What force accelerates a liquid moving in a narrowing pipe? Say there is a liquid which behaves like an incompressible fluid and is flowing steadily through a pipe which is moving from a cross section of area $A_1$ to the cross section of area $A_2$, where $A_2$ is less than $A_1$. As per the continuity equation, $v_2>v_1$ and so the liquid seems to be accelerating. What force is causing this acceleration?
 A: There is more mass per area behind than ahead of the constriction so since ppressure is force divided by area there develops a pressure difference
A: The simple answer is pressure.
As you state, from the continuity equation you can see that the velocity $v_2>v_1$. The next step is a momentum balance (like any balance in fluid dynamics: $\frac{d}{dt}=in-out+production$). The momentum flowing into the system is smaller than the momentum flowing out of the system ($\rho A_1 v_1^2 < \rho A_2 v_2^2 = \rho A_1 v_1^2 \frac{A_1}{A_2}$, $\frac{A_1}{A_2}>1$).
The actual force is not the pressure itself, but the pressure difference, or, actually, the difference in force, because on the left the force is $p_1 A_1$ and on the right $p_2 A_2$.
From another conceptual point. Suppose you have a garden hose. Than you have a fixed pressure drop. Now, if you squeeze it, you create a contraction. Part of the pressure drop is now needed to accelerate the fluid at the exit. The overall flowrate decreases, because you also need pressure to overcome frictional forces.
A: 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)
$$\frac{1}{2}\rho v_1^2 + p_1 = \frac{1}{2}\rho v_2^2 + p_2$$
