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A particle of mass m is initially at rest

  1. Feb 8, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m, initially at rest at x=0, is accelerated by a force that increases in time as F=Ct2. Determine its velocity v as a function of time.

    2. Relevant equations
    x = vt
    v = at

    3. The attempt at a solution
    The correct method makes sense, but my method has no error. However, my answer is wrong. Why?
    http://photo1.ask.fm/726/204/412/-69996997-1sha200-dqjla0edmhpamrf/original/IMG_4066.jpg [Broken]
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Feb 8, 2015 #2

    haruspex

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    v=at only works for constant acceleration. Integration works for constant or varying acceleration.
     
  4. Feb 11, 2015 #3
    So because force is a function of time (f = Ct2), acceleration is also a function of time and therefore not constant?
    Thanks
     
  5. Feb 11, 2015 #4

    haruspex

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    Yes.
     
  6. Feb 11, 2015 #5
    Thank you so much
     
  7. Mar 3, 2016 #6
    I don't see how you got 3m at the bottom? does it have to do with the fact that t is to the third power?
     
  8. Mar 3, 2016 #7

    haruspex

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    Yes. What is ##\int t^2.dt##?
     
  9. Mar 3, 2016 #8
    I figured it out, it is the antiderivative I believe
     
  10. Mar 4, 2017 #9
    I solved this questions like;

    $$ F=m.a $$ $$ F=Ct^2 $$ => $$ m.a = Ct^2 $$ $$ a= \frac {Ct^2} {m} $$
    and if we integrate "a" respect to time(t), we obtain velocity(v);
    $$ \int \frac {Ct^2} {m} dt = \frac {Ct^3} {3m} = V $$
    and if we integrate "v" respect to time(t), we obtain position(x);
    $$ \int \frac {Ct^3} {3m} dt = \frac {Ct^4} {12m} = X $$

    I'm not sure, tell me if it's correct.
     
    Last edited: Mar 4, 2017
  11. Mar 4, 2017 #10

    haruspex

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    It is, but technically you should go through the step of allowing for a constant of inrtegration each time, then using the initial conditions to show it is zero.
     
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