The problem asks to write an equation modeling the diver's vertical position $y$ (in feet) as a function of time $x$ (in seconds), where the diver descends at a rate of $\frac{15}{4}$ feet per second.

This is a linear motion problem since the descent is described at a constant rate, which gives a linear relationship between time and depth. The general form of a linear equation is:

$y = mx + b$

Where:

• $m$ is the slope (rate of change, or how fast the diver descends),
• $b$ is the y-intercept (initial position),
• $x$ is time in seconds, and
• $y$ is the vertical position in feet.

From the information provided:

• The diver descends at $\frac{15}{4}$ feet per second, so $m = -\frac{15}{4}$. The negative sign indicates a downward motion.
• Initially, the diver is at the surface, so at $x = 0$, $y = 0$. Thus, $b = 0$.

The equation becomes:

$y = -\frac{15}{4}x$

This equation models the diver’s vertical position as they descend over time.

Would you like further details or explanations? Here are five related questions that expand on this topic:

1. How would the equation change if the diver were initially at a depth of 10 feet?
2. How can this situation be modeled using a quadratic equation?
3. What would the diver's position be after 5 seconds?
4. How can you calculate the diver's velocity using this equation?
5. What happens to the graph if the diver ascends instead of descending?

Tip: When modeling real-world problems, always pay attention to whether the motion is constant (linear) or changing (quadratic, cubic, etc.).