A Ferris wheel is 25 meters in diameter and boarded from a platform that is 4 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 21 meters above the ground?
2 answers
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the height is at the bottom at the start, so, since cos(x) is a maximum at x=0, you want something of the form
y = a - b*cos(kx)
25 meters in diameter, so
y = a - 25/2 cos(kx)
platform that is 4 meters above the ground
y = (4 + 25/2) - 25/2 cos(kx)
1 full revolution in 6 minutes, so 2π/k = 6
y = 33/2 - 25/2 cos(π/3 x)
now, when does the height = 21?
when x = 1.85 and 4.43
so it is above 21 for 4.43-1.85 = 2.58 minutes each cycle
y = a - b*cos(kx)
25 meters in diameter, so
y = a - 25/2 cos(kx)
platform that is 4 meters above the ground
y = (4 + 25/2) - 25/2 cos(kx)
1 full revolution in 6 minutes, so 2π/k = 6
y = 33/2 - 25/2 cos(π/3 x)
now, when does the height = 21?
when x = 1.85 and 4.43
so it is above 21 for 4.43-1.85 = 2.58 minutes each cycle