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Probability distribution

  1. Dec 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Supose that a traffic study measures the speed at which people drive on the highway, and determines that the situation is well modeled by the probability distribution P(v)=Cv4e-v/vo. (a) If we are to measure speeds in mph, give the appropriate units for C and vo. (b) In order to perform an analysis of damages in crashes, transform that distribution into a distribution for kinetic energy.


    2. Relevant equations
    the product p(v)*dv is a dimensionless quantity.


    3. The attempt at a solution
    I understand that the units for dv are miles/hour, so p(v) is hour/miles.
    therefore C must be in units (hour/miles)5

    as for part B
    K meaning kinetic energy
    P(K)dK, dimensionless
    i thinking that dK would be equal to .5*m*v*dv
    and that P(K) would be P(v) multiplied by a constant 'A' (A having units hour/kg*mile)

    so P(K)=ACv4e-v/vo units, (hour2/kg*mile2)
    and dK=.5*m*v*dv units, (kg*mile2/hour2)

    so if we were to intergrate,

    fraction=p(k)*dk=p(k)*.5*m*v*dv=[tex]\frac{1}{2}[/tex]ACm[tex]\int[/tex]v5e-v/v0 dv

    i think most people in the class just substituted 'v' with (2K/m)1/2, but i feel like that is incorrect, as you still need v to integrate.

    is this the answer? is the constant 'A' necessary?
     
  2. jcsd
  3. Dec 5, 2009 #2

    kuruman

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    Gold Member

    The probability distribution should really be

    [tex]\frac{dP}{dv}=Cv^4e^{-v/v_0}[/tex]

    To convert to kinetic energy, use the chain rule

    [tex]\frac{dP}{dK}=\frac{dP}{dv}\frac{dv}{dK}[/tex]

    You already have dP/dv but you need to calculate dv/dK.

    Don't forget to change all occurrences of v into K, otherwise you will not have p(K). You do not need a separate constant A. When you are done, do dimensional analysis to make sure everything is in order.
     
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