Asked by linda
At one time, Maple Leaf Village (which no longer exists) had North
America’s largest Ferris wheel. The Ferris wheel had a diameter of
56 m, and one revolution took 2.5 min to complete. Riders could
see Niagara Falls if they were higher than 50 m above the ground.
Sketch three cycles of a graph that represents the height of a rider
above the ground, as a function of time, if the rider gets on at a height
of 0.5 m at t=0 min.Then determine the time intervals when the
rider could see Niagara Falls
America’s largest Ferris wheel. The Ferris wheel had a diameter of
56 m, and one revolution took 2.5 min to complete. Riders could
see Niagara Falls if they were higher than 50 m above the ground.
Sketch three cycles of a graph that represents the height of a rider
above the ground, as a function of time, if the rider gets on at a height
of 0.5 m at t=0 min.Then determine the time intervals when the
rider could see Niagara Falls
Answers
Answered by
Reiny
I will start with a sine curve
The period must be 2.5 min, .... 2.5 = 2π/k, k = 4π/5
so we start with sin 4πt/5
the amplitude must be 56
so ... 56 sin 4πt/5
but we want the min to be .0 so we have to raise it all up by 56.5
so far we have h = 56 sin 4πt/5 + 56.5
testing what I have so far
t = 0, h = 56.5
t = .625, h = 112.5 .. the max
t = 1.25 , h = 56.5
t = 1.875 , h = .5 .. the min
but you wanted the min to be when t = 0, so I will shift the curve .625 units to the right
y = 56 sin 4π/5(t - .625) + 56.5
when is h above 50 m ?
50 = 56 sin 4π/5(t - .625) + 56.5
56 sin 4π/5(t - .625) = -6.5
sin 4π/5(t - .625) = -.11607
4π/5(t - .625) = 3.2579 or 6.16685
t-.625 = 1.2963 or 2.4537
t = 1.9213 or 3.0787 minutes
so the time interval where the person would be 50 m is 3.0787 - 1.9213 = 1.157 minutes, and thus the time he would be above the 50 m is
2.5 - 1.157 = 1.343 minutes
The period must be 2.5 min, .... 2.5 = 2π/k, k = 4π/5
so we start with sin 4πt/5
the amplitude must be 56
so ... 56 sin 4πt/5
but we want the min to be .0 so we have to raise it all up by 56.5
so far we have h = 56 sin 4πt/5 + 56.5
testing what I have so far
t = 0, h = 56.5
t = .625, h = 112.5 .. the max
t = 1.25 , h = 56.5
t = 1.875 , h = .5 .. the min
but you wanted the min to be when t = 0, so I will shift the curve .625 units to the right
y = 56 sin 4π/5(t - .625) + 56.5
when is h above 50 m ?
50 = 56 sin 4π/5(t - .625) + 56.5
56 sin 4π/5(t - .625) = -6.5
sin 4π/5(t - .625) = -.11607
4π/5(t - .625) = 3.2579 or 6.16685
t-.625 = 1.2963 or 2.4537
t = 1.9213 or 3.0787 minutes
so the time interval where the person would be 50 m is 3.0787 - 1.9213 = 1.157 minutes, and thus the time he would be above the 50 m is
2.5 - 1.157 = 1.343 minutes
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