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

  1. Sep 3, 2010 #1
    Hey guys,

    I'm looking to calculate the initial velocity of a large object sliding to a stop over a surface. This is easy without the drag force of the air, but I'd like to include that to prove a point in my research. As a quick background, I haven't done a differential equation in over a decade since I was an undergrad, so I'm trying to relearn on the fly.

    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. 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 want to find the velocity V with respect to position x. The solution I came up with is:

    [itex]V(x) = \sqrt{-\frac{Mfg}{k}(1-e^{\frac{-2k(x-D)}{M}})}[/itex]

    This uses the fact that when x = D, V = 0. Does this look correct? My calculations make sense, but I want to be sure. I can include the intermediate steps if necessary, but that's a lot of typing!

    Thanks from an old out of practice former physics nerd who now only uses this stuff on occasion.
  2. jcsd
  3. Sep 4, 2010 #2

    Filip Larsen

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  4. Sep 4, 2010 #3


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    Writing A= -fmg and B= -k the equation is A+ Bv= mVV' so that (mVV')/(A+ BV)= 1 and (mVdV)/(A+ BV)= dt.

    To solve that, integrate both sides: [itex]\int (MV)/(A+ BV) dV= \int dt= t+ C[/itex].

    To integrate on the left, let, u= A+ BV so that du= Bdv and (1/B)du= dv. Also BV= u- A so that V= (u- A)/B. The integral becomes
    [tex]\frac{M}{B^2}\int \frac{u- A}{u} du= \frac{M}{B^2}\int 1- \frac{A}{u}du[/tex]
    [tex]\frac{M}{B^2}(u- Aln(u))= \frac{M}{B}(A+ BV- Aln(A+ BV))= t+ C[/tex].
  5. Sep 4, 2010 #4

    V(x) is on the right side because the equation is in the form of forces = mass times acceleration. Since acceleration is V'(t) but I don't want it as a function of t, I want it as a function of position (x), I used these steps....

    a = dV/dt

    dV/dt = (dV/dx)(dx/dt) - chain rule

    therefore, since velocity is just the derivative of position w/ resp to t...

    a = dV/dx V, or as I have it written in my eqn., a is V'(x)V(x)

    I attached my soltion to the eqn in .doc format with explanations...if someone could look at it and tell me if the logic/steps are good, I'd appreciate it.

    Thank you for the assistance.

    Attached Files:

  6. Sep 4, 2010 #5

    Filip Larsen

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    Ok, I missed that you have used the chain rule.

    So, it looks like you have the differential equation

    (1) [tex] -f - k v^2 = \frac{dv}{dx}v [/tex]

    where I have combined the constants into just f and k. Separating and integrating I get

    (2) [tex] dx = - \frac{v dv}{f+kv^2} [/tex]


    (3) [tex] x_1 - x_0 = \frac{1}{2k}\left[ \log\left( f+kv^2 ) \right]_{v_0}^{v_1} [/tex]

    which with [itex] x_0 = 0, x_1 = D, v_1 = 0 [/itex] inserted and rearranged gives

    (4) [tex] v_0^2 = \frac{f}{k} \left( e^{2kD} -1 \right ) [/tex]

    [STRIKE]I haven't looked at your attachment yet, but if you still get the same result as in your first post we then seem to disagree on the square root[/STRIKE].
    Last edited: Sep 4, 2010
  7. Sep 4, 2010 #6
    Yeah, if you wouldn't mind looking at it, I'd appreciate it. Like I said, it's been over 10 years since I've done this stuff!

    The square root in my solution came from the fact that r v2 was in the equation and that after solving for that, I just wanted v.

    You've made it simpler by combining the constants more that I did...I made it harder on myself. But, hey, this whole thing's been a challenge anyway!
  8. Sep 4, 2010 #7

    Filip Larsen

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    Ah, my mistake, sorry. I can see that the square of v0 slipped out of my calculation somewhere between equation (3) and (4). You are correct in having a square root.

    It looks a slight bit more complicated than necessary, I think.

    In your step 3 you have dx on both sides. I would instead use that, by definition, v'(x) = dv/dx, so that you can replace v'(x) by dv/dx, that is left-hand side gets dv and right-hand side gets dx, resulting in

    (Step 3) [tex]\int \frac{v}{-f-kv^2}dv = \int dx[/tex]

    In step 6 the two constants of integration can be combined into one constant (that would be valid for any sum of indefinite integrals). Alternatively, you can use definite integrals inserting the boundary values early on. In this case step 3 would have looked like

    (Step 3') [tex] \int_{v}^{0} \frac{v}{-f-kv^2}dv = \int_{x}^{D} dx [/tex]
  9. Sep 4, 2010 #8
    Thanks, I'll try simplifying a bit. 10 years out of practice, I guess I'm glad that I got an answer at all! Thanks for your input.
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