Solution
= Solution
A particle released at the origin at time $s$ satisfies
$$
\dot x=2t,\qquad \dot y=xt,
\qquad x(s)=y(s)=0.
$$
Integration gives
$$
x(t)=t^2-s^2,
\qquad
y(t)=\frac{(t^2-s^2)^2}{4}.
$$
Its distance from the origin is therefore
$$
\boxed{
\sqrt{x^2+y^2}
=(t^2-s^2)
\sqrt{1+\frac{(t^2-s^2)^2}{16}}}.
$$