A ferris wheel is 25 meters in diameter and boarded from a platform that is 2 meters above the ground

Question

A ferris wheel is 25 meters in diameter and boarded from a platform that is 2 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. The function h = f(t) gives your height in meters above the ground t minutes after the wheel begins to turn. Write an equation for h = f(t).

Answer 1

h(t)=12.5*cos[(π/3)*(t-3)]+14.5

t=0, h(t)=2

t=3 h(t)=27

t=6 h(t)=2

π/3 is frequency (=2π/6=one cycle of circle/period, so 2π cycle/6minutes)

12.5 is amplitude of the height function displacemnt from the midline

14.5 is midline (center of wheel)

3 minutes is the offset (since the point starts at the bottom=0 minutes and takes 3 minutes to reach top, top is same as cos being 1)

and after 3 minutes it is at top (move graph positve 3 minutes, means minus in t-3)

Answer 2

radius = 12.5

Imagine the wheel moves from six o'clock position to 8 o'clock. The height as a function of the angle between..., h = f(theta) =

[r - r*cos(theta)] + 2 ... you really have to draw this to see... take away cos(theta)-part from r and the 2 is from the base height.

theta as a function of time: (2*pi)/6*t ==> 1 revolution is 2*pi per 6 minutes. this is the rate at which the angle is changing with time.

now sub in pi/3 * t into f(theta) to get f(t) ==> [r - r*cos(pi/3*t)] + 2

Answer 3

You want a function of the form:

h(t) = A cos(Bt - C) + D

where:

1. A = amplitude = radius of the ferris wheel = -12.5. We want the minus sign because the cosine starts at the 12 o'clock position which would mean we load at the top. We want to load at the bottom or 6 o'clock starting position. The minus sign flips the graph across the x-axis so it starts at the 6 o'clock position at t = 0
2. B = Angular Frequency = 2π/period = 2π/1 = 2π
3. C = Phase shift = 0
4. D = Midline height = amplitude + 2 meters = 14.5 m

Plug the numbers into the general formula above.