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Homework Help: Finding x(t) from a time dependant force

  1. Mar 14, 2012 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m is subject to a force
    F(t) = mae-bt

    The initial position and speed are zero. Find x(t)
    2. Relevant equations

    3. The attempt at a solution

    how would i begin to start this??
    Last edited: Mar 14, 2012
  2. jcsd
  3. Mar 14, 2012 #2


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  4. Mar 14, 2012 #3

    this is what i tried

    F(t) = m(dv/dt) e-bt

    F(t)ebtdt = mdv

    ∫F(t)ebtdt = ∫mdv (with bounds t=0 to t=t, and v=0 to v=v)

    ∫F(t)ebtdt = mv (only integrating right side yields)

    ∫F(t)ebtdt = m(dx/dt)

    [∫F(t)ebtdt]dt = dx (integrating both sides again)

    ∫ [∫F(t)ebtdt]dt = x(t)

    is this correct? can it be simplified further?
  5. Mar 15, 2012 #4


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    You mean "a" is the acceleration?

    No, that makes no sense.

    If F(t) is the force, and m is the mass, then F(t) = ma,

    so F(t) cannot be mae-bt. :redface:

    What exactly was the question? :confused:
  6. Mar 15, 2012 #5
    I think I get it:

    F(t)=ma(t) in which a(t)=a0e-bt

    If you integrate you get:


    And again:


    Since x0 and v0 are both zero.

  7. Mar 15, 2012 #6
    touche sir. That does not make sense. ahh i see now

    Question was F(t) = ma0e-bt
    now i know why it is a0

    so in this case acceleration a = a0e-bt

    but i'm still stuck! How do i get to x(t)?
  8. Mar 15, 2012 #7
    wait..you can just integrate like that without the dt on the right hand side?
    Last edited: Mar 15, 2012
  9. Mar 15, 2012 #8
    I simply skipped the integration steps :tongue:
  10. Mar 15, 2012 #9
    alright ill take your word for it!
  11. Mar 15, 2012 #10


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    Please do not do the student's homework for them. That is against the PF rules.
  12. Mar 15, 2012 #11
    I'm sorry, I let myself go :shy:
    It won't happen again :approve:
  13. Mar 15, 2012 #12
    haha yeah stop doing my homework for me!

    Ok i figured it out properly

    F(t) = ma0e-bt

    m(dv/dt) = ma0e-bt

    ∫mdv = ∫ma0e-btdt (from v=0 to v=v and t=0 to t=t)

    mv = m(a0e-bt)/-b

    (dx/dt) = (a0e-bt)/-b

    ∫dx = ∫(a0e-bt)/-b dt (from x=0 to x=x and t=0 to t=t)

    x(t) = a0e-bt/b2
  14. Mar 16, 2012 #13


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    Hi Keshroom! :smile:

    Yes, that's fine except for the constants. :wink:

    When you got down to here …
    … although you said "from … t=0 to t=t", you didn't do it! :rolleyes:

    (to put it another way: always check your solutions just by looking at them … does this satisfy the initial condition of v = 0 ? :wink:)

    Try again. :smile:
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