Velocity and position as functions of time

[Talltunetalk]

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 the faster it is moving, the more collisions it makes per unit time. Another model of the 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 the 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.

 

[Spinnor]

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

 

[Talltunetalk]

im sorry where did you get the lna -lnb cause i am adding bt +(m/v0) inside the ln and i don't think you can separate the two into ln(bt) + ln(m/v0)..

 

[Spinnor]

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?

 

[Talltunetalk]

I understand now thank you