#### Answer

$\dfrac{\pi}{2}$

#### Work Step by Step

We know that the velocity and acceleration is defined as: velocity is given as: $v(t)=r'(t)$ and acceleration is given as: $a(t)=v'(t)$
Now,
$v(t)=\lt 2t(t^2+1)^{-1},(t^2+1)^{-1},t(t^2+1)^{-1/2} \gt \\ \implies v(0)=\lt 0,1,0 \gt$
and $a(t)=\lt 2t(t^2+1)^{-1}-4t^2(t^2+1)^{-2},-2t(t^2+1)^{-2},(t^2+1)^{-1/2} -t^2(t^2+1)^{-3/2}\gt \\ \implies a(0)= \lt 2,0,1 \gt$
As we know that $\theta =\cos ^{-1}\dfrac{v(0) \cdot a(0)}{|v(0)||a(0)|}$
$\theta =\cos ^{-1}(\dfrac{0}{(\sqrt{1})(\sqrt{5}}))=\dfrac{\pi}{2}$