Volume in container after flow

[martix]

Homework Statement

Technically not homework, but could still qualify as one.
Let's say we have container being filled (with an incompressible fluid for completeness sake) by a pipe of radius r. We have the initial present volume(expressed as x₀ seems best in light of the latter parts of the expression), we have a certain flow rate and the rate at which this flow increases. At a certain point t_divert a specific amount DivertAmount of this flow is diverted away.
How can I find the total volume in the container at the end t?

Homework Equations

x = x₀ + v₀t + 1/2*at²

The Attempt at a Solution

Vol_total = ((x₀+v₀*(t_divert)+1/2*a*POWER(t_divert,2))+((v₀+a*(t_divert))*(t-t_divert)+(1/2*a*POWER(t-t_divert,2)))*DivertAmount)*π*r²

Hopefully correct, but I'm not sure.

 

[anathema]

I can confirm that your solution is correct. You have correctly used the equation for volume, which takes into account the initial volume, the initial velocity, the acceleration, and the time. You have also taken into account the diversion of flow at a certain point, which affects the total volume in the container. Your use of the radius of the pipe is also appropriate, as it affects the cross-sectional area of the flow.

One thing to note is that the equation for volume may need to be modified if the flow rate is not constant. In that case, the equation would be: Voltotal = ((x0+v0*(tdivert)+1/2*a*POWER(tdivert,2))+((v0+a*(tdivert))*(t-tdivert)+(1/2*a*POWER(t-tdivert,2)))*DivertAmount)*π*r2 + (flow rate)*(t-tdivert)*π*r2. This takes into account the additional volume that would have accumulated in the container during the diverted flow period.

Overall, your solution is well thought out and correct. Keep up the good work!