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Kinetics question

  1. Aug 24, 2014 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=72422&stc=1&d=1408927480.png

    2. Relevant equations


    3. The attempt at a solution
    v = ds/dt

    ∫v dt = ∫ds

    (put all inital values for V)
    ∫70-70e^t dt = ds

    70t - 70t*e^t = s

    There is something wrong with my methodology. can you guide me here?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

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  2. jcsd
  3. Aug 24, 2014 #2
    Looks like you did not integrate [itex]e^{-bt}[/itex] correctly. Check you last eqution. You have something extra there.
     
  4. Aug 24, 2014 #3

    Nathanael

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    Homework Helper

    Are you sure that is the correct integral?


    Edit:
    Sorry, I didn't see Mr-R's post when I posted this
     
  5. Aug 24, 2014 #4
    ∫v0 (1-e^(-b*t))

    ∫v0 - v0*e^(-b*t)

    = v0*t - V0*e^(-b*t) + c

    is that correct?
     
  6. Aug 24, 2014 #5

    Well the constant c is nice :smile:. Missing something from [itex]e^{-bt}[/itex] when you intagrated it though.

    Edit: are you writing b explicitly or just its value? If you just substitute its value there then your integral is correct. (looks like you did not)
     
    Last edited: Aug 24, 2014
  7. Aug 24, 2014 #6
    Can you show me the integration?
     
  8. Aug 24, 2014 #7
    Of course I can, but I know you can do it by yourself so I will give an example.

    [itex]∫e^{Cx}dx=\frac{1}{C}e^{Cx}+c[/itex]

    Can you see you mistake and fix your integral now?
     
  9. Aug 24, 2014 #8
    v0*t + (V0*e^(-b*t))/b + c

    Yes. That should be it? Now I sub in all the values and find the constant C and my final answer for distance is -1126m. The answer is correct but it should be a positive value.
    Is this because my answer is displacement and the question requires distance (just magnitude only)?
     
  10. Aug 25, 2014 #9
    Well the function describes a train moving with a velocity that has a negative sign(if b=-1 see below). If they defined the positive direction to be the right direction. Then the function gives you a backwards moving train.

    See the attached file,

    The first graph is your given equation (velocity). The second graph is the distance which you found by integrating.

    Examine the given equation(velocity) and you will notice that "b" is the constant which makes the velocity to be positive or negative. If b=1 then the train will travel in the positive direction and you will end up with positive distance.
     

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