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 $\boxed{{\displaystyle d=5\sin \left(\frac{2\pi t}{7}\right)}}$.