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
The velocity and acceleration in terms of $u_r$ and $u_{\theta}$ can be expressed as: $v=\dfrac{dr}{dt}u_r+r \dfrac{d\theta}{dt}u_{\theta}$ and $a=(\dfrac{d^2r}{dt^2}-r\theta^{.2})u_r+(r\theta^{..} +2r^{.}\theta^{.})u_{\theta}$
Now, $v=\dfrac{d(a(1-\cos \theta))}{dt}u_r+3ru_{\theta}$
Now, we will use the chain rule.
$v=\dfrac{d(a(1-\cos \theta))}{dt} \dfrac{d\theta}{dt} u_r+3ru_{\theta}$
or, $v=a \sin \theta (3 u_r)+3ru_{\theta}=3a \sin \theta u_r+3ru_{\theta}$
Substitute $r=1-\cos \theta$
So, $v=3a \sin \theta \ u_r+3a (1-\cos \theta) \ u_{\theta}$
and $a=(\dfrac{d^2r}{dt^2}-r\theta^{.2})u_r+(r\theta^{..} +2r^{.}\theta^{.})u_{\theta}$
So, $a=9a (2\cos \theta -1)u_r+18a \sin \theta u_{\theta}$
