Water is discharged from a pipeline at a velocity v(in ft/sec) given by v=1240p^(1/2), where p is the pressure (in psi). I f the water pressure is changing at a rate of 0.406 psi/sec, find the acceleration(dv/dt) of the water when p=33.0 psi.

Question

I know the answer is suppose to be 43.8ft/sec^2 but i have no clue how this is possible! Help!!!!!!

Answer 1

You know that acceleration is the derivative of velocity.

v = 1240 p^(1/2)
dv/dt = 620 p^(-1/2) dp/dt

= 620(33^(-1/2))(.406
= 620/√33)(.406) = 43.82

Answer 2

Thank you so much Reiny you rock! I was putting that .406 in for V but i did have 620 with the 33.0^-1/2. Thanks you Rock!!!!!