Solving dv/dt - (k/m)v² = g for air resistance

[$id]

Hi i posted last time about solving an equation regarding air resistance being proportional to the speed

This time i need it to be proportional to the speed squared

I have managed to get an equation but haven't managed to solve it yet

dv/dt -(k/m)v^2 = g


Any ideas on how to solve this. I have tried all the methods i know, I think i need to substitute in. I will need more than just a hint here guys

Thanks a lot

sid

 

[Crosson]

The equation is seperable (you can separate v and t). Any first order seperable equation can be reduced to a problem of indefinite integration:

$$\int \frac{dv}{g - \frac{k}{m} v^2} = t + Const.$$

 

[Data]

$$v^\prime = g + \frac{k}{m}v^2 \Longrightarrow \frac{v^\prime}{\frac{k}{m}v^2 + g} = 1$$

$$\Longrightarrow \int \frac{dv}{\frac{k}{m}v^2 + g} = \frac{m}{k} \int \frac{dv}{v^2 + \frac{gm}{k}} = \sqrt{\frac{m}{kg}} \arctan \left(v \sqrt{\frac{k}{mg} \right) = t + C$$

$$\Longrightarrow \arctan \left(v \sqrt{\frac{k}{mg}} \right) = \sqrt{\frac{kg}{m}}(t+C) \Longrightarrow v \sqrt{\frac{k}{mg}} = \tan \left( \sqrt{\frac{kg}{m}}(t+C)\right)$$

$$\Longrightarrow v = \sqrt{\frac{mg}{k}} \tan \left( \sqrt{\frac{kg}{m}}(t+C)\right)$$

 

[dextercioby]

Hi i posted last time about solving an equation regarding air resistance being proportional to the speed

This time i need it to be proportional to the speed squared

I have managed to get an equation but haven't managed to solve it yet

dv/dt -(k/m)v^2 = g


Any ideas on how to solve this. I have tried all the methods i know, I think i need to substitute in. I will need more than just a hint here guys

Thanks a lot

sid

A bit weird not to include among "all the methods" the separation of variables,which is the simplest possible.

Anyway,i hope you understood the solution Data gave you.

Daniel.

 

[$id]

Hmm i can follow Data logic despite never have seen anything like that before,

I guess now the main problem is that equation gives me the velocity. for the model i am making the data readings are distances at a given time. Hence I will most probably to integrate that beast.

All i know is that integral of tanx = - ln(cos x) + c or - ln (sec^2 x) or something like that

sid

 

[dextercioby]

$$\int \tan x \ dx=-\ln|\cos x|+C$$ it can be proven using the definition of tangent & the substitution $\cos x=u$.

Daniel.