pre-calc

posted by ikora

A ferris wheel is 28 meters in diameter and makes one revolution every 10 minutes. For how many minutes of any revolution will your seat be above 21 meters?

  1. Reiny

    fit your data into a sine curve
    period of sine curve = 10 minutes
    2π/k = 10
    k = 2π/10 = π/5
    so start with
    y = 14 sin π/5 t + 14

    testing:
    t = 0: y = 14sin0 + 14 = 14
    t = 2.5 , y = 28
    t = 5 , y = 14
    t = 7.5 , y = 0
    t = 10 , y = 14 , looks good

    solve for
    14 sin πt/5 + 14 = 21
    sin πt/5 = 1/2
    πt/5 = π/6 or πt/5 = 5π/6
    t = 5/6 or t = 25/6 minutes

    so time over 21 metres = 25/6 - 5/6 = 20/6 minutes
    or 2 minutes and 20 seconds or 200 seconds

  2. Steve

    I'm sure the answer is unchanged, but I'd have set up the equation as

    y = 14(1 - cos pi/5 t)

    That way, y(0) = 0, when the person gets on the wheel, rising to y(5) = 28 when it's 5 minutes later, at the top. And so on.

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