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Motion under gravity

  1. Apr 7, 2004 #1
    I would really appreciate some help on this please:

    Distance in metres
    0.1
    0.2
    0.3
    0.4
    0.5
    0.6
    0.7
    0.8
    0.9
    1

    Time talen to travel distance (in seconds)

    0.72
    1.28
    1.67
    1.85
    2.08
    2.50
    2.83
    3.01
    3.13
    3.25

    What would you expect the accelartion to be and also what would you expect the velocity to be?

    Mass = 0.164 Kg

    Inital velocity = 0.

    The experiment was letting a cylinder roll down the ramp and the data is above.

    I just cant seem to calculate the acceleration and velocity could someone please help me thank you very much (much appreciated).
     
  2. jcsd
  3. Apr 7, 2004 #2
    You know that:
    [tex]x(t) = x_0 + v_0t + \frac{1}{2}at^2[/tex]
    You have three variables, x0, v0 and a. You already know that x0 and v0 are zero, so that leaves you with:
    [tex]x(t) = \frac{1}{2}at^2[/tex]
    To find the acceleration just use one of the measurements, for example 1m and 3.25s:
    [tex]1m = \frac{1}{2}a(3.25s)^2[/tex]
    As for calculating the velocity, that is just:
    [tex]v(t) = v_0 + at[/tex]
    The value of t varies depending on when you want to find the velocity.
     
  4. Apr 7, 2004 #3
    At the last point the final velocity, so what would the acceleration be, not to clude up on the equation(s).

    Thank you very much
     
  5. Apr 7, 2004 #4
    Anybody please tyty.
     
  6. Apr 7, 2004 #5
    I don't understand your questions.
     
  7. Apr 7, 2004 #6

    turin

    User Avatar
    Homework Helper

    In a discrete set of data, acceleration will require two time intervals, since it is a second order derivative.

    (note: position is zeroth order, and does not require an interval, just a point; velocity is first order, and therefore requires one time interval, that is, two points; every order of time derivative, requires that many time intervals to specify)

    You have 9 time intervals, so, you have 8 meaningful accelerations. I'll give you an example:

    (t,x)0 = (0.1,0.72)
    (t,x)1 = (0.2,1.28)
    (t,x)2 = (0.3,1.67)

    This gives you three positions (three null intervals), two velocities (2 x 1 interval), and one acceleration (1 x 2 intervals). The velocity is the space interval divided by the time interval from one point to the next:

    v0 = (x1 - x0)/(t1 - t0) = (1.28 - 0.72)/(0.2 - 0.1) = 5.6
    v1 = (x2 - x1)/(t2 - t1) = (1.67 - 1.28)/(0.3 - 0.1) = 3.9

    Notice that the velocity is changing, thus, there is an acceleration.

    a0 = (v1 - v0)/(t1 - t0) = (3.9 - 5.6)/(0.2 - 0.1) = -17

    There are also other ways to do it.


    Notice the similarity in:

    x'(t) = lim(as dt -> 0){x(t+dt)-x(t)}/{(t+dt)-t}

    The discrete treatment given above is a consequence of not taking the limit.
     
    Last edited: Apr 7, 2004
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