Answer
$v=3a \sin \theta \ u_r+3a (1-\cos \theta) \ u_{\theta}$
and $a=9a (2\cos \theta -1)u_r+18a \sin \theta u_{\theta}$
Work Step by Step
Write the expressions for the velocity and acceleration in terms of $u_r$ and $u_{\theta}$ . $v=\dfrac{dr}{dt}u_r+r \dfrac{d\theta}{dt}u_{\theta}$ and $a=(\dfrac{d^2r}{dt^2}-r\theta^{.2})u_r+(r\theta^{..} +2r^{.}\theta^{.})u_{\theta}$
We will apply the chain rule.
$v=\dfrac{d(a(1-\cos \theta))}{dt}u_r+3ru_{\theta} \\=\dfrac{d(a(1-\cos \theta))}{dt} \dfrac{d\theta}{dt} u_r+3ru_{\theta} \\=a \sin \theta \times (3 u_r)+3ru_{\theta} \\=3a \sin \theta \times u_r+(3r) \times u_{\theta}$
Set $1-\cos \theta =r$
Thus, $v=3a \sin \theta \ u_r+3a (1-\cos \theta) \ u_{\theta}$
Next, $a=(\dfrac{d^2r}{dt^2}-r\theta^{.2})u_r+(r\theta^{..} +2r^{.}\theta^{.})u_{\theta} \\=9a (2\cos \theta -1)u_r+18a \sin \theta u_{\theta}$
