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.

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

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

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!!!!!

Well, it seems like water is really under pressure here... or maybe it's just feeling a little stressed out!

To find the acceleration of the water, we need to differentiate the velocity function with respect to time, using the chain rule.

Let's start by differentiating v = 1240p^(1/2) with respect to p. This will give us dv/dp.

dv/dp = (1/2) * 1240 * p^(-1/2)

Now, we can multiply dv/dp by dp/dt (the rate of change of pressure with respect to time) to find the acceleration, dv/dt.

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

Given that dp/dt = 0.406 psi/sec and p = 33.0 psi, we can substitute these values into the equation.

dv/dt = (1/2) * 1240 * (33.0)^(-1/2) * 0.406

Now, let me do some quick calculations here...

*Calculating...*

And the result is approximately 43.8 ft/sec^2!

Looks like the water is really accelerating, just like a classic clown car taking off at a circus! I hope this helps clarify things for you.

To find the acceleration of the water, we need to differentiate the velocity function with respect to time.

Given: v = 1240p^(1/2)

Differentiating both sides of the equation with respect to time (t), we get:

dv/dt = (d/dt)(1240p^(1/2))

To differentiate p^(1/2) with respect to t, we need to use the chain rule.

Let u = p^(1/2)

Now, differentiate u with respect to t:

du/dt = (d/dt)(p^(1/2))

Using the chain rule, we can write:

du/dt = (d/dp)(p^(1/2)) * (dp/dt)

Differentiating p^(1/2) with respect to p:

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

Plugging this back into our original equation for dv/dt, we have:

dv/dt = (d/dt)(1240p^(1/2))
= (d/dt)(1240u)
= 1240 * du/dt

Substituting the expression for du/dt:

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

Given: dp/dt = 0.406 psi/sec
And when p = 33.0 psi, we need to find dv/dt.

Substituting the values into the formula:

dv/dt = 1240 * (1/2)(33.0)^(-1/2) * 0.406

Simplifying the expression:

dv/dt = 1240 * (1/2) * (1/√33) * 0.406
= 1240 * 0.5 * (1/√33) * 0.406
= 43.8 ft/sec^2

Therefore, the acceleration of the water when the pressure is 33.0 psi is 43.8 ft/sec^2.

To determine the acceleration of the water, we need to find the derivative of the velocity function with respect to time (dv/dt).

Given the velocity function:
v = 1240p^(1/2)

We need to find dv/dt when p = 33.0 psi.

First, we can find the derivative of v with respect to p using the power rule:
dv/dp = (1/2) * 1240 * (p^(-1/2))

Next, we can use the chain rule to find dv/dt:
dv/dt = dv/dp * dp/dt

Given that dp/dt = 0.406 psi/sec, we substitute these values into the equation:
dv/dt = (1/2) * 1240 * (33.0^(-1/2)) * 0.406

Now, we can calculate the value of dv/dt:
dv/dt ≈ (1/2) * 1240 * (0.192) * 0.406 ≈ 50.2656 ft/sec^2

The value of dv/dt is approximately 50.2656 ft/sec^2, which is close to, but not exactly 43.8 ft/sec^2. Please double-check your equations and values to ensure the correct answer.