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Acceleration as a function of velocity - HW problem help

  1. Jan 24, 2008 #1
    1. The problem statement, all variables and given/known data

    The acceleration of a particle is defined by the relation a = -0.05v^2, where a is expresssed in m/s^2 and v in m/s. The particle starts at s=0 m with a velocity of 5 m/s. Determine (a) the velocity v of the particle after it travels 10m, (b) the distance s the particle will travel before its velocity drops to 2 m/s, (c) the distance s the particle will travel before it comes to rest.


    2. Relevant equations

    a = v (dv/ds)

    3. The attempt at a solution

    [​IMG]

    I don't think this is right because when I integrated the right side, it ends up being undefined?
     
  2. jcsd
  3. Jan 24, 2008 #2
    not sure if what i'm thinking is correct but its an idea:

    a= dv/dt so we know dv/dt = -.05v^2. We can integrate that to get the velocity function.

    once we have the velocity function we can integrate again since v(t) = ds/dt. we have initial conditions. its just an idea.
     
  4. Jan 24, 2008 #3
    No it isn't!

    [tex]\alpha=-0.05\,v^2 \Rightarrow v\,\frac{d\,v}{d\,s}=-0.05\, v^2\Rightarrow \int_{v_0}^v\frac{d\,v}{v}=-\int_o^s 0.05\,d\,s\Rightarrow \ln v\Big|_{v_0}^v=-0.05\,s\Rightarrow v=v_o\,e^{-0.05\,s}[/tex]
     
    Last edited: Jan 24, 2008
  5. Aug 20, 2008 #4
    How can you just cancel "v" on both sides in second step , where it can take a value zero...
     
  6. Nov 2, 2008 #5
    Just to set the record straight on Rainbow child's answer:
    This slightly wrong, it should be a == d(0.5v2)/ds not d(v2)/ds.
     
  7. Sep 9, 2010 #6
    Hi.. just to go one step further, would the distance (with respect to time) equation be :-

    [tex]\int{{{v}{}_{0}\over{e^{0.05\,t}}}}\,dt = -{{20\,v{}_{0}}\over{e^{{{t}\over{20}}}\,\log e}}[/tex]
     
  8. Sep 9, 2010 #7
    Easy, you just say that the step was valid for all non-zero values of v.
    [tex]a=v\frac{dv}{dx}=-kv^2[/tex] (In this case, k=0.05)
    From this step to the next, you have two options. Either v is always 0, in which case you have a solution for v, and you are not allowed to divide by v. Or v is different from 0, and you are allowed to divide by it.

    suchara, the velocity as a function of time is not what you posted. Note that you used the expression for the velocity as a function of distance and replaced the x with t (A mistake).

    [tex]v=\frac{v_0}{1+kv_0 t}[/tex]
     
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