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
Why did the Ferris wheel go out of business? It couldn't handle the pressure of being North America's biggest "wheel"ationship! 🎡💔
Anyway, let's talk math. Since the Ferris wheel completes one revolution in 2.5 minutes, we can divide the time axis into intervals of 2.5 minutes. The height of the rider will vary as the wheel rotates.
Now, since one revolution is completed in 2.5 minutes, let's divide each 2.5-minute interval into four equal parts (2.5/4 = 0.625 minutes). This will allow us to sketch three cycles.
Using these intervals, we can plot the height of the rider above the ground on the y-axis, with time on the x-axis. Here's a sketch of three cycles:
| _
| _/
| _/
| _/
|/
0 2.5 5 0 2.5 5 0 2.5 5
min min min min min min min min min
Please note that this is just a simplified sketch, but it gives you an idea of the height of the rider at different times during the cycles.
Now, to determine when the rider could see Niagara Falls, we need to find the time intervals when the height is above 50 meters. From the sketch, we can see that the rider is above 50 meters during the second half of each cycle (approximately from 2.5 to 5 minutes). So, the rider could see Niagara Falls during those time intervals.
Hope that clears things up! Let me know if you have any more questions.
To sketch the graph representing the height of a rider above the ground as a function of time, we can start by determining the equation for the height.
The diameter of the Ferris wheel is given as 56 m, which means the radius is 28 m (diameter/2). The circumference of the wheel can be found using the formula:
Circumference = π * Diameter = π * 56m = 176 m
Since one revolution takes 2.5 minutes, the period of the function is 2.5 minutes or 150 seconds. The equation for the height of the rider above the ground, as a function of time, can be written as:
h(t) = A * sin((2π * t) / T)
Where:
h(t) = height above the ground at time t
A = amplitude, which is equal to the radius (28 m in this case)
t = time
T = period, which is 150 seconds in this case
Given that the rider gets on at a height of 0.5 m at t=0 min, we can modify the equation to incorporate this initial condition:
h(t) = A * sin((2π * t) / T) + 0.5
Now let's determine the intervals when the rider could see Niagara Falls. The rider can see Niagara Falls if the height above the ground (h(t)) is greater than 50 m.
So, we need to solve the inequality:
A * sin((2π * t) / T) + 0.5 > 50
Now, let's sketch the graph for the height of the rider above the ground as a function of time and identify the intervals when the rider could see Niagara Falls.
(Note: Since it's difficult to accurately sketch a graph using plain text, I can describe how the graph would look like)
Graph Sketch:
1. The graph is a sinusoidal wave that starts at 0.5 m (the initial height).
2. The maximum height is A + 0.5 m, which is 28 + 0.5 = 28.5 m.
3. The minimum height is -A + 0.5 m, which is -28 + 0.5 = -27.5 m.
4. The period is 150 seconds, and the graph completes three cycles within this period.
5. The values of t for the three complete cycles will be between 0 and 450 seconds.
Intervals when the rider could see Niagara Falls:
Based on the graph, the rider could see Niagara Falls when the height is above 50 m. From the graph sketch, this occurs during the crest portions of the cycle.
Therefore, the time intervals when the rider could see Niagara Falls are the intervals where the graph is above 50 m.
To sketch the graph that represents the height of a rider above the ground, as a function of time, we need to understand a few key concepts and equations.
1. The formula for the height of a point on a Ferris Wheel at any given time can be represented by the equation: h(t) = R + r * sin(w * t), where:
- h(t) is the height of the rider above the ground at time t
- R is the radius of the Ferris Wheel (half the diameter)
- r is the amplitude of the vertical motion (maximum displacement from the midline)
- w (omega) is the angular frequency, which can be calculated using the formula: w = 2 * π / T, where T is the period of one revolution
- t is the time
2. Given values:
- Radius (R) = 56 m / 2 = 28 m
- Amplitude (r) = maximum height - minimum height = 28 m - 0.5 m = 27.5 m
- Period (T) = 2.5 min = 2.5 * 60 seconds = 150 seconds
Now, let's determine the angular frequency (w):
w = 2 * π / T = 2 * π / 150 seconds
To sketch three cycles of the graph, we need to plot the height of the rider over the range of time that covers three periods or revolutions. Let's assume one period completes at t = 0 seconds, t = T seconds, and so on.
1. Calculate the angular frequency (w):
w = 2 * π / 150 seconds
2. Determine the time intervals for three cycles:
- Cycle 1: 0 seconds to T seconds
- Cycle 2: T seconds to 2T seconds
- Cycle 3: 2T seconds to 3T seconds
3. Calculate the height at various times within each cycle:
- For each cycle, plug in the time values into the equation: h(t) = R + r * sin(w * t)
- Calculate the height (h) at each time point.
4. Sketch the graph:
- On the y-axis, represent the height in meters.
- On the x-axis, represent the time in seconds.
- Using the calculated height values, plot points on the graph corresponding to each time point within each cycle.
- Connect the points smoothly to form a wave-like graph.
Once we have the graph, we can determine the time intervals when the rider could see Niagara Falls. The rider can see Niagara Falls when their height is above 50 m.
To find these time intervals:
- Observe the graph and determine the time points when the height is above 50 m.
- Identify the time intervals where the rider's height stays above 50 m.
Note: Due to the limitations of text-based format, I am unable to create and display visual content like graphs directly. However, I have provided a step-by-step explanation of how to sketch the graph. You can use a graphing tool or software to plug in the equation and values to get a visual representation.
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