Vlad is playing on a swing set.

His horizontal distance \[D(t)\] (in \[\text{m}\]) from the center (where being behind the center means a negative distance) as a function of time \[t\] (in seconds) can be modeled by a sinusoidal expression of the form \[a\cdot\cos(b\cdot t)+d\].

At \[t=0\], when he pushes off, he is \[1\text{ m}\] behind the center, which is as far back as he goes. The swing reaches the center \[\dfrac{\pi}{6}\] seconds later.

**Find \[D(t)\].**
*\[\text{\it{t}}\] should be in radians.*

\[D(t) = \] 

To model Vlad's horizontal distance \(D(t)\) from the center, we use the cosine function of the form:

\[
D(t) = a \cdot \cos(b \cdot t) + d
\]

### Step-by-Step Solution:

1. **Determine the parameters \(a, b,\) and \(d\):**

   - **Amplitude \(a\):**  
     The amplitude of the cosine function represents half the difference between the maximum and minimum distances from the center.  
     Since Vlad starts \(1 \, \text{m}\) behind the center (as far back as he goes), the amplitude \(a\) is \(1\).

     \[
     a = 1
     \]

   - **Vertical shift \(d\):**  
     In this scenario, since the swing oscillates back and forth around the center (which corresponds to \(0 \, \text{m}\)), the vertical shift \(d\) is \(0\). The center of the swing's motion is the origin, so:

     \[
     d = 0
     \]

   - **Period and Frequency \(b\):**  
     The swing reaches the center \( \frac{\pi}{6} \) seconds after pushing off. This means it moves from the maximum point to the center, which corresponds to a quarter of the period of the cosine function.

     If \( T \) is the period, then \( \frac{T}{4} = \frac{\pi}{6} \), so:

     \[
     T = 4 \cdot \frac{\pi}{6} = \frac{2\pi}{3}.
     \]

     The period of a cosine function is given by \( T = \frac{2\pi}{b} \), so solving for \( b \):

     \[
     \frac{2\pi}{b} = \frac{2\pi}{3} \implies b = 3.
     \]

2. **Construct the function \(D(t)\):**

   Now we substitute these values into the general form:

   \[
   D(t) = 1 \cdot \cos(3 \cdot t) + 0.
   \]

   Simplifying:

   \[
   D(t) = \cos(3t).
   \]

### Final Answer:

\[
\boxed{D(t) = \cos(3t)}
\]

This is the function that models Vlad's horizontal distance \(D(t)\) from the center over time \(t\) in seconds, with \(t\) in radians.

Would you like more details or have any questions?

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### Follow-up Questions:

1. What happens to \(D(t)\) if the swing starts \(2 \, \text{m}\) behind the center instead of \(1 \, \text{m}\)?
2. How would the function change if the swing reached the center after \( \frac{\pi}{4} \) seconds instead of \( \frac{\pi}{6} \) seconds?
3. If Vlad's swing took a full second to return to the starting point, what would be the new value of \(b\)?
4. What is the significance of the parameter \(d\) in the function \(D(t)\)?
5. How would the graph of \(D(t)\) change if the frequency \(b\) were doubled?

### Tip:
Understanding how changes in parameters \(a, b,\) and \(d\) affect sinusoidal functions can help in modeling a wide variety of oscillatory motions, such as swings, pendulums, or even waves.

Learn how to model Vlad's horizontal distance from the center on a swing set using a cosine function. Detailed step-by-step solution with amplitude, period, and frequency calculations included. Suitable for high school students studying trigonometry and sinusoidal functions.

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