Time evolution of a free wave packet
Let Ψ(x,t) = (2π)^-1/2∫g(k) exp(i(kx - ω_kt)) dk
and assume that |g(k)| is centered at k₀ and has width Δk.
Let g(k) = |g(k)|exp(iα(k)).  Then
Ψ(x,t) = (2π)^-1/2∫|g(k)|exp(i(α(k) - ω_kt)) exp(ikx) dk = (2π)^-1/2∫|g(k)|exp(iβ(k)) exp(ikx) dk.
Here β(k) = α(k) - ω_kt.
If we again use the use a Taylor series expansion,
β(k) = β(k₀) + dβ(k)/dk|_k0 (k - k₀) + ... .
then we already know that Ψ(x,t) peaks at 
-dβ(k)/dk|_k0 = -dα(k)/dk|_k0 + dω(k)/dk|_k0t = x₀ + dω(k)/dk|_k0t = x₀ + v_gt.
The peak of the wave packet moves with the group velocity, v_g = dω/dk|_k=ko.
The velocity of a plane wave exp(i(kx - ωt) is v_p = ω/k and is called the phase velocity.  We have
ħω = ħ²k²/(2m),   ω/k =  ħk/(2m),   dω/dk =  ħk/m,  v_g =  ħk₀/m = 2v_p(k₀).
A wave for which Δx << x and Δp << p can represent a classical particle moving with velocity v = v_g.