#### Answer

$v=-8 \sin 4t \ u_r+4 \cos 4t \ u_{\theta}$
$a=-40 \cos 4t u_r-32 \sin 4t u_{\theta}$

#### Work Step by Step

The velocity and acceleration in terms of $u_r$ and $u_{\theta}$ can be computed as:
$v=r_t u_r+r \theta_t u_{\theta}$ and $a=(\dfrac{d^2r}{dt^2}-r\theta^{.2})u_r+(r\theta^{..} +2r^{.}\theta^{.})u_{\theta}$
Now, $\dfrac{d \theta}{dt}=\theta^{.}=2$ and $\theta^{..}=0$
Thus, $v=-8 \sin 4t \ u_r+4 \cos 4t \ u_{\theta}$
Next, $a=(\dfrac{d^2r}{dt^2}-r\theta^{.2})u_r+(r\theta^{..} +2r^{.}\theta^{.})u_{\theta}$
So, $a=-40 \cos 4t u_r-32 \sin 4t u_{\theta}$