A ferris wheel at an amusement park has a diameter of 18metres, and travels at a rate of 5 revolutions every 2 minutes.

Question

Extra information: At the bottom of the ride, the passenger is 2 metres above ground.

 

A) Determine a function the represents height, h, above the ground at time, t, if h=20 and t=0.

 

B) Determine when the passenger is 15metres above the ground during the first two revolutions.  

 

Grade 11 weird problem that is supposed to be study question for test. There is no answer, so I don't know if I'm doing it right. 

Answer 1

the basic equation is called a "sinusoid", and is either a sine or cosine function, depending on the starting point.

A sine wave starts at the mean value at t = 0, a cosine wave starts at the peak value.

 

We're told that we start at a height of 20, which is the top, so it's a cosine function

 

The position of the wheel can be stated in radians ( one revolution = 360 degrees = 2*pi radians )

 

we're told that the wheel turns 5 revolutions in 2 minutes.  5 revolutions is 10*pi radians, so we have the rate = 5*pi radians / minute

 

(angular rotation speed is typically represented as ω (omega))

 

SO--the height of a point on the wheel is 11 + 9*cos ωt, where ω = 5*pi

h = 11 + 9cos (5*pi*t)

 

(the cosine wave is centered at the midpoint of the wheel, which is at a height of 11 meters)

 

To find the time when h = 15, we plug into the formula and rearrange:

 

15 = 11 + 9cos (5*pi*t)

4/9 = cos (5*pi*t)

therefore:

 

5*pi*t = cos-1 4/9 = pi/2

t = pi/2 * 1/5pi = 2/5

t = 0.4 minutes

 

To find more values of t, recognize that h has the same value every 1/2 revolution.  5 revolutions takes 2 minutes, so 1/2 revolution is 0.2 minutes