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Velocity/Accelration/Displacment all in one question.

  1. Sep 19, 2007 #1
    1. The problem statement, all variables and given/known data

    A student drives a moped along a straight road as described by the velocity versus time graph in Figure P2.14. The divisions along the horizontal axis represent 1.5 s and the divisions along the vertical axis represent 3.5 m/s.
    Sketch this graph in the middle of a sheet of graph paper. (Do this on paper. Your instructor may ask you to turn in your work.)

    Figure P2.14
    <img src=http://a425.ac-images.myspacecdn.com/images01/103/l_3fca2200153cb8eb21ea540316125000.gif>
    (a) Directly above your graph, sketch a graph of the position versus time, aligning the time coordinates of the two graphs. (Do this on paper. Your instructor may ask you to turn in your work.)

    (b) Sketch a graph of the acceleration versus time directly below the vx-t graph, again aligning the time coordinates. (Do this on paper. Your instructor may ask you to turn in your work.)
    On each graph, show the numerical values of x and ax for all points of inflection.

    (c) What is the acceleration at t = 9.0 s?
    m/s2
    (d) Find the position (relative to the starting point) at t = 9.0 s.

    m
    (e) What is the moped's final position at t = 13.5 s?
    m


    2. Relevant equations
    all of the kinematics and equations dealing with velocity,time,positon,accerlation
    will help out


    3. The attempt at a solution

    let see i figured the part looking for accelration out by taking the a tanget of the line near 9s using equation for slope form. vf-vi/tf-ti= 0-14m/s/12s-9s=-4.67m/s^2

    now the rest im totaly lost ive tried to intergrate the function of the graph but im not sure how to come up with the equation of the function. can anyone help me out?
     
  2. jcsd
  3. Sep 19, 2007 #2
    heres the graph[​IMG]
     
  4. Sep 19, 2007 #3

    learningphysics

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    The acceleration is just the slope of the lines...

    The change in position is the area under the v-t graph...
     
  5. Sep 19, 2007 #4
    ok the area under the curve, now this where im getting stuck. do i use intergration?
     
  6. Sep 19, 2007 #5

    learningphysics

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    You don't have to... suppose I'm looking at the first segment of the curve (the positive slope part)... so 0 up to some time t... what is the area under the graph ?
     
  7. Sep 19, 2007 #6
    lets see the area under the first part. looks like two triangles. so area of the first triangle is bXhX1/2 =1/2 4.5X(14)=31.5 and the 2nd 1/2X4.5X(14)=31.5 so 31.5+31.5= 63
     
    Last edited: Sep 19, 2007
  8. Sep 19, 2007 #7

    learningphysics

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    Why are you getting 2 triangles? I just see one...

    suppose you aren't going all the way to t = 4.5. you're calculating the area from t = 0... to some t = tf... what is the area of the triangle?

    Actually maybe it is better to take the integral... so for the first segment... what is the equation for v(t) ? ie: for t = 0 to t = 4.5...
     
  9. Sep 19, 2007 #8
    equation for v(t)= absx
     
  10. Sep 19, 2007 #9

    learningphysics

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    what is abs?

    You want an equation for v(t) in terms of time...
     
  11. Sep 19, 2007 #10
    v(t)=│t│
    t=0
    v(t)=│0│=0
    t=4.5
    v(t)=│4.5│=4.5
     
    Last edited: Sep 19, 2007
  12. Sep 19, 2007 #11

    learningphysics

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    Hmmm... how do you get that? I see the slope as 14/4.5 = 3.111
     
  13. Sep 19, 2007 #12

    learningphysics

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    Isn't v = 14 at t = 4.5?

    You don't need to use absolute value... since time is positive anyway.
     
  14. Sep 19, 2007 #13
    yea v= 14 at t= 4.5 , my mistake on that.
     
    Last edited: Sep 19, 2007
  15. Sep 19, 2007 #14
    so if i were to intergrate that. it be ∫v(t) dt
     
    Last edited: Sep 19, 2007
  16. Sep 19, 2007 #15

    learningphysics

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    dt not dx.

    But that equation for v(t) only holds for t = 0 to t = 4.5... What is the equation of each segment of v(t) ?

    When you get that, then you can get the integrals...
     
  17. Sep 19, 2007 #16
    t=0 to t= 4.5
    v(t)=t
    t=4.5 to t=9
    v(t)=14
    t=9 to t= 15
    v(t)=(-t-12)
     
  18. Sep 19, 2007 #17

    learningphysics

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    But the slope of the line is 14/4.5 = 3.11 so this can't be right.

    Yes, good.

    you calculated the slope of this part in your original post... it's not -1
     
  19. Sep 19, 2007 #18
    ok so first one must be v(t)=3.11t
    and the last is v(t)=(-4.67t-12)
     
  20. Sep 19, 2007 #19

    learningphysics

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    yes.

    The slope is right, but the y-intercept is wrong.
     
  21. Sep 19, 2007 #20
    hmm well i figured it was a reflection of the f(t) and it shifted to the right to 12. but i guess i didnt do it right. should i extend the line to the y axis to get the intercept?
     
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