= Solution
The identity
$$
\frac{\ddot a}{a}=H^2+\dot H=H^2(1-\epsilon)
$$
shows that <accelerating expansion> has $H>0$ and $\epsilon<1$, whereas <accelerated contraction> has $H<0$, $\ddot a<0$, and $\epsilon>1$.
For the <monomial inflation potential> $V=\lambda\phi^p$, the two <potential slow-roll parameters> are
$$
\epsilon_V=\frac{p^2M_{\rm Pl}^2}{2\phi^2},
\qquad
\eta_V=\frac{p(p-1)M_{\rm Pl}^2}{\phi^2}.
$$
Thus $\epsilon_V<1$ requires $|\phi|>pM_{\rm Pl}/\sqrt2$, while $|\eta_V|<1$ requires $|\phi|>\sqrt{p|p-1|}\,M_{\rm Pl}$.
At $\phi_*$, the specified velocity gives kinetic energy
$$
\frac12\dot\phi_*^2=2\lambda\phi_*^p=2V(\phi_*).
$$
Hence $\rho_\phi=3V$, $P_\phi=V$, and the <first Hubble slow-roll parameter> is
$$
\boxed{\epsilon=\frac32\left(1+\frac{P_\phi}{\rho_\phi}\right)=2}.
$$
The universe is therefore \b[decelerating], despite the flatness of the potential: the chosen initial kinetic energy violates the <slow-roll approximation>.
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