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Question 1133000: At the fair the Ferris Wheel has a diameter of 32m and its centre is 18m above the ground. The wheel completes one revolution every 30 seconds.
a) determine the domain and range of the function.
b) Find the equation for the height of the person in terms of time.
c) How high would the person be after being on the ride for 3.5 minutes?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! your equation will be y = 18 - 16 * sin(12 * (x - 22.5)), as best i can determine.
the period of one complete cycle of the ferris wheel is 30 degrees, where each degree represents one second of travel.
at 0 seconds, the person is 2 meters above the ground.
at 7.5 seconds, the person is 18 meters above the ground.
at 15 seconds, the person is 34 meters above the ground.
at 22.5 seconds, the person is again at 18 meters above the ground.
at 30 seconds, the person is again at 2 meters above the ground.
this repeats every 30 seconds.
you can graph the equation as shown below for one complete cycle.
if the person was on the ferris wheel for 3.5 minutes, that means that the person was on the feris wheel for 3*60 + 30 = 210 seconds.
when x = 210, the formula of y = 18 - 16 * sin(12 * (x - 22.5)) becomes equal to 2 meters above the ground.
this stand to reason because the person on the ferris wheel is 2 feet above the ground at 0 seconds and at every 30 seconds after that.
the can be shown graphically below:
a sine wave was used to model this.
the basic formula of the sine wave is y = a * sin(b * (x - c)) + d
a is the amplitude.
b is the frequency.
c is the horizontal displacement.
d is the vertical displacement.
frequency is equal to 360 divided by period.
the period of the sine wave needed to be 30 degrees, which allowed each degree to represent one second of time.
frequency = 360 / 30 = 12, therefore the frequency of the sine wave was 12.
the formula became y = a * sin(12 * (x - c) + d.
the diameter of the ferris wheel was 32 meters, therefore the radius of the ferris wheel became 32 / 2 = 16 meters.
the radius of the ferris wheel was the amplitude of the sine wave.
the formula became y = 16 * sin(12 * (x - c) + d
the vertical displacement for the center line of the sine wave was 0.
the formula became y = 16 * sin(12 * (x - c))
the center of the ferris wheel was 18 meters above the ground level.
the formula of y = 16 * sin(12 * (x -c) gave the distance between the point on the circumference of the ferris wheel and the center line of the ferris wheel which was 18 meters above the ground.
therefore, the distance from the person on the ferris wheel to the ground became 18 minus the distance from the person on the ferris wheel to the center line of the ferris wheel which became 18 - 16 * sin(12 * (x - c))
without any horizontal displacement, the person would be 18 meters above the ground when x = 0.
it needed to be 2 feet abovee the ground when x = 0.
without any horizontal displacement, the person was 2 feet above the ground when x equaled 7.5 or when x equaled -22.5
this required the graph to be shifted to the 7.5 units or to the right 22.5 units.
i chose shifting to the right 22.5 units.
the formula became y = 18 - 16 * sin(12 * (x - 22.5)).
now, when x = 0, the person was 2 feet above the ground, which is where the person would be when the person got on or off the ferris wheel.
that became the final equation.
a pictorial representation of one calculation is shown below.
the circled number 1 is the distance between the person and the center line of the ferris wheel.
the circled number 2 is 18 minus the distance between the person and the center line of the ferris wheel.
that was the distance from the person to the ground.
at 3.25 seconds on the ferris wheel, the person was calculated to be 5.565664617 meters above the ground.
this was confirmed graphically by the picture showon below:
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