Question

Water is discharged from a pipeline at a velocity v (in ft/sec) given by v = 1306p(1/2), where p is the pressure (in psi). If the water pressure is changing at a rate of 0.354 psi/sec, find the acceleration (dv/dt) of the water when p = 36.0 psi

Answer 1

Answer:

$a=38.5 ft/sec^{2}$

Explanation:

Note that acceleration is the rate change of velocity i.e

$acceleration=\frac{change in velocity}{change in time}\\a=\frac{dv}{dt}$.

Since the velocity is giving as a variable dependent on the pressure, we have to differentiate implicitly both side with respect to time,i.e

$\frac{dv}{dt}=1306*(1/2)p^{-1/2}\frac{dp}{dt}$

if we substitute value for the pressure as giving in the question and also since the rate change of pressure is 0.354psi/sec, we have

$a=653*0.1667*0.354$

$a=\frac{dv}{dt}=653(36)^{-1/2}*0.354\\ a=38.5 ft/sec^{2}$