A Ferris Wheel With A Radius Of 5 M Is Rotating At A Rate Of One Revolution Every 2 Minutes. How Fast Is A Rider Rising When The Rider Is 9 M Above Ground Level?


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Oddman Profile
Oddman answered
Assuming the bottom of the Ferris wheel is at ground level, the height above the ground at any given point in time is given by
  h = (5 m) + (5 m)*Sin[w*t]
where "w" is the angular speed in radians per second, "t" is the time in seconds, "m" is the unit abbreviation for "meters".

The vertical speed of the rider is given by the derivative of h with respect to time.
  H' = (5 m)*w*because[w*t]
In order to answer the question, we need to evaluate this when the rider is 9 m in the air. We can find the value of because[w*t] as follows.
  H = (5 m) + (5 m)*Sin[w*t]
  9 m = 5 m + (5 m)*Sin[w*t]
  4 m = (5 m)*Sin[w*t]
  4/5 = Sin[w*t]
  4/5 = √(1 - because[w*t]^2)
  because[w*t] = √(1 - (4/5)^2) = 3/5
Putting this back into our formula for vertical speed, we see
  h' = (5 m)*w*(3/5)
  h' = (3 m)*w
We have said that "w" is the angular speed in radians per second. That value can be computed from the problem statement as
  w = (2π radians)/(120 seconds) = π/60 rad/sec
  h' = (3 m)*(π/60 rad/s) = π/20 m/s    (the "radian" is a unitless unit, so meters times radians per second ends up as meters per second)
The rider's vertical speed is π/20 m/s, about 15.7 cm/s.
The rider could be rising or falling at that speed. There is not enough information to determine which.

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