= Solution
The <free particle> <dispersion relation> is $\hbar\omega=\hbar^2k^2/(2m)$. Hence
$$
\boxed{v_{\rm ph}=\omega/k=\hbar k/(2m),\qquad v_g=d\omega/dk=\hbar k/m}.
$$
The <phase velocity> follows constant-phase surfaces of a <plane wave>. The <group velocity> follows the envelope of a narrow-band <wave packet> and equals its classical velocity $p/m$. Dispersion can spread the packet; the phase speed is not the particle's transport velocity. In three dimensions $\mathbf v_g=\hbar\mathbf k/m$ and the phase speed quoted is along $\mathbf k$.
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