Answer
$a=0.528m/s^2$
Work Step by Step
We can determine the required acceleration as follows:
$A=\frac{\pi}{4}d^2=\frac{\pi}{4}(0.75)^2=0.4418m^2$
and the flow rate is $Q=v_BA=14\times 0.4418=6.185m^3/s$
The mass flow rate is $m=\rho_a Q=1.22\times 6.185=7.5457Kg/s$
We know that:
$\Sigma F_x=m(v_B-v_A)$
$\implies -F=m(v_B-v_A)$
We plug in the known values to obtain:
$-F=7.5457-(-14-0)=105.64N$
Now we can calculate the acceleration as
$F=ma$
We plug in the known values to obtain:
$105.64=200a$
$\implies a=0.528m/s^2$
