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Uniform Accelerated Motion - Ticker Tape Experiment

  1. Aug 18, 2007 #1
    Ok, I'm currently trying to do an experiment of a cart coasting down an incline plane using ticker tape. The ticker tape is set up so that the space between each dot represents 0.02 s. That being said, between every 5th dot would represent 0.1 second.

    I know that the point of this experiment is to prove that the cart has Uniform Accelerated Motion, however, using the data that I have obtained, my Velocity-Time graphs say otherwise. Below is the data I have:

    Position (cm) | Time (s)
    0.000 cm 0.000 s
    0.5 cm 0.100 s
    1.5 cm 0.200 s
    3.3 cm 0.300 s
    5.7 cm 0.400 s
    9.2 cm 0.500 s
    13.6 cm 0.600 s

    From that information I am able to find the Interval Displacement
    d=p2-p1 so:

    First = 0.5 cm - 0.0 cm = 0.5 cm
    Second = 1.5 cm - 0.5 cm = 1.0 cm
    Third = 3.3 cm - 1.5 cm = 1.8 cm
    Fourth = 5.7 cm - 3.3 cm = 2.4 cm
    Fifth = 9.2 cm - 5.7 cm = 3.5 cm
    Sixth = 13.6 cm - 9.2 cm = 4.4 cm

    Using the Interval Displacement information, I can then find the Average Velocities: V=change of D/ change of T

    First = 0.5cm / 0.1 s = 5 cm/s
    Second = 1.0 cm / 0.1 s = 10 cm/s
    Third = 1.8 cm / 0.1 s = 18 cm/s
    Fourth = 2.4 cm / 0.1 s = 24 cm/s
    Fifth = 3.5 cm / 0.1 s = 35 cm/s
    Sixth = 4.4 cm / 0.1 s = 44 cm/s

    Knowing that the Vi at the midpoints of the intervals is equal to the Average velocity, they would be representative of the same answers.

    Mid-point intervals:
    First = 0.05 s
    Second = 0.15 s
    Third = 0.25 s
    Fourth = 0.35 s
    Fifth = 0.45 s
    Sixth = 0.55 s

    Now after calculating all this information and plotting into a Velocity Time Graph both by hand, and also from using Data Studio and Graphical Analysis, I come up with the same answers. They are not representative of an object with Uniform Acceleration. If they were, then the V-T graph would be a straight line moving upwards to the right. My line is more or less upwards to the right but not at a constant velocity, they change at every second.

    Does anyone know what I might have done wrong with this experiment?? I'm not looking for anyone to give me the correct answers, as I would like to find them on my own, but any suggestions as to where I messed up, miscalculated, used wrong information etc.. PLEASE HELP,
     
  2. jcsd
  3. Aug 18, 2007 #2

    mgb_phys

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    Science Advisor
    Homework Helper

    You have calculated the velocity correctly but then you need to take the differences between the velocity at each tinterval,
    First-second = 5-10 = 5cm/s
    Second-third = 10-18 = 8cm/s
    Third-forth = 18 - 24 = 6cm/s
    Fourth-fifth = 24-35 = 11cm/s
    Fifth-sixth = 35 - 44 = 9cm/s
    So accelaration is reasonably constant, at slow speeds you have a bit of friction.
     
  4. Sep 12, 2010 #3
    You haven't done any calculations wrong its just that you must have had an incline that was too steep to acheive uniform acceleration. The gradient must be slight, and when the trolly is on the slope it must only be at the point of tipping. (so its not moving) just enough of a gradient to counteract the friction.

    hope this helped.
     
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