A buoy floating in the ocean is bobbing in simple harmonic motion with amplitude $5 \mathrm{ft}$ and period 7 seconds. Its displacement $d$ from sea level at time $t=0$ seconds is $0 \mathrm{ft}$, and initially it moves upward. (Note that upward is the positive direction.) Give the equation modeling the displacement $d$ as a function of time $t$. \[ d= \]

Step 1 ：The displacement of the buoy can be modeled using the sine function, as it represents simple harmonic motion. The general form of the sine function is \(y = A \sin(B(x - C)) + D\), where \(A\) is the amplitude, \(B\) determines the period, \(C\) is the phase shift, and \(D\) is the vertical shift.

Step 2 ：In this case, the amplitude \(A\) is 5 ft, the period is 7 seconds, the phase shift \(C\) is 0 (since the displacement is 0 at \(t=0\)), and the vertical shift \(D\) is also 0 (since the displacement is measured from sea level).

Step 3 ：The period of the sine function is given by \(2\pi/B\), so we can find \(B\) by setting \(2\pi/B = 7\) and solving for \(B\).

Step 4 ：Let's calculate \(B\) and write down the equation for the displacement \(d\) as a function of time \(t\).

Step 5 ：\(B = 2\pi/7\)

Step 6 ：\(d = 5\sin(2\pi t/7)\)

Step 7 ：Final Answer: The equation modeling the displacement \(d\) as a function of time \(t\) is \(\boxed{d = 5 \sin\left(\frac{2\pi t}{7}\right)}\).