To determine when the height above ground of a person sitting on a Ferris wheel reaches 25 meters, given the height function:
[tex]\[ h(t) = 18.8 - 16.7 \cos \left(\frac{2\pi}{5} t\right) \][/tex]
we need to solve the following equation for [tex]\( t \)[/tex] within the first 5 minutes of the ride:
[tex]\[ 18.8 - 16.7 \cos \left(\frac{2\pi}{5} t\right) = 25 \][/tex]
### Step-by-step Solution:
1. Set up the equation:
[tex]\[ 18.8 - 16.7 \cos \left(\frac{2\pi}{5} t\right) = 25 \][/tex]
2. Isolate the cosine term:
[tex]\[ - 16.7 \cos \left(\frac{2\pi}{5} t\right) = 25 - 18.8 \][/tex]
[tex]\[ - 16.7 \cos \left(\frac{2\pi}{5} t\right) = 6.2 \][/tex]
3. Solve for cosine:
[tex]\[ \cos \left(\frac{2\pi}{5} t\right) = -\frac{6.2}{16.7} \][/tex]
[tex]\[ \cos \left(\frac{2\pi}{5} t\right) = -0.3713 \][/tex]
4. Determine the possible values for [tex]\( t \)[/tex]:
We need to find the values of [tex]\( t \)[/tex] within the first 5 minutes where this cosine equation holds true.
5. Use the inverse cosine function to find [tex]\( t \)[/tex]:
[tex]\[ \frac{2\pi}{5} t = \cos^{-1}(-0.3713) \][/tex]
This gives us two primary solutions for the cosine function within one period [tex]\([0, 2\pi]\)[/tex].
6. Calculate the principal value:
[tex]\[ \cos^{-1}(-0.3713) \approx 1.94 \ \text{radians} \][/tex]
However, since we are dealing with a Ferris wheel and the general behavior of cosine, the solutions occur symmetrically in the period [tex]\( 0 \)[/tex] to [tex]\( 2\pi \)[/tex].
7. Find the second solution within the same period:
The cosine function is symmetric, so the second solution is:
[tex]\[ 2\pi - \cos^{-1}(-0.3713) \approx 4.35 \ \text{radians} \][/tex]
8. Convert the angles back to [tex]\( t \)[/tex]:
[tex]\[ t = \frac{5}{2\pi} \times 1.94 \approx 1.55 \ \text{minutes} \][/tex]
[tex]\[ t = \frac{5}{2\pi} \times 4.35 \approx 3.45 \ \text{minutes} \][/tex]
Thus, within the first 5 minutes, the person will be 25 meters above the ground at [tex]\( t = 1.55 \)[/tex] minutes and [tex]\( t = 3.45 \)[/tex] minutes.
Therefore, the person will be 25 meters above the ground at [tex]\( t = 1.55 \)[/tex] minutes or [tex]\( t = 3.45 \)[/tex] minutes.
