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.
Given, velocity of water from the pipe, `v = 1064p^(1/2)`
Differentiate both sides with respect to t,
Put the given values:
Therefore, acceleration (dv/dt) of the water when p = 32.0 psi is 25 ft/sec^2.