1. ## ODE help

Hey friends, I found this site searching for ode help and I think I might have found a home.. I do need help though and would appreciate any assistance you can give.

I have an object that weighs 10kg and the drag force is proportional to the square of the velocity and the limiting velocity is 49 m/s. The equation of motion given is $\displaystyle m(dv/dt) = mg - 2v$
or
$\displaystyle dv/dt = 9.8 - v/5$ after putting in the drag force constant, gravity and weight. So my objective is to prove how the equation of motion can be written as $\displaystyle dv/dt = [49^2 - v^2] / 245$.

If any of you happen to have Elementary Diff EQs and Boundry value problems its number 11 in section 1.2

And this is the coolest site ever, I love math. T"hanks for your help.

2. Originally Posted by DinoJockey
I have an object that weighs 10kg and the drag force is proportional to the square of the velocity and the limiting velocity is 49 m/s. The equation of motion given is $\displaystyle m(dv/dt) = mg - 2v$
or
$\displaystyle dv/dt = 9.8 - v/5$ after putting in the drag force constant, gravity and weight. So my objective is to prove how the equation of motion can be written as $\displaystyle dv/dt = [49^2 - v^2] / 245$.
your DE statement and your DE, $\displaystyle m(dv/dt) = mg - 2v$, do not match up.

if the drag force is proportional to the square of velocity, then the DE should be

$\displaystyle m\frac{dv}{dt} = mg - kv^2$

3. Hmm I thought I missed something but that still leaves me at $\displaystyle \frac{dv}{dt} = 9.8 - \frac{v^2}{5}$ right?

4. at limiting speed, $\displaystyle \frac{dv}{dt} = 0$

$\displaystyle mg = kv^2$

$\displaystyle \frac{mg}{v^2} = k$

$\displaystyle \frac{98}{49^2} = k$

$\displaystyle \frac{2}{49} = k$

$\displaystyle \frac{dv}{dt} = g - \frac{k}{m} v^2$

$\displaystyle \frac{dv}{dt} = 9.8 - \frac{v^2}{245}$

5. ah ha! you are a scholar indeed. I appreciate your help.