Solution

From $F_x=F_0\sin(\omega t)$ and $F_y=F_0\cos(\omega t)$ and the initial conditions $v_x=v_y=0$ and $x=y=0$ at , we have

\begin{eqnarray} a_x&=&\frac{F_0}{m}\sin(\omega t),\ a_y&=&\frac{F_0}{m}\cos(\omega t),\ v_x&=&\frac{F_0}{m\omega}-\frac{F_0}{m\omega}\cos(\omega t)\ v_y&=&\frac{F_0}{m\omega}\sin(\omega t)\ x&=&\frac{F_0}{m\omega}t-\frac{F_0}{m\omega^2}\sin(\omega t)\ y&=&\frac{F_0}{m\omega^2}-\frac{F_0}{m\omega^2}\cos(\omega t) \end{eqnarray}

The last two equations are parametric equations of a cycloid, which is the trajectory of a point on the rim of a disk of radius $R=F_0/(m\omega^2)$ that rolls without slipping on the $x$-axis (shown in blue). Its average velocity is $F_0/(m\omega)$ in the positive $x$ direction. If the initial velocity is $v_0=F_0/(m\omega)$ in the negative $x$ direction, which is perpendicular to the initial force, then the time-dependent force will keep it in a circular orbit (shown in red) as the equation $F_0=mv_0^2/R$ for centripetal force is satisfied.