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A bit OT, math related

Discussion in 'Electronic Repair' started by JURB6006, Aug 10, 2003.

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  1. JURB6006

    JURB6006 Guest

    We try to keep our brains sharp around here. Those who know physics can see
    that this problem, which I plan to present to my friends' son who is attending
    college right now is just a little brain exercise, but I would also like to
    know if I came up with the right answer. Part of the calculation bears
    similarity to figuring out paralell resistance and series capacitance, so,
    while a bit OT it is not out in left field.:
    ________________
    An amusement park ride accelerates in a horizontal direction to a speed of 120
    mph in 4 seconds. What is the average G effective force experienced by the
    rider during that time ?

    What is the force vector, using that figure, as compared to a line
    perpendicular to the ground ?
    _________

    I came up with 2 2/3rd Gs. If necessary, I'll "show my work" in a followup.

    The second part I have not yet calculated, but my guess is around 52 degrees.

    To me, this is much better than discussing the last episode of "Friends" or
    some dumb crap like that.

    JURB
     
  2. Franc Zabkar

    Franc Zabkar Guest

    v = u + a*t

    where v = final velocity
    u = initial velocity
    a = acceleration
    t = duration

    Substitution gives:

    120 * 1610 / 3600 = 0 + a * 4

    Therefore a = 1610 / 30 /4 = 13.4 m/s2

    Now 1G = 9.8 m/s2, so a = 13.4/9.8 = 1.37G
    |\
    | \
    1G | \
    | \
    V \
    --------->
    1.37G

    The force vector's magnitude is sqrt( 1 + 1.37 * 1.37) = 1.70G.

    The angle relative to the vertical is arctan(1.37) = 54 deg

    If you perform the same calculation in imperial units (yuck), then you
    have:

    120 * 5280 / 3600 = a * 4

    So a = 5280 / 30 / 4 = 44 ft/sec2

    Now 1G = 32 ft/sec2 , so a = 44/32 = 11/8 = 1.375G


    - Franc Zabkar
     
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