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Distance time graph

  1. Oct 23, 2012 #1
    1. The problem statement, all variables and given/known data
    A plank is raised at one end in order to form an inclined plane. A light freely running trolley travels from rest down the plane. The times taken to travel from different points to the lower end of the plank are measured and the following results obtained
    Distance (cm) 0 50 100 150 225 325

    Time (s) 0 1.4 2.0 2.5 3.0 3.6

    Draw a graph to show that the acceleration is uniform and calculate its value. Use this value to calculate the velocity which the trolley would acquire if it travelled 400 cm from rest along the same inclined plane.



    2. Relevant equations



    3. The attempt at a solution

    I have actaully drawn the graph but the answer I got is not ryming with the answer provided for the question.

    The scale I used was:

    2cm for 25 units on distance axis (vertical or y axis).

    2cm for 0.5 unit on time axis (horizontal axis or x axis).

    The slope I got was 109.1cm. Another thing that suprised me was that the straight line I got when I joined the points did not pass through the origin; it went a cut the horizontal axis at exctaly 1 sec i.e intercept =1 second . That affected all my calculation badly. What could have been done, am confused.
     
  2. jcsd
  3. Oct 23, 2012 #2
    What is the shape of the graph? Is it linear or parabolic?
     
  4. Oct 23, 2012 #3
    It is a linear graph.
     
  5. Oct 23, 2012 #4

    haruspex

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    Doesn't look linear to me. Note that it does not say "draw a distance-time graph". Could it mean some other graph?
     
  6. Oct 23, 2012 #5
    We all know that the slope of a velocity time graph gives the acceleration. What if it is a velocity time graph? Am saying that is distance time graph because of what I saw on the table. But the work they did suggest that it could be displacement time graph or velocity time graph.

    Am saying this because the values of the distance s on the table was plotted against the square of the time t^2 (s^2) to obtain a straight line passing through the origin.

    The slope was then given as (1/2)a.

    And the acceleration a was obtained from (1/2)a = change in y axis/change in horizontal axis.

    I don't just understand the principle guiding that computation.
     
  7. Oct 23, 2012 #6

    haruspex

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    Ok, I'm not entirely clear who did what, but we agree that what they're after is a distance v. time-squared graph.
    With constant acceleration from a stationary start (time zero), what equation relates distance to time? If you plot distance against time-squared for that you should get a straight line, right? Would you get a straight line if the acceleration were not constant?
     
  8. Oct 24, 2012 #7
    [/QUOTE]Would you get a straight line if the acceleration were not constant?[/QUOTE]
    No, I will only get straight line if the acceleration is costant.

    This actually what they did:
    s = ut + at^2/2

    But initial velocity, u = 0

    therefore s = at^2/2

    slope = a/2
    since graph s t^2 is a striaght line, it means that acceleration is uniform.

    I think they are making use of the fact that for an object to have constant acceleration, the distance of it travel must be directly proportional to the square of the time of travel.

    But am yet to figure out how slope = a/2
     
  9. Oct 24, 2012 #8

    haruspex

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    You have s = at2/2, and you plot y = s against x = t2. So the graph should look like y = ax/2. What slope would such a graph have?
     
  10. Oct 26, 2012 #9
    If s the distance = ut+(1/2)at^2
    ut = 0
    s = at2/2
    we are plotting s against time t2
    that means the slope
    = at2/2/t2

    = a/2
    I think we are making some sense here.
     
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