What is the kinetic energy distribution given the probability distribution?

[jderm]

Homework Statement

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/v_o. (a) If we are to measure speeds in mph, give the appropriate units for C and v_o. (b) In order to perform an analysis of damages in crashes, transform that distribution into a distribution for kinetic energy.

Homework Equations

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

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/v_o 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=\frac{1}{2}ACm\intv⁵e^-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?

 

[kuruman]

The probability distribution should really be

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

To convert to kinetic energy, use the chain rule

\frac{dP}{dK}=\frac{dP}{dv}\frac{dv}{dK}

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.