Physics
posted by mymi .
Water at room temperature flows with a constant speed of 8 m/s through a nozzle with a square cross section, as shown in the figure. Water enters the nozzle at point A and exits the nozzle at point B. The lengths of the sides of the square cross section at A and B are 64 cm and 44 cm, respectively.
(a) What is the volume flow rate at the exit?
(b) What is the acceleration at the exit? The length of the nozzle is 2 m.
(c) If the volume flow rate through the nozzle is increased to 7 m3/s, what is the acceleration of the fluid at the exit?
I got the correct answer for a which is 3.27 m^3/s, help me on part b and c please.

(b) and (c) are poorly worded questions. Once the water reaches the exit, acceleration stops. Prior to that, while water is still in the channel,
V*A = b^2*V = constant
= volume flow rate, Q
b is the side length of the square water channel
2 ln b + ln V = 0
(1/V)dV/dt = (1/V)*acceleration
= 2 (db/dt)(1/b)
= 2 (db/dx)*(1/b)*V
acceleration = V^2*(1/b)*db/dx
To get an answer, you need to know the distance over which the side length b decreases from 0.64 to 044 m.