pre-calc

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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?

  • pre-calc -

    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

  • pre-calc -

    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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