Question

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

Answer 1

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

Answer 2

Answer 3

Diameter of the wheel \(=56 \mathrm{~m}\)

radius of the wheel \(=28 \mathrm{~m}\)

Time periond, \(T=2.5 \mathrm{~min}\)

Angualer speed \(\omega=\frac{2 \pi}{T}=\frac{2 \pi}{2.5}=\frac{4 \pi}{5} \mathrm{rad} / \mathrm{min}\)

There is \(1 \mathrm{~m}\) gap from the bottom of the wheel to the ground

(a)

Let us take the \(y\) -axis along vertically up passing through the

center of the wheel and \(x\) - axis along horizontal on the

ground as shown in the image.

The centre of the wheel is at \((0,29)\)

Equation of the wheel is

\(x^{2}+(y-29)^{2}=28^{2}--(1)\)

at \(t=0\) rider is at the bottom, \(\quad\) i.ex \(=0\) and \(y=1 \mathrm{~m}\)

Let us take the parametric equations of the above circle

\(x=28 \sin (\omega t)--(1 a)\)

\(y-29=-28 \cos (\omega t)\)

\(y=29-28 \cos (\omega t)--(1 b)\)

\((1 a)\) and

(1b) satisfy the intial condition \(x=0\) and \(y=1 \mathrm{~m}\) at \(t=0\)

see the image for plot of \(y\) ves \(t\)

(b)

Niagar falls is ssen if \(y>50 \mathrm{~m}\)

\(50=29-28 \cos (\omega t)\)

\(\cos (\omega t)=\frac{29-50}{28}=-\frac{3}{4}\)

\(\omega t=2.4188\)

\(t=\frac{2.4188}{\frac{4 \pi}{5}}=0.96 \mathrm{~min}\)

Time period \(=T=2.5 \mathrm{~min}\)

\(\cos (\omega t)\) will get the same value after \(2.5 \mathrm{~min}\)

he can view the Niagar falls from \(0.96\) to \(T-0.96\)

Viewing duration in 1st rotation is \(0.96\) min \(<t<1.54\)min

In the second cycle \(0.96+2.5<t<1.54+2,5\)

Viewing duration in2nd rotation is \(3.46\) min \(<t<4.04\)min

(c)

Avrage rate of change of height during 1st 2 min

$$ y_{a v}=\frac{1}{2} \int_{0}^{2}(29-28 \cos (\omega t)) d t=\frac{1}{2} \times 29 \times 2-14\left[\frac{\sin (\omega t)}{\omega}\right]_{0}^{2}=29-\frac{14}{\frac{4 \pi}{5}} \sin \left(2 \times \frac{4 \pi}{5}\right) $$

\(y_{a v}=34.30 \mathrm{~m}\)