Answer
$v=a \cos t u_r +a(1+\sin t) e^{-t} u_{\theta}$
$a=-a(\sin t +(1+\sin t) e^{-2t})u_r-ae^{-t} (1+\sin t -2 \cos t) u_{\theta}$
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, $\dfrac{d \theta}{dt}=\theta^{.}=e^{-t}$ and $\theta^{..}=-e^{-t}$
Thus, $v=a \cos t u_r +a(1+\sin t) e^{-t} u_{\theta}$
Next, $a=(\dfrac{d^2r}{dt^2}-r\theta^{.2})u_r+(r\theta^{..} +2r^{.}\theta^{.})u_{\theta}$
So, $a=-a(\sin t +(1+\sin t) e^{-2t})u_r-ae^{-t} (1+\sin t -2 \cos t) u_{\theta}$
