The angle relative to the positive direction of the x axis is $\theta = 270^{\circ}$
$a = 0.30~m/s^2~~$ and $~~\theta = 270^{\circ}$
