Answer
$v=2a e^{a \theta}u_r+2e^{a \theta}u_{\theta}$
and $a=4e^{a \theta}(a^2 -1)u_r+8a e^{a \theta} u_{\theta}$
Work Step by Step
The velocity in terms of $u_r$ and $u_{\theta}$ can be found as:
$v=r_t u_r+r \theta_t u_{\theta}$
The acceleration in terms of $u_r$ and $u_{\theta}$ can be found as: $a=(\dfrac{d^2r}{dt^2}-r\theta^{.2})u_r+(r\theta^{..} +2r^{.}\theta^{.})u_{\theta}$
$\dfrac{d \theta}{dt}=\theta^{.}=2$ and $\theta^{..}=0$
So, $v=2a e^{a \theta}u_r+2e^{a \theta}u_{\theta}$
$a=(\dfrac{d^2r}{dt^2}-r\theta^{.2})u_r+(r\theta^{..} +2r^{.}\theta^{.})u_{\theta}=4e^{a \theta}(a^2 -1)u_r+8a e^{a \theta} u_{\theta}$
