Question

1. the maximum height of a ferris wheel is 42m and the wheel takes 1.5 mins to make one full revolution. passengers aboard the ferris wheel at 6m above the ground at the bottom of its rotation. t-5 marks a) sketch two cycles of the ferris wheels height in metres with respect to time in seconds. (2 marks) b) write a cosine equation to represent the position of a passenger at any time t, in seconds. (2 marks) c) how high is the passenger after 35 s? round your answer to two decimal places. (1 mark)

Explanation:

Step1: Determine the amplitude, vertical shift, and period

• Amplitude \(A=\frac{42 - 6}{2}=18\)
• Vertical shift \(D=\frac{42 + 6}{2}=24\)
• Period \(T = 1.5\times60=90\) seconds, so \(B=\frac{2\pi}{T}=\frac{2\pi}{90}=\frac{\pi}{45}\)
• Since the passenger starts at the minimum height, the equation is \(h(t)=-A\cos(Bt)+D\)

Step2: Write the cosine equation

Substitute \(A = 18\), \(B=\frac{\pi}{45}\), and \(D = 24\) into the equation:
\(h(t)=- 18\cos(\frac{\pi}{45}t)+24\)

Step3: Calculate the height at \(t = 35\)

Substitute \(t = 35\) into \(h(t)\):

$$
LATEXBLOCK0
$$

Using a calculator, \(\cos(\frac{7\pi}{9})\approx - 0.7660\)

$$
LATEXBLOCK1
$$

Answer:

b) The cosine equation is \(h(t)=-18\cos(\frac{\pi}{45}t)+24\)
c) The height of the passenger after \(35\) seconds is \(37.79\) meters.