Water is discharged from a pipeline at a velocity v (in ft/sec) given by v = 1064p(1/2), where p is the pressure (in psi). If the water pressure is changing at a rate of 0.266 psi/sec, find the...

Water is discharged from a pipeline at a velocity v (in ft/sec) given by v = 1064p(1/2), where p is the pressure (in psi). If the water pressure is changing at a rate of 0.266 psi/sec, find the acceleration (dv/dt) of the water when p = 32.0 psi.

Asked on by pphok

1 Answer | Add Yours

llltkl's profile pic

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

Given, velocity of water from the pipe, `v = 1064p^(1/2)`

Differentiate both sides with respect to t,

`(dv)/(dt)=1064*1/2*p^(-1/2)*(dp)/(dt)`

`=532* p^(-1/2)*(dp)/(dt)`

Put the given values:

`(dv)/(dt)= 532*32^(-1/2)*0.266`

`=25` ft/sec^2

Therefore, acceleration (dv/dt) of the water when p = 32.0 psi is 25 ft/sec^2.

Sources:

We’ve answered 318,976 questions. We can answer yours, too.

Ask a question