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Short question

  1. Aug 2, 2004 #1
    A drag racer experiences air resistance equal to -cv². Assuming the racer is designed for maximum acceleration, I am to find v[t] and v[x]. If I can just get the equation of motion I'll be set.

    I think it is F = ma - μmg - cv² = m dv/dt. According to the answer, the book thinks it is F = - μmg - cv².

    Of course, when I evaluate our answers with the assigned numbers for μ, v, and vt (terminal velocity), we both get imaginary answers. :( I have no clue where we are wrong, but I suppose the books values could be the problem there.
  2. jcsd
  3. Aug 2, 2004 #2


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    It's definitely not F = ma - μmg - cv². a = dv/dt, so you'd have:

    ma - μmg - cv² = m dv/dt
    ma - μmg - cv² = ma
    -μmg - cv² = 0

    Fnet = ma, as you know, so your equation would also be wrong because it's saying:

    Fnet = Fnet - μmg - cv²

    And you run into the same problem. ma is not one of the forces acting on the body, it is the NET FORCE, so you can't add it with other forces to get the net force.

    Anyways, what is it that you need to find?
  4. Aug 2, 2004 #3
    The way I see it, the net force, m dv/dt, is equal to the driving force of the engine (ma) minus the frictional force (μmg) minus the air resistance (cv²). Since the racer is designed for optimal acceleration, I assumed that "ma" was this maximal driving force, a constant.

    Anyway, the problem is asking to obtain v[t] and v[x]. Then I am to "eliminate the coefficient of friction and solve the resulting equation numerically for the terminal velocity that can produce the 1988 world record of v=129.1 m/s, with t=4.99s for x=0.4 km."

    Funny thing is that the v[t] that the book lists has the term ln[(u+v)/(u-v)] where u is the terminal velocity, which is given as 123.6 m/s in the back. Using the values of v and u, the log is imaginary.

    Oh, thanks for your help, as well.
  5. Aug 2, 2004 #4


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    Okay, well that's just confusing. In that way, "a" is not the acceleration of the vehicle at all, just some constant optimal acceleration. You should replace ma with "Fapplied" or "m*a_optimal" to avoid the confusion.
    Just to clarify, you mean speed (v) as a function of time (t) and distance (x)? I would expect to see this as v(t) and v(x), just not sure exactly what it is. I'll look over that question to see if I can get v(t) and v(x).
  6. Aug 2, 2004 #5
    Yeah, that was confusing. Sorry 'bout that.

    v[t] and v[x] are indeed velocity as a function of time and distance. I use the brackets since I've been using mathematica a lot. What I have obtained is:

    t = u/(a-g μ) ArcTanh(v/u)

    x = -u²/(a-g μ) Ln(v²-u²).
  7. Aug 2, 2004 #6


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    Okay, let A represent the optimal acceleration
    Let B represent [itex]\mu g[/itex]
    Let C represent [itex]\frac{c}{m}[/itex]

    [tex]m\frac{dv}{dt} = mA - mB - cv^2[/tex]

    [tex]\frac{dv}{dt} = A - B - Cv^2[/tex]

    [tex]\frac{dv}{dx}\frac{dx}{dt} = A - B - Cv^2[/tex]

    [tex]v\frac{dv}{dx} = A - B - Cv^2[/tex]

    [tex]\frac{dv}{dx} = \frac{A - B - Cv^2}{v}[/tex]

    [tex]\frac{v}{A - B - Cv^2} dv = dx[/tex]

    [tex]\int _{v(x=0)} ^{v(x=x)} \frac{v}{A - B - Cv^2} dv = x[/tex]

    Let [itex]u = A - B - Cv^2[/itex]

    [tex]\int _{A - B - C[v(x=0)]^2} ^{A - B - C[v(x=x)]^2} \frac{-1}{2C} \frac{du}{u} = x[/tex]

    [tex]-\frac{1}{2C} \ln u |_{A - B - C[v(x=0)]^2} ^{A - B - C[v(x=x)]^2} = x[/tex]

    [tex]-\frac{1}{2C} \ln \left ( \frac{A - B - C[v(x=x)]^2}{A - B - C[v(x=0)]^2} \right )= x[/tex]

    [tex]\frac{A - B - Cv(x)^2}{A - B - Cv_0^2} = e^{-2Cx}[/tex]

    Isolate v(x), and you should be okay from there, I think.
    Last edited: Aug 2, 2004
  8. Aug 2, 2004 #7


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    Assuming what you have is right:

    v(t) = u Tanh [t(a - gμ)/u]

    Also, now that you have v in terms of t, you can replace the v's in the second equation, "x = -u²/(a-g μ) Ln(v²-u²)" with a function of t. You'll have x in terms of t. Isolate t, to get t in terms of x. Take that equation and plug it into your equation for v in terms of t, so you'll have v in terms of x, v(x).
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