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Homework Help: Finding equation of the position

  1. May 17, 2009 #1
    1. The problem statement, all variables and given/known data
    hi everyone ,, this is my last question before exam :

    1) The figure shows the velocity versus time graph. If a particle starts motion at t=0.0
    from x0 = 10.0 m:
    a) Write the equation describing the position of the particle as a function of time t.
    b) What are the positions of the particle at t= 2 s and t = 4 s.?
    c) What is the position of the particle at the instant when it momentarily stops?
    d) Find the distance and displacement of the particle between t = 2 s and t = 5 s.

    2. Relevant equations

    3. The attempt at a solution
    I know that a=0 ,, and at t=0 > x=10m and at the same time Vo=12 m/s ,, and the velocity is decreasing so I made this equation :x=10-12t but the problem at t=0 Vo=0 and when I apply x(3) I don't get 18m which it's what i have to get by the graph .. can someone tell me what to do to get the equation ..?

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  3. May 17, 2009 #2


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    Consider how you find displacement from a velocity time graph.

    If V = dX/dt, then isn't

    X = ∫ V dt ?

    And if you integrate V on your graph, what is the easiest way to do that?

    Can you think of something that rhymes with area?
  4. May 17, 2009 #3

    I didn't understand you quite well ,, I tried by doing X = ∫ V dt from x(0,3) x=18 and i get v=6 m/s ,, and from x(0,5) v=2 m/s but the problem is at x(0,0) I should get x=10 and I dont get it,, V changes ...
  5. May 17, 2009 #4


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    Xo is the constant of integration isn't it?
  6. May 17, 2009 #5
    yea Xo is constant ,, should I include it in the integral or X = Xo + ∫ V dt ??
    and BTW when i integrate ∫ V dt ,, I'll get Vt right ??
  7. May 17, 2009 #6


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    V(t) = 12 - 4t isn't it?

    So ∫ V(t) dt = ∫ (12 - 4t) dt = X(t)

    That means that X(t) = 12t - 1/2*4*t2 + C

    Where C = Xo
  8. May 17, 2009 #7
    aha ,, so that what you meant X = ∫ V ,, I didn't know that first I should get V(t) equation then integrate .. thanks very much :D
  9. May 17, 2009 #8


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    You can also solve through visual inspection, by simply calculating the area under the curve. Each square represents 4 m.
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