Ferris Wheel

HAHA.. umm.. while doing this post I thought no one did our group assignment so I started to do one. Except when i finished Bryan posted his up while I was working on mine so now theres two.

This is Miller Representing group 3 I believe. There has been some communication error within the group. I lost your hotmails. anyways...

Ferris Wheel Function

As you are waiting in line to ride the Ferris wheel at the state fair, you begin to be hypnotized by the repetitive motion of the seats. The seat is at the bottom; it's halfway up, it's at the top; it's halfway down; it's at the bottom; it's halfway up; it's at the top...

You decide to time the wheel and observe that the ferris wheel makes 4 complete revolutions every minute. You also assume that the diameter of the wheel is 100 feet because a sign beside the wheels says "Climb 100 feet into the sky on our Ferris wheel". However, you note that when you get into a seat, the seat is about 3 feet off the ground, so technically the maximum height of a person would reach would be...?

a) Write two equations, one sine and one cosine, that would represent the distance above the ground as a function of time. Let t = 0 when the seat is at the lowest point on the wheel.

b) How high will a person be 10 seconds after they begin riding the ferris wheel?

c) How long after a person gets on the ride will they actually be 100 feet in the air?

d) How long, during each revolution, is a person more than 60 feet in the air?

This is what the graph looks like:




A) To solve the two equations first we need the ABCD.
Sin-A=50 Cos-A=-50
B=(2π/15) B=(2π/15)
C=15/4 C=0
D=53 D=53

Sine : d(t)= 50sin((2π/15)(t-(15/4))+53
Cos : d(t)= -50cos(2π/15)(t)+53

B: To find out how high the person would be after 10 sec on the ferris wheel we must use one of our equations and plug in 10 for T. I will use the cosine equation.

t=10
d(t)= -50cos(2π/15)(t)+53
d(10)= -50cos(2π/15)(10)+53
= 78feet

C: To find out how high a person will be at 100 feet in the air we have to y{d(t)} equal 100. Again i will use the Cosine equation.

d(t)=100
d(t)= -50cos(2π/15)(t)+53
100=-50cos(2π/15)(t)+53

to make things easier lets let (2π/15)(t)= θ
100 =-50cosθ+53
cos=47/50
θ=2.7934,3.4898

(2π/15)(t)= 2.7934
t=2.7934/(2π/15)
t=6.6688s

D) To find out how long a person is more than 60 feet in the air during one revolution we let d(t)=60

d(t)=60
d(t)= -50cos(2π/15)(t)+53
60= -50cos(2π/15)(t)+53

again we let (2π/15)(t)=θ

60=-50cosθ+53
7=-50cosθ
cosθ=(7/(-50))
θ = 1.71126 , 4.5719

(2π/15)(t)=1.71126
t=1.71126(15/2π)
t=4.0853s

(2π/15)(t)=4.5719
t=4.5719(15/2π)
t=10.9146

=10.9146-4.0853s=6.8293s

Group members msg me with your names so i can add them to this post.. Thanks