Velocity and position as functions of time

  • #1

Homework Statement


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??

Homework Equations


F=-bv^2


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.
 

Answers and Replies

  • #2
Spinnor
Gold Member
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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
 
  • #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)..
 
  • #4
Spinnor
Gold Member
2,176
384
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?
 
  • #5
I understand now thank you
 

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