How does the gas density in the piston change over time?

[stunner5000pt]

Homework Statement

A gas filled pneumatic piston of a strut in a car suspension behaves like a piston apparatus. At one instant the piston is L from the closed end of the cylinder and the gas density is ρ. The piston is moving away from the closed end at v. The gas velocity varies linearly from zero at the closed end to velocity V at the piston. Find the rate of change of gas density at this instant. Also find the average density as a function of time.

2. The attempt at a solution
since there is a linear profile of the velocity, u=kx
u=0,y=0 and
u=V,y=L

we can use continuity equation
\frac{d \rho}{dt}=-\rho\frac{du}{dx}
\frac{d\rho}{dt}=-\rho k

For some reason the LaTex is not working...
dp/dt = -p du/dx
dp/dt = -pk

i used p for rho

Does this work till this step solve the first question?

can we just integrate and use the above conditions to get our density as a function of time?

Am i right? Please help!

Thank you for your input, it is greatly appreciated

 

[stunner5000pt]

any suggestions??

I just need to know if I'm right or wrong that's all...

 

[stunner5000pt]

<br /> \frac{d \rho}{dt}=-\rho\frac{du}{dx}
<br /> \frac{d\rho}{dt}=-\rho k <br />

when i rearrange this equation i get

\int\frac{d\rho}{\rho}=-k\int dt

this presents a problem to integrate because I do not have limits the density or the time

all i am given is velocity, density and the distance of the piston to the cylinder head. The resulting expression from the above is

\rho\left(t\right)=C\exp\left(-kt\right)

Rate of change of hte density is \frac{d\rho}{dt} =-k\rho
using u=kx and u=V when x=L we can solve for k.

For the average density as a function of time...
How to solve for C??

Please help!