Respuesta :
The motion of the hour hand of a clock is a periodic motion that repeats every 12 hours.
- The equation that can be used to models the height as a function of time, based on the given options, is presented as follows; [tex]\displaystyle \underline{ H = 0.5 \cdot cos \left(\frac{\pi}{6} \cdot t \right) + 9.5}[/tex]
Reasons:
The given parameters are;
The variation of the height of the tip of the hour hand = 9 feet to 10 feet
At t = 0, the time = 12.00 a.m.
Required: The equation that can be used to model the height as a function of time.
Solution:
The general form of the cosine function is presented as follows;
y = a·cos(b·x - c) + d
The amplitude = a
[tex]\displaystyle a = \mathbf{ \frac{Maximum \ value -Minimum \ value}{2}}[/tex]
Therefore;
[tex]\displaystyle a = \frac{10 -9}{2} = 0.5[/tex]
b = The cycle speed
The period, T, is given as follows;
[tex]\displaystyle T = \mathbf{ \frac{2 \cdot \pi}{b}}[/tex]
The period of a block, T = 12 hours
Therefore;
[tex]\displaystyle b = \frac{2 \cdot \pi}{T}[/tex]
[tex]\displaystyle b = \frac{2 \cdot \pi}{12} = \mathbf{ \frac{ \pi}{6}}[/tex]
[tex]\displaystyle d = \mathbf{ \frac{Maximum \ value +Minimum \ value}{2} = \frac{10 + 9 }{2}} = 9.5[/tex]
At t = 0, H = The highest point = 10 ft.
Which gives;
[tex]\displaystyle H = \mathbf{0.5 \cdot cos \left(\frac{\pi}{6} \times 0 -c\right) + 9.5} = 10[/tex]
[tex]\displaystyle cos \left( -c\right) = \frac{0.5}{0.5} = 1[/tex]
-c = arcos(1) = 0
-c = 0
c = 0
The equation is therefore;
[tex]\displaystyle H = \mathbf{0.5 \cdot cos \left(\frac{\pi}{6} \cdot t \right) + 9.5}[/tex]
The above equation is the same as; H = 0.5cosine (StartFraction pi Over 6 EndFraction t) + 9.5
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