A Ferris wheel is 25 meters in diameter and the bottom of the Ferris wheel is 8

Question

A Ferris wheel is 25 meters in diameter and the bottom of the Ferris wheel is 8 meters above the ground. You board the Ferris wheel at the 3 o'clock position a. The wheel completes one full revolution every 3.2 minutes. What is the angular speed (in radians per minute) that the Ferris wheel is rotating? pi/1.6 radians per minute Preview b. Write a formula that gives the angle measure (in radians) swept out from the 3 o'clock position, a, in terms of the number of minutes elapsed since you boarded the Ferris wheel, t. a(t) (T205) syntax ok a(t)=((pi/1.6//12.5) Preview syntax ok ( 125 c. Define a function fthat gives your height above the ground (in meters) in terms of the number of minutes elapsed since you boarded the Ferris wheel, t. ft)=sin(12.5(pi/1.6))t)+8 se Preview

Explanation / Answer

comparing with f(t)= (A*sin(B*t)) +C

a)

completes one revolution every 3.2 minutes

=>period =3.2 minutes

=>2/B =3.2

=>B=(2/3.2)

=>B=(/1.6) radians per minute

angular speed is (/1.6) radians per minute

b)

angle swept =angular speed * time elapsed

a(t)=(/1.6)*t

c)

for 3 O'clock position,there wont be any phase shift for sine function

amplitude A =25/2 =12.5 meters

mean value(height of center of wheel above the ground) ,c =8+(25/2) =20.5

so the function is f(t)=(12.5*sin((/1.6)*t)) +20.5

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