To determine when the object is 6 inches above the rest position, we start with the given equation:
[tex]\[ d = -8 \cos \left(\frac{\pi}{6} t\right) \][/tex]
We are given that [tex]\( d = 6 \)[/tex], so we need to solve for [tex]\( t \)[/tex].
1. Set up the equation:
[tex]\[ 6 = -8 \cos \left(\frac{\pi}{6} t\right) \][/tex]
2. Isolate the cosine term:
[tex]\[ \cos \left( \frac{\pi}{6} t \right) = \frac{6}{-8} \][/tex]
[tex]\[ \cos \left( \frac{\pi}{6} t \right) = -\frac{3}{4} \][/tex]
3. Solve for the angle that satisfies this cosine value. To find [tex]\( \frac{\pi}{6} t \)[/tex]:
[tex]\[ \frac{\pi}{6} t = \arccos(-\frac{3}{4}) \][/tex]
4. Calculate [tex]\( \arccos(-\frac{3}{4}) \)[/tex]:
Since [tex]\( \cos^{-1}(-\frac{3}{4}) \)[/tex] produces the principal value in the range [tex]\([0, \pi]\)[/tex], we determine:
[tex]\[ \frac{\pi}{6} t \approx 2.42 \][/tex]
5. Solve for [tex]\( t \)[/tex] by dividing both sides by [tex]\( \frac{\pi}{6} \)[/tex]:
[tex]\[ t = \frac{2.42}{\frac{\pi}{6}} \][/tex]
6. Carry out the division:
[tex]\[ t \approx 4.62 \text{ seconds} \][/tex]
Now, the cosine function is periodic, meaning the same value for cosine will occur again by adding the period of the cosine wave to the first solution. The period of the function [tex]\( \cos \left( \frac{\pi}{6} t \right) \)[/tex] is found by:
[tex]\[ \frac{2\pi}{\frac{\pi}{6}} = 12 \text{ seconds}\][/tex]
Thus, another time when the object will be 6 inches above the rest position:
[tex]\[ t_2 = 4.62 + 12 = 16.62 \text{ seconds} \][/tex]
The principal solution closest to the given options is [tex]\( t \approx 4.62 \)[/tex] seconds.
Therefore, the object will be 6 inches above the rest position at approximately 4.62 seconds.
Among the given choices:
- 0 seconds
- 1.38 seconds
- 4.62 seconds
- 8 seconds
The closest and correct time is 4.62 seconds.
