In the form you've stated it ($A_1 v_1 = A_2 v_2$), the continuity equation is only holds for incompressible fluids. So what you've found is that the type of accelerated flow you're describing cannot happen for an incompressible fluid in a pipe of uniform cross-section. The more general form for the continuity equation is based on conservation of mass (i.e., mass per time entering = mass per time exiting), and states
$$
Q_m = \rho_1 A_1 v_1 = \rho_2 A_2 v_2,
$$
where $Q_m$ is the mass flow rate (i.e., mass per time). This means that if the fluid increases in velocity, it must decrease in density.
An analogy here would be cars on a highway. Suppose you have a highway that leads from the center of a small town out into the country. Suppose further that the drivers are all perfectly safe drivers and obey the two-second rule, i.e., the cars pass a given point in the highway at a rate of one car every two seconds. If the speed limit in the town is low, then the cars will be more closely spaced, since two seconds corresponds to less distance. Thus, the density of cars is higher at this point. When the cars get out of town and the speed limit increases, the cars get further apart in distance (since two seconds now corresponds to a longer distance), and so the density decreases.
Bernoulli's equation, meanwhile, doesn't hold so simple a form for compressible fluids. Rather, you have to define a pressure potential $w(P)$ from the equation of state $\rho(P)$:
$$
\frac{1}{2} v^2 + gy + w(P) = \mathrm{const.},
$$
where
$$
w(P) = \int \frac{\mathrm dP}{\rho(P)}.
$$
For the case of an "incompressible" fluid, $\rho(P) = \rho$ is a constant, $w(P) = P/\rho$, and the familiar form of Bernoulli's Law is recovered. But for a compressible fluid, the equation may look quite different.
Given your assumptions, this flow satisfies bernoulli. From the continuity equation, you can see that velocity will vary as the cross sectional area varies. Assuming the vertical distance change is very minor, it can be dropped from the bernoulli equation. From the continuity equation, v2 is less than v1, so that p2 is greater than p1. For a given cross section which is perpendicular to the flow, the magnitude of the velocity will be the same, since there are no friction forces acting on the fluid.
Best Answer
You are right. From continuity of the incompressible fluid you have $$A_1 v_1 = A_2 v_2.$$ So obviously the velocity is changing. Thus the fluid is accelerated, and therefore there must be a force causing this acceleration. In this case the force comes from the pressure difference between the wide and the narrow part of the pipe.
(image from ResearchGate - Diagram of the Bernoulli principle)
This can be described by Bernoulli's equation ($p$ is pressure, $\rho$ is density) $$\frac{1}{2}\rho v_1^2 + p_1 = \frac{1}{2}\rho v_2^2 + p_2$$