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Acceleration, Velocity and distance

  1. Nov 18, 2008 #1
    1. The problem statement, all variables and given/known data

    An engineer designing a system to control a router for a machining process models the system so that the router's acceleration during an interval of time is a=-1.8vm/s . When t=0 , its position is s=0 and its velocity is 2.4 m/s. Determine the router's position (unit: m) at s.

    2. Relevant equations

    a=dv/dt, v=ds/dt

    3. The attempt at a solution

    First i did -1.8v=dv/dt did some integration and work and found v = e^(-1.8t)

    I then used e^(-1.8t) = ds/dt and solved for s and subsituted t=0(lower bound) and t=0.75(upper bound) and found s = 0.412m

    I am not sure if the first step I did is allowed since I never used the v=2.4. Please tell me my mistakes. Thanks!
     
  2. jcsd
  3. Nov 19, 2008 #2

    tiny-tim

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    Hi Oblivion77! :smile:

    You integrated twice, but you only put in a constant of integration the second time.

    The v = 2.4 comes in the first time! :wink:
     
  4. Nov 19, 2008 #3
    Ok, using the v=2.4 with the first integration I get -1.8v=dv/dt, doing some integration I get
    t=(-1/1.8)ln(v)+0.49 (does this look right so far?). Now I want to solve for v and use v=ds/dt to get the function for position. so v=e^(0.88-1.8t), e^(0.88-1.8t)=ds/dt .Then to sub t=0.75s for the upper bound and t=0 for the lower bound. Does this look right now? Or did I make a mistake somewhere? Thanks!
     
  5. Nov 19, 2008 #4

    tiny-tim

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    oooh :cry:

    No, you integrated dv/v - -1.8dt to get v = Ce-1.8t.

    And so, when t= 0, e-1.8t = … ? , and so C = … ? :smile:
     
  6. Nov 20, 2008 #5
    Ok, so when i do that I get v=Ce^(-1.8t). And the intial conditions are t=0, v=2.4. So would i do 2.4=Ce^0, C=2.4. Then v=2.4e^(-1.8t), or do i not use the velocity?
     
  7. Nov 20, 2008 #6

    tiny-tim

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    What a long question! :wink:

    Yes, v = 2.4e-1.8t

    why does that worry you? :smile:
     
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