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Velocity and position as functions of time

  1. Feb 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose that a neutron with with mass M and initial speed v0 is traveling through a material, such as graphite. As the neutron moves, it makes collisions with atoms in the material, so teh faster it is moving, the more collisions it makes per unit time. Another model of teh net effect of this is that the neutron experiences a resistive force proportional to the square of its speed, e.g., the force exerted by the material on teh neutron goes like F=-bv^2. Determine the velocity and position of the neutron as a function of time??

    2. Relevant equations
    F=-bv^2


    3. The attempt at a solution
    I calculated the velocity and got v(t)= m/[bt+(m/v0)] and this makes sense dimensionally at t=0 v=v0 but when i calculated the position I got x(t)= m/b [ln(bt+(m/v0))]+C .. The problem is this does not make sense dimensionally and conceptually.
     
  2. jcsd
  3. Feb 10, 2012 #2
    Hopefully others will correct this if it is wrong. After you integrated you had limits of integration, 0 to x and 0 to t. You should be subtracting a term ln(m/v0) and as lna - lnb = ln[a/b] the units will cancel inside the ln. I think your function is OK?

    Good luck!

    See,

    http://www.sciforums.com/showthread.php?p=1331961
     
  4. Feb 11, 2012 #3
    im sorry where did you get the lna -lnb cause i am adding bt +(m/v0) inside the ln and i dont think you can seperate the two into ln(bt) + ln(m/v0)..
     
  5. Feb 11, 2012 #4
    I think you are integrating between limits, on the left side from 0 to x and on the right side from time = 0 to t. There will be a difference of two terms on the right hand side, x(t) - x(0) = something like

    ln(bt+(m/v0)) - ln(b*0+(m/v0)) = ln[(bt+(m/v0))/(m/v0)]

    The units cancel, that was what you were confused about?
     
  6. Feb 11, 2012 #5
    I understand now thank you
     
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