Your question is too short, and BTW only applies to free particles. I don't know what you are starting with. Or even whether you mean experimental verification. As Dextrecioby points out, if you already know expressions for p and E, the result is trivial. Another possible starting point is that you know only
[tex]E^2=p^2c^2-m_0^2c^4[/tex]
and Hamilton's equations. Then use H=E(p,x), and
[tex]v=dx/dt=\partial H/\partial p[/tex]
or the vector generalization.
If a particle is in a potential field [itex]\Phi(x,y,z)[/itex], then
[tex](E-\Phi)^2=p^2c^2-m_0^2c^4[/tex]
so v/c is not equal to pc/E. One can define "kinetic energy" as [itex]K=E-\Phi(x,y,z)[/itex], then v/c=pc/K. That's just one example of non-free.
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