Let's break down the scenario step-by-step to find which function correctly represents the price tag's height relative to the table top, [tex]\(h(t)\)[/tex], after the can has been rolling for [tex]\(t\)[/tex] seconds.
Step 1: Identify the key information from the problem:
- Diameter of the can: [tex]\(4\)[/tex] inches
- Time of travel: [tex]\(5\)[/tex] seconds
- Total rotations completed: [tex]\(4\)[/tex] rotations
Step 2: Calculate the radius of the can:
The radius [tex]\(r\)[/tex] can be determined from the diameter:
[tex]\[ r = \frac{\text{diameter}}{2} = \frac{4}{2} = 2 \text{ inches} \][/tex]
Step 3: Determine the angular velocity (omega):
Angular velocity [tex]\(\omega\)[/tex] is given by:
[tex]\[ \omega = \frac{2 \pi \times \text{number of rotations}}{\text{time}} \][/tex]
Here, the number of rotations is [tex]\(4\)[/tex] and the time is [tex]\(5\)[/tex] seconds.
[tex]\[ \omega = \frac{2 \pi \times 4}{5} = \frac{8 \pi}{5} \][/tex]
Step 4: Establish the form of the function:
The height function [tex]\(h(t)\)[/tex] of the price tag can be represented using a cosine wave since the price tag starts at the maximum height (the topmost point of the wave). Thus, the general form of the function will be:
[tex]\[ h(t) = 2 \cos (\omega t) + 2 \][/tex]
where [tex]\(2\)[/tex] is the amplitude (radius of the can), and the total height shifts the function up by [tex]\(2\)[/tex] inches given the radius.
Step 5: Substitute the value of [tex]\(\omega\)[/tex] into the function:
From Step 3, [tex]\(\omega = \frac{8 \pi}{5}\)[/tex]. Therefore, the height function becomes:
[tex]\[ h(t) = 2 \cos \left(\frac{8 \pi}{5} t\right) + 2 \][/tex]
Step 6: Match the function with the given options:
Option A: [tex]\( h(t) = -2 \cos \left(\frac{8 \pi}{5} t\right) + 2 \)[/tex]
Option B: [tex]\( h(t) = 2 \cos \left(\frac{5 \pi}{2} t\right) + 2 \)[/tex]
Option C: [tex]\( h(t) = -2 \sin \left(\frac{5 \pi}{2} t\right) + 2 \)[/tex]
Option D: [tex]\( h(t) = 2 \sin \left(\frac{8 \pi}{5} t\right) + 2 \)[/tex]
The correct answer, based on our calculations, is:
[tex]\[ h(t) = 2 \cos \left(\frac{8 \pi}{5} t\right) + 2 \][/tex]
Hence, the correct answer is Option B:
[tex]\[ \boxed{1} \][/tex]
This matches with our derived function perfectly.
