\[t\]
by \[x=\sqrt{t^2+1}, \; y= \sqrt{2t}\]
then the distance of the particle from the origin is \[r= \sqrt{x^2+y^2}= \sqrt{t^2+1+2t}=t+1\]
.The velocity is
\[\begin{pmatrix} \ dx/dt\\ dy/dt\end{pmatrix} = \begin{pmatrix} \frac{t}{\sqrt{t^2+1}} \\ \frac{1}{\sqrt{2t}} \end{pmatrix}\]
and the acceleration is \[\begin{pmatrix} \ d^2x/dt^2 \\ d^2y/dt^2\end{pmatrix} = \begin{pmatrix} \frac{}{(\sqrt{t^2+1})^3} \\ - \frac{1}{2 \sqrt{2t}} \end{pmatrix}\]
.The speed is
\[\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} = \sqrt{\frac{2t^3+t^2+1}{2t(t^2+1)}}\]
and the magnitude of the acceleration is \[\sqrt{(\frac{d^2x}{dt^2})^2+(\frac{d^2y}{dt^2})^2} = \sqrt{\frac{8t+(t^2+1)^3}{8t(t^2+1)^3}}\]
.