The thrust (T$T$) of an engine is T = Q(V.out-V.in)$T = Q(V_{out}-V_{in})$ where Q$Q$ is the mass flow through the engine, V.out$V_{out}$ is the exit velocity of the exhaust gas and V.in$V_{in}$ is the speed at which the air flow enters the engine inlet.
The mass flow Q$Q$ is a product of volume of the air sucked by the engine and air density.
When engine is producing considerable thrust, the exhaust speed (V.out$V_{out}$) is the speed of sound. So for the sake of simplicity, we can consider that more or less a constant value. When aircraft is stationary, like at the beginning of the takeoff roll, V.in$V_{in}$ is zero. As the aircraft begins to accelerate, V.in$V_{in}$ increases.
Air is compressible gas, but at low speeds compressibility is negligible. As the forward speed increases, the Q$Q$ is constant for a given engine RPM, the V.out$V_{out}$ is constant and V.in$V_{in}$ increases, the net result is that engine begins to lose thrust.
But as the speed increases past about Mach 0.2, which equals approximately 130kts at sea level at standard day conditions, the air being packed (rammed) in the engine inlet begins to compress due to increasing pressure. That means that the density of air in the inlet increases. As density increases, and if we assume constant RPM (and thus volume) the mass of the air flowing through the engine increases.
At some point the increased mass flow Q$Q$ has compensated the loss of velocity change in the engine and thrust has reached its stationary engine value. This is the ram recovery point.
