A Ferris wheel is 50 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 44 meters above the ground?

4 answers

A Ferris wheel is 50 meters in diameter

amplitude is 25, so

y = 25sin(kt)+k

The boarding platform is 1 meter from the ground, so the axle is 26 feet up

y = 26+25sin(kt)

The period is 6 minutes, so

2π/k = 6 ==> k = π/3

y = 26+25sin(π/3 t)

If we assume that at t=0 we are boarding, then that's the minimum value, so we ought to use a cosine function:

y = 26 - 25cos(π/3 t)

so now, just solve

26 - 25cos(π/3 t) = 44

and figure the interval when y is above that line.
radius = 25

height of center = 26

T is angle from vertical
height = 26 + 25 cos T
so
44 = 26 + 25 cos T
25 cos T = 18
cos T = .72
T = 44 degrees from vertical
so 88 degrees out of 360 are above 44 meters
(88/360)6 = 1.47 minutes
no
no