= Solution
Assume an air-filled or evacuated lossless <electromagnetic waveguide>, and a pulse bandwidth narrow enough to use its carrier's local <group velocity>. The longitudinal <dispersion relation> is
$$
\omega^2=c^2k_z^2+\omega_c^2.
$$
It gives
$$
v_{\mathrm p}=\frac\omega{k_z}=\frac c{\sqrt{1-(f_c/f)^2}},
\qquad
v_{\mathrm g}=\frac{d\omega}{dk_z}=c\sqrt{1-(f_c/f)^2}.
$$
Here $f_c/f=20/40=1/2$, so
$$
\boxed{v_{\mathrm p}=\frac{2c}{\sqrt3}=3.46\times10^8\,\mathrm{m\,s^{-1}},\qquad
v_{\mathrm g}=\frac{\sqrt3c}{2}=2.60\times10^8\,\mathrm{m\,s^{-1}}.}
$$
The <phase velocity> exceeds $c$, but the pulse envelope travels at the subluminal <group velocity>; a moving constant-phase surface does not by itself transport information. Their product is $v_{\mathrm p}v_{\mathrm g}=c^2$. A broad pulse can distort because this <waveguide> is dispersive.
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