6. The Ferris wheel at a carnival has a diameter of 16 m and descends to 1 m above the ground at its lowest point.
Assume that a rider enters a car at this point and rides the wheel for one revolution that takes 2 minutes.
a) Model the rider’s height above the ground as a function of time in seconds in an equation. [3A]
b) Sketch the graph and label points [3C]
c) Use the equation to predict the height at 100s. [1A]
d) Use the equation to determine between what times (in the first cycle) the rider is above 12 m. [2A]
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amplitude = 8
for period: 2π/k = 2, k=π
starting equation: height = 8sin πt
the min of that is -8,we want it to be 1, so
height = 8sin πt + 9
the min of that occurs at t = 1.5 min, so we need a horizontal shift
height = 8sin π(t + 1.5) + 9 , where t is in minutes
check
when t = 0, h = 1
when t = .5, h = 9
when t = 1 , h = 17
when t = 1.5 , h= 9
when t = 2, h = 1 <---- back to the start
my equation is correct
so at 100 seconds, t = 100/60 = 5/3
h = 8sin π(5/3t + 1.5) + 9 = 5 m
we want to be > 12
8sin π(5/3t + 1.5) + 9 > 12
consider 8sin π(t + 1.5) + 9 = 12
8sin π(t + 1.5) = 3
sin π(t + 1.5) = 3/8 , take arcsin
π(t + 1.5) = 3/8 = .384396.... or π(t + 1.5) = π-.384396...
solving for t, I got -1.3776, but our period is 2min, so t = .622 min
or π(t + 1.5) = π-.384396... = 2.757
...
t = -.622 or 1 period over adding 2 minutes, t = 1.378 min
confirmed this by graphing the equation an y = 12 on DESMOS website