A Ferris wheel is 25 meters in diameter and boarded from a platform that is 1 meters above the ground.

Question

A Ferris wheel is 25 meters in diameter and boarded from a platform that is 1 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 6 minutes. How many minutes of the ride are spent higher than 17 meters above the ground?

Answer 1

We can model this using a negative cosine function since the "start" is at the bottom and it goes up and back down from there.

Where A is the amplitude, the period is 2π/B, the phase shift is C, and the midline is y = D

The center of the Ferris Wheel is 12.5 + 1 or 13.5 meters above the ground. From there, it goes up 12.5 m above this centerline and 12.5 m below this centerline or midline. So D = 13.5 and A = 12.5

We are told that the period is 6 minutes therefore B = 2π/6 = π/3

There is no phase shift.

So the modeling function is f(t) = -12.5cos(π/3t) + 13.5

To find the answer to the question, we can set f(t) = 17 to find the time that the wheel reaches the 17 foot mark.

17 = -12.5cos(π/3t) + 13.5

17 - 13.5 = -12.5cos(π/3t)

3.5/-12.5 = cos(π/3t)

-0.28 = cos(π/3t)

π/3t = cos^-1(-0.28)

π/3t = 1.855

t = 1.855(3/π) = 1.771 minutes

Since the Ferris Wheel reaches its peak at 3 minutes (half of the 6 minute period), the time it takes to go from the 17 meter mark to the top is 3 - 1.771 = 1.229 minutes. It will also take another 1.229 minutes to from the top back down to the 17 meter mark on the other side of the peak. So the total time above the 17 meter mark is 1.229•2 = 2.458 minutes