Question

A Ferris wheel is 15 meters in diameter and boarded from a platform that is 1 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 2 minutes. How many minutes of the ride are spent higher than 15 meters above the ground?

Answer 1

Answer:

The number of minutes of the ride that are spent higher than 15 meters above the ground is 18 minutes.

Step-by-step explanation:

We will use the sin function for the height of the Ferris wheel.

$y=A\ sin(B(x+C))+D$

A = amplitude

C = phase shift

D = Vertical shift

2π/B = period

From the provided information:

A = 15/2 = 7.5 m

$Period=2\\\\\frac{2\pi}{B}=2\\\\B=\pi$

Compute the Vertical shift as follows:

D = A + Distance of wheel from ground

   = 7.5 + 1

   = 8.5

The equation of height is:

$h(t)=7.5\cdot sin(\pi (t+C))+8.5$

Now at t = 0 the height is, h (t) = 1 m.

Compute the value of C as follows:

$h(t)=7.5\cdot sin(\pi (t+C))+8.5$

$1=7.5\cdot sin(\pi (0+C))+8.5\\\\-7.5=7.5\cdot sin\ \pi C\\\\sin\ \pi C=-1\\\\\pi C=sin^{-1}(-1)\\\\\pi C=\frac{3\pi}{2}\\\\C=\frac{3}{2}$

So, the complete equation of height is:

$h(t)=7.5\cdot sin(\pi (t+\frac{3}{2}))+8.5$

Compute the number of minutes of the ride that are spent higher than 15 meters above the ground as follows:

h (t) ≥ 15

$h(t)=15\\\\7.5\cdot sin(\pi (t+\frac{3}{2}))+8.5=15\\\\7.5\cdot sin(\pi (t+\frac{3}{2}))=6.5\\\\sin(\pi (t+\frac{3}{2}))=\frac{6.5}{7.5}\\\\(\pi (t+\frac{3}{2}))=sin^{-1}[\frac{13}{15}]\\\\\pi (t+\frac{3}{2})=60.074\\\\t+\frac{3}{2}=19.122\\\\t=17.622\\\\t\approx 18$

Thus, the number of minutes of the ride that are spent higher than 15 meters above the ground is 18 minutes.