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Posted 3 months ago

Vlad is playing on a swing set.
His horizontal distance D(t)D(t) (in m\mathrm{m} ) from the center (where being behind the center means a negative distance) as a function of time tt (in seconds) can be modeled by a sinusoidal expression of the form acos(bt)+da \cdot \cos (b \cdot t)+d.

At t=0t=0, when he pushes off, he is 1 m1 \mathrm{~m} behind the center, which is as far back as he goes. The swing reaches the center π6\frac{\pi}{6} seconds later.

Find D(t)D(t).
tt should be in radians.

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Answer from Sia
Posted 3 months ago
Solution by Steps
step 1
Given the initial condition at t=0t=0, Vlad is 1 meter behind the center, we can determine the vertical shift dd
step 2
Since the farthest distance behind the center is 1 meter, d=1d = -1
step 3
The swing reaches the center π6\frac{\pi}{6} seconds later, which corresponds to the cosine function being zero. This allows us to find the horizontal stretch bb
step 4
The cosine function is zero at π2\frac{\pi}{2}, so bπ6=π2b \cdot \frac{\pi}{6} = \frac{\pi}{2}. Solving for bb gives b=3b = 3
step 5
There is no information given about a vertical stretch or compression, so we assume the amplitude aa is 1
step 6
The function D(t)D(t) is therefore D(t)=acos(bt)+d=cos(3t)1D(t) = a \cdot \cos(b \cdot t) + d = \cos(3t) - 1
D(t)=cos(3t)1D(t) = \cos(3t) - 1
Key Concept
Modeling motion with sinusoidal functions
The horizontal distance of a swing can be modeled by a sinusoidal function, where the amplitude represents the maximum distance, the horizontal stretch/compression factor adjusts the period of the swing, and the vertical shift represents the initial position relative to the center.

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