Home Leaving Cert Mathematics Trigonometry A Ferris wheel has a diameter of 120 m
A Ferris wheel has a diameter of 120 m - Leaving Cert Mathematics - Question 8 - 2022 Question 8
View full question A Ferris wheel has a diameter of 120 m.
When it is turning, it completes exactly 10 full rotations in one hour.
The diagram above shows the Ferris wheel before it st... show full transcript
View marking scheme Worked Solution & Example Answer:A Ferris wheel has a diameter of 120 m - Leaving Cert Mathematics - Question 8 - 2022
Complete the table below. The value of h(1) is given. Only available for registered users.
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To calculate h(t) for t = 0 to 8:
For t = 0:
h ( 0 ) = 72 − 60 cos ( 0 ) = 72 − 60 ⋅ 1 = 12 h(0) = 72 - 60 \, \text{cos}(0) = 72 - 60 \cdot 1 = 12 h ( 0 ) = 72 − 60 cos ( 0 ) = 72 − 60 ⋅ 1 = 12
For t = 1:
h(1) is given as 42.
For t = 2:
h ( 2 ) = 72 − 60 cos ( π 3 ) = 72 − 60 ⋅ 0.5 = 42 h(2) = 72 - 60 \, \text{cos}\left( \frac{\pi}{3} \right) = 72 - 60 \cdot 0.5 = 42 h ( 2 ) = 72 − 60 cos ( 3 π ) = 72 − 60 ⋅ 0.5 = 42
For t = 3:
h ( 3 ) = 72 − 60 cos ( π 2 ) = 72 − 60 ⋅ 0 = 72 h(3) = 72 - 60 \, \text{cos}\left( \frac{\pi}{2} \right) = 72 - 60 \cdot 0 = 72 h ( 3 ) = 72 − 60 cos ( 2 π ) = 72 − 60 ⋅ 0 = 72
For t = 4:
h ( 4 ) = 72 − 60 cos ( 2 π 3 ) = 72 − 60 ⋅ ( − 0.5 ) = 102 h(4) = 72 - 60 \, \text{cos}\left( \frac{2\pi}{3} \right) = 72 - 60 \cdot (-0.5) = 102 h ( 4 ) = 72 − 60 cos ( 3 2 π ) = 72 − 60 ⋅ ( − 0.5 ) = 102
For t = 5:
h ( 5 ) = 72 − 60 cos ( 5 π 6 ) = 72 − 60 ⋅ ( − 3 / 2 ) ≈ 110.4 h(5) = 72 - 60 \, \text{cos}\left( \frac{5\pi}{6} \right) = 72 - 60 \cdot (-\sqrt{3}/2) \approx 110.4 h ( 5 ) = 72 − 60 cos ( 6 5 π ) = 72 − 60 ⋅ ( − 3 /2 ) ≈ 110.4
For t = 6:
h ( 6 ) = 72 − 60 cos ( 2 π ) = 72 − 60 ⋅ 1 = 12 h(6) = 72 - 60 \, \text{cos}(2\pi) = 72 - 60 \cdot 1 = 12 h ( 6 ) = 72 − 60 cos ( 2 π ) = 72 − 60 ⋅ 1 = 12
For t = 7:
h ( 7 ) = 72 − 60 cos ( 7 π 6 ) = 72 − 60 ⋅ ( − 0.5 ) = 102 h(7) = 72 - 60 \, \text{cos}\left( \frac{7\pi}{6} \right) = 72 - 60 \cdot (-0.5) = 102 h ( 7 ) = 72 − 60 cos ( 6 7 π ) = 72 − 60 ⋅ ( − 0.5 ) = 102
For t = 8:
h ( 8 ) = 72 − 60 cos ( 4 π 3 ) = 72 − 60 ⋅ ( − 0.5 ) = 110.4 h(8) = 72 - 60 \, \text{cos}\left( \frac{4\pi}{3} \right) = 72 - 60 \cdot (-0.5) = 110.4 h ( 8 ) = 72 − 60 cos ( 3 4 π ) = 72 − 60 ⋅ ( − 0.5 ) = 110.4
The completed table values are:
t | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
h(t) | 12 | 42 | 42 | 72 | 102 | 110.4 | 12 | 102 | 110.4 |
Find the period and range of h(t). Only available for registered users.
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The function h(t) is a cosine function, which has a period calculated as:
e x t P e r i o d = 2 π π 6 = 12 minutes . ext{Period} = \frac{2\pi}{\frac{\pi}{6}} = 12 \text{ minutes}. e x t P er i o d = 6 π 2 π = 12 minutes .
The range values based on the minimum and maximum heights are:
Minimum height: 12 m
Maximum height: 132 m
Therefore, the range is:
Range = [ 12 , 132 ] . \text{Range} = [12, 132]. Range = [ 12 , 132 ] .
During a 50-minute period, what is the greatest number of minutes for which the point A could be higher than 42 m? Only available for registered users.
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To find how long point A is above 42 m, we can determine that:
From the table, point A is above 42 m in the intervals:
This gives us:
For 50 minutes:
50 minutes = 4 minutes + 2 periods ⋯ = 32 minutes 50\text{ minutes} = 4\text{ minutes} + 2\text{ periods} \cdots = 32\text{ minutes} 50 minutes = 4 minutes + 2 periods ⋯ = 32 minutes
By solving the following equation, find the second time (value of t) that the point A is at a height of 110 m, after it starts turning. Only available for registered users.
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To solve the equation:
72 − 60 cos ( π 6 t ) = 110 72 - 60 \, \text{cos}\left( \frac{\pi}{6} t \right) = 110 72 − 60 cos ( 6 π t ) = 110
Rearranging gives:
60 cos ( π 6 t ) = − 38 ⇒ cos ( π 6 t ) = − 19 30 . 60\text{cos}\left( \frac{\pi}{6} t \right) = -38 \quad \Rightarrow \quad \text{cos}\left( \frac{\pi}{6} t \right) = -\frac{19}{30}. 60 cos ( 6 π t ) = − 38 ⇒ cos ( 6 π t ) = − 30 19 .
Using the cosine function:
t = \frac{6}{\pi} \text{cos}^{-1}\left(-\frac{19}{30}\right)\approx 1.3\text{ and}: 3.8\text{ minutes}.
So, the second time is:
( t \approx 3.85 \text{ minutes}. )
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