# Math Help - Initial speed of object sliding to a stop w/ air resistance

1. ## Initial speed of object sliding to a stop w/ air resistance

Hello all,

First, a brief intro...I haven't done a differential equation since I was an undergrad (over 10 years ago!). I was tasked with finding the initial velocity of an object skidding to a stop over a surface including the frictional forces and air resistance.

Basically, I want to find a solution for V(x). Treating the vehicle as beginning at x=0 and moving in the postitive x direction a known distance D, the equation I came up with is

-fMg - kV2(x) = MV'(x)V(x)

The left side are the forces and the right side is mass times acceleration (acceleration was found by using the chain rule to put everything in terms of position x rather than time). f is the adjusted coeff. of friction, M is the object's mass, g is gravitational acceleration, k is a constant encompassing the other constants in front of v2 in the drag force.

I have attached my steps and final solution as a word document. Someone has already pointed out that I could have simplified it by combining constants more...so that I know. But, I'd like to see if my logic and my solution makes sense.

Thank you to anyone who can help.

2. It's kind of an unwritten rule around here not to post Word documents. They can have nasties in them. Could you please post a pdf? Thanks!

3. Sorry about that! I converted it and attached again.

4. Ok, so just to be sure, your DE is as follows:

$-f\,m\,g-k\,v^{2}(x)=m\,v'(x)\,v(x).$ Is this correct?

You're using a quadratic air resistance, correct?

5. From $fMg- kV^2(x) = MV'(x)V(x)$ we can get $\frac{MV}{fMg- kV^2}V'= 1$ or $
\frac{MV dV}{fMg- kV^2}= dt$
. You should be able to integrate the left side by letting $u= fMg- kV^2$ so that [tex]du= -2kV dV[tex] so that $V dV= -\frac{2u}{2k}$ and $\frac{MVdV}{fMg- kV^2}= -\frac{M}{2k}\frac{du}{u}$.

6. Ackbeet-

Yes, that's correct. I know (if my memory serves me correctly) that at some point the object will slow down enough to not cause turbulent flow, but I'm ignoring that here. The equation you have is correct and my solution is for v(x) (which I, of course, can use to get the velocity at any point in its travel, including its initial velocity at x = 0).

7. Everything in your solution looks good until you get to Step 6. There, I'd say you should have had

$\displaystyle{-\frac{m}{2k}\,\ln|u|=x+C.}$

There's no need for two constants of integration, since you're only integrating once (it's a first-order DE). Also, a technical point: you need to have absolute magnitudes around the argument of the logarithm function when you do that integral. It doesn't end up affecting your final answer, since you exponentiate everything.

So, I would carry those corrections through and see what you get.

Side note: isn't LaTeX better than Word?

8. Right. You mentioned the constants thing in the OP. Sorry about duplicate mentioning there.

9. Everything looks fine to me, except that I'm not sure you can just choose $C_{1}=-fg.$ As I said before, and as someone else has apparently told you, I think you're better off not using two constants of integration.

10. Point taken on the absolute magnitude. Why would I not be able to choose an arbitrary constant for C1? I put that constant in there to simplify the equation so I could easily find the value of my other constant (C2). I figured it would be okay since $\frac{d}{du}ln (u)$ is the same as $\frac{d}{du}ln (\frac{u}{any\: constant})$ ...namely, 1/u....right? Or am I making up my own math there?

Thanks for your continued help and patience.

11. Ok, you've convinced me you can do that.

Good work!

Have a good one.

12. Thanks! Glad all that undergrad math didn't go completely to waste after all. I appreciate your help.

13. You're welcome. Have a good one!