Answer
$v (t)=-8 \sin 4t \ u_r+4 \cos 4t \ u_{\theta}$
and $a (t)=-40 \cos 4t u_r-32 \sin 4t u_{\theta}$
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 $\theta^{..}=0$
$\implies v (t)=-8 \sin 4t \ u_r+4 \cos 4t \times \ u_{\theta}$
so, $a (t)=(\dfrac{d^2r}{dt^2}-r\theta^{.2})u_r+(r\theta^{..} +2r^{.}\theta^{.})u_{\theta} =-40 \cos 4t \times u_r-32 \sin 4t u_{\theta}$
