Answer
$\dfrac{\pi}{2}$
Work Step by Step
$v(t)=\dfrac{dr}{dt}=i[e^t(\cos t -\sin t) ]+j [e^t(\sin t +\cos t) ]$
and $a(t)=\dfrac{dv(t)}{dt}=-2 e^t ( \sin t i- \cos t j)$
Let us suppose that $\theta$ be the angle between $r$ and $a(t)$
Thus, $\theta =\cos^{-1} [\dfrac{r \cdot a(t)}{|r| |a(t)|}]$
or, $=arccos [\dfrac{ e^t \cos t \times (-2 e^t \sin t)
+(e^t \sin t) (2e^t \cos t)}{\sqrt {(e^t \cos t )^2+(e^t \cos t )^2 \cdot \sqrt {(-2 e^t \sin t)^2 +( e^t \cos t)^2}}}]$
or, $\theta=\dfrac{\pi}{2}$
