= Solution
Each Fourier mode has zero initial velocity, so
$$
\phi(x,t)=\int_{-\infty}^{\infty}a(k)e^{ikx}\cos(\omega(k)t)\,dk.
$$
Along $x=Vt$ this is
$$
\phi(Vt,t)=\frac12\int_{-\infty}^{\infty}a(k)
\left[e^{it(kV+\omega(k))}+e^{it(kV-\omega(k))}\right]dk.
$$
Put
$$
s=\sqrt{c^2-V^2},
\qquad
k_0=\frac{AV}{s}.
$$
The two phases have stationary points at $-k_0$ and $k_0$, respectively. At those points,
$$
k_0V-\omega(k_0)=-As,
\qquad
\omega''(k_0)=\frac{s^3}{Ac^2}.
$$
Because the initial field is real, $a(-k)=a(k)^*$. Applying the <stationary phase method> to the two conjugate contributions gives the <stationary-phase asymptotic of a Klein-Gordon wave along a subluminal ray>:
$$
\boxed{
\phi(Vt,t)\sim
\left(\frac{2\pi Ac^2}{t s^3}\right)^{1/2}
\operatorname{Re}\left\{
a(k_0)e^{-i(As t+\pi/4)}
\right\}.}
$$
Equivalently, if $a(k_0)=|a(k_0)|e^{i\delta}$,
$$
\phi(Vt,t)\sim
|a(k_0)|\left(\frac{2\pi Ac^2}{t s^3}\right)^{1/2}
\cos\left(As t+\frac\pi4-\delta\right).
$$
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