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Linear Motion

  1. Jul 2, 2013 #1
    Hi All,

    A question to calculate the speed at constant velocity during a Trapezoid move.

    An object moves from standstill at a known acceleration (1m/s/s) to a constant velocity (unknown) for a period (unknown) and then decelerates at a known deceleration (2m/s/s) to standstill. The total displacement is known (10m) and the total time is known (25s). How can I calculate the constant velocity and therefore the speed at constant velocity during the trapezoid move?

    Thanks!
    Rob
     
  2. jcsd
  3. Jul 2, 2013 #2

    tiny-tim

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    welcome to pf!

    Hi Rob! Welcome to PF! :wink:

    Call the times t1 t2 and t3

    then write out all the equations …

    show us what you get. :smile:
     
  4. Jul 2, 2013 #3
    Which equation?

    Thank you for the quick reply Tim!

    I don't really know where to start due to the amount of unknowns...

    If I begin with finding x for t1:

    x = v0t1 + ½at12
    x = 0*t1 + ½*1*t12
    x = ½*t12

    Is this in the right direction for t1?
     
  5. Jul 2, 2013 #4

    tiny-tim

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    yes :smile:

    now find the speed at t1, which will be the starting speed for the next part,

    and then write the equation for t2 :wink:
     
  6. Jul 2, 2013 #5
    Constant Acceleration - Final Velocity

    Final Velocity
    v2 = v1 + at
    v2 = at

    This gives me final velocity after the constant acceleration...but I am at a bit of a loss as to what to do here. Thanks for the help so far, you are very good!
     
  7. Jul 2, 2013 #6
    Constant Velocity Displacement

    Does this mean that the displacement during the constant velocity is calculated:
    x2 = (at)2
     
  8. Jul 2, 2013 #7

    tiny-tim

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    (been out all evening! :biggrin:)
    (what is ?)

    yes v1 = v2 = t1

    t2 is unknown

    now work backwards from the end, to find the speed at the start of the deceleration, the same way you worked forwards from the start to find the speed at the end of the acceleration :smile:
     
  9. Jul 4, 2013 #8
    Decel Part

    OK, so working backwards for t3:

    x = v0t3 + ½at32
    x = 0*t3 + ½*2*t32
    x = t32

    The equation is simplified to this only because of the value that I passed in, had the decel not been 2m/s/s, the equation would look different, obviously!
     
  10. Jul 4, 2013 #9

    tiny-tim

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    ok, now add all the x's, and all the t's :smile:
     
  11. Jul 4, 2013 #10
    OK, so adding...

    ACCEL: x = ½*t12
    C.VEL: x = (at2)2
    DECEL: x = t32

    x = ½*t12 + (at2)2 + t32
     
    Last edited: Jul 4, 2013
  12. Jul 5, 2013 #11

    tiny-tim

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    (just got up :zzz:)
    no, x = vt2

    (and then put the total displacement = 10, and write out total time = 20)
     
  13. Jul 5, 2013 #12
    Simplifying

    Thank you for the info so far!

    x = (½*t12) + (vt2) + (t32)
    10 = (½*t12) + (vt2) + (t32)

    But I only have total time, and each of these time periods are separate, so how would I "put it" total time=20 into this equation?
     
  14. Jul 5, 2013 #13

    tiny-tim

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    you have to learn to translate from english to maths! :biggrin:

    "total time" -> t1+ t2 + t3 !​

    (and now you'll need those two equations for v)
     
  15. Jul 5, 2013 #14
    Sure, well I know:

    t = t1 + t2 + t3
    x = 10

    So,

    10 = (½*t12) + (vt2) + (t32)

    But I am stuck here. If during static velocity (t2) v=x2/t2 then v = tx2. But this doesn't really help does it?
     
  16. Jul 5, 2013 #15

    tiny-tim

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    you have t1+ t2 + t3 = 25

    and (½*t12) + (vt2) + (t32) = 10

    but you don't know what v is, sooo …

    now you use your two equations for v, the one working forwards from the start, and the other working backwards from the finish! :smile:

    (i'm going out now for the rest of the day)
     
  17. Jul 5, 2013 #16
    Confused...

    But I only need to find v, which is not part of the acceleration period or deceleration period, so why do the first two equations have anything to do with v (the constant velocity period)?
     
  18. Jul 5, 2013 #17

    tiny-tim

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    you need v for the total displacement equation,

    and you find v (as a function of t1 or t3) from the acceleration and deceleration equations :smile:
     
  19. Jul 5, 2013 #18
    This equation?
    v=v0+at
     
  20. Jul 5, 2013 #19

    tiny-tim

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    yes, use it forwards from the start, and backwards from the finish
     
  21. Jul 6, 2013 #20
    Im really stuck...
    I can't seem to figure out how v=v0+at for the acceleration deceleration period will help me find v? Am i being stupid? I have been looking at this for hours and cant seem to slot the final pieces together.

    t1+ t2 + t3 = 25
    (½*t12) + (vt2) + (t32) = 10

    How does using v=v0+at help me from here?
     
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