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I have a trig question regarding a ferris wheel, radius 35m, axles 37m, clockwise direction, rotates twice every 10 mins. I have calculated the following function:
h = -35cos(2pi*X/5)+37
I have calculated at what height I would be after 3 minutes by substituting X=3 in the above equation. However, I need to calculate at what times I would be 30m off the ground. Can you please tell me if I am on the right track and how to calculate the time?
Many thanks

  • Maths - ,

    I agree with your equation, so let' set it equal to 30

    30 = -35cos(2pi*X/5)+37
    -7 = -35cos(2pi*X/5)
    .2 = cos(2pi*X/5)
    take the inverse cosine to get
    2pi*X/5 = 1.369438
    x = 1.09 minutes

    check it by repeating the same steps you did finding the height at 3 minutes.

  • Maths - ,

    Many thanks, I understand that now.
    I have another question I am stuck on also. How do I get an equation for the height of the car on the ferris wheel over a 10 minute period if we start the analysis one minute into the ride.
    Can I still use the original formula
    h = -35cos(2pi*X/5)+37
    I don't understand what part changes to make it one minute into the ride. Appreciate any help.

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