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Homework Help: Function of force with respect to v & t; find final v and distance

  1. Sep 13, 2013 #1
    1. The problem statement, all variables and given/known data
    A particle ( m = 1 kg ) moves on the x axis. The initial position is x = 0 and the initial velocity is v(0) = 5 m/s. The particle experiences a force,

    F(v,t) = − βv^2 e^(−αt)

    (a) Determine the final velocity of the particle.
    (b) Determine the distance it has moved at t = 10 seconds.
    [DATA: α = 0.27 s−1 ; β = 1.19 kg/m .]

    2. Relevant equations

    3. The attempt at a solution

    This is part of the calculus review of my physics homework. What is confusing me is the function which is dependent on two variables.

    For part (a) I divided the F function by the mass 1 kg which should then mean that a(v,t)=F(v,t)/m, correct? Beyond that I'm not sure how to differentiate the acceleration function to get velocity. If I differentiate with respect to t or with respect to v I still have a v in the equation which means I cannot find the limit of the resulting function as t approaches infinity to find the final velocity.

    Once I figure out how to find the acceleration function I can probably differentiate that to find the position function for part (b).

    Any suggestions? Thank you!
    Last edited: Sep 13, 2013
  2. jcsd
  3. Sep 13, 2013 #2


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    Velocity is not equal to the derivative of acceleration. Check your definitions of position, velocity, and acceleration again.
  4. Sep 13, 2013 #3


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    What is the interpretation of v2? Is it v*2 or v^2 or something else?


    Note that the acceleration can be written as a = dv/dt. Use this for the left hand side of a(v,t)=F(v,t)/m. See if you can somehow integrate this to find v(t). Hint: separation of variables.
  5. Sep 13, 2013 #4
    *facepalm* Thanks for pointing that out. That would explain the confusion I suppose.

    Yes, TSny, I missed the carrot and have adjusted the original. It is supposed to be v^2.

    I'll get back after going in the -correct- direction.
    Last edited: Sep 13, 2013
  6. Sep 13, 2013 #5
    So, here it goes:

    dv/dt=-1.19(v^2)e^(-.27t) =>

    v^(-2)dv=-1.19e^(-.27t)dt (integrate both sides) =>

    -v^(-1)=4.40741e^(-.27t)+C =>

    How do I do this without ending up with a limit depending on C? I could integrate the right half from 0 to infinity but what about the left half?

    EDIT: Going to try to solve for C by leaving it on the left and plugging in the initial conditions.

    EDIT 2: Plugging in 5 for v and 0 for t I get a C of ~4.6074. If I then solve the equation for v and take the limit as t approaches infinity with that C value the result is -.217 m/s.

    Taking the integral of the above equation solved for v to find x(t) and plugging in t=10 seconds I get x(10)=-2.22

    I'm not sure which, if not both, are incorrect because the solution is dependent on both being correct and it is not accepting my answers. Suggestions?
    Last edited: Sep 13, 2013
  7. Sep 13, 2013 #6
    At Infiniti the exponential goes to zero therefore, 1/v = + c
  8. Sep 13, 2013 #7
    To find c apply the given initial condition.
    After applying
    1/5 =-β/αm + c
  9. Sep 13, 2013 #8
    Thank you, yands, yes. That should be a positive 0.217 m/s.
  10. Sep 13, 2013 #9
    Just keep calm while solving problems dear:p
  11. Sep 13, 2013 #10


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    Note the negative sign on the left side of the equation. Should C be a positive or negative number?

    [EDIT: Sorry, I see you found the mistake.]
  12. Sep 13, 2013 #11
    It should be negative. So the resulting velocity would be 0.217 m/s. That is still not correct, though.

    EDIT: I mean to say, that the combination of the two results is not correct even if 0.217 m/s is. I have to try the integral again with the negative.
  13. Sep 13, 2013 #12
    You are making signs mistakes dude
    V^-2 dv = -β/me^-αt dt
    -V^-2 dv = β/me^-αt dt
    V^-1 = -β/αm e^-αt +c
    Last edited: Sep 13, 2013
  14. Sep 13, 2013 #13
    Try to substitute the numerical values after you done manipulations
  15. Sep 13, 2013 #14
    Now again apply nicely the initial condition
  16. Sep 13, 2013 #15
    C is a constant to be determined so negative or positive it does not matter
  17. Sep 13, 2013 #16
    Just be sure to take the reciprocal of the answer
  18. Sep 13, 2013 #17
    Yeah, that's what I did. C=4.6074. Are you interpreting the ~ as a negative?
  19. Sep 13, 2013 #18
    I got 0.217 too . Why are you so sure it is not correct
  20. Sep 13, 2013 #19
    The answer is in two parts. Both are required for it to be correct. I need to find the distance traveled at t=10.
  21. Sep 13, 2013 #20
    Why are you so sure the answer is incorrect
  22. Sep 13, 2013 #21
    Can you show us the second part solution
  23. Sep 13, 2013 #22
    When I submit the answers it says incorrect, haha.

    At this point I'm not even sure what to try. We just solved the equation for v so I should be able to integrate that, solve for the new C by plugging in x=0 and t=0 and then solving for t=10 but there is a log of a negative number in the answer so I'm getting imaginary numbers.
  24. Sep 13, 2013 #23
    I doubt that the first answer is incorrect. And I do think you have done some mistake in the second part
  25. Sep 13, 2013 #24
    I've solved the second part and the log seems fine
  26. Sep 13, 2013 #25
    Well, v=-1/(4.40741(0.763379)^t+4.60741).

    If I integrate that to find x(t) I get x(t)=-.80386*ln(1.30996^(-.27t)+.956594)+C.

    Plugging in t=0 and t=0 to find C I find that C=.539555

    Plugging in t=10 and C=.539555 to find x(t) I find that x(t)=-1.68.

    One of the answers is incorrect.
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