The problem asks for an equation modeling the diver's vertical position, $y$, in terms of time, $x$, with a given rate of descent.

Information from the problem:

• The diver descends at a rate of $\frac{7}{2}$ feet per second.
• There is a parabolic path (indicating quadratic motion).
• At $x = 0$, $y = 0$ (since the diver starts at the surface).
• There are two key points from the diagram:
  • $(3, 0)$, when the diver starts descending from the surface at 3 seconds.
  • $(17, 0)$, when the diver is at 0 feet again at 17 seconds.

Equation of the parabola:

Since the diver returns to the same vertical level (0 feet) after 17 seconds, this suggests a quadratic equation in standard form:

$y = a(x - 3)(x - 17)$

Now, we need to determine the constant $a$. We know that the rate of descent is $\frac{7}{2}$ feet per second at $x = 1$ second after the diver starts descending. Let's use this information to calculate $a$.

At $x = 4$ seconds, the diver has descended by $\frac{7}{2} \times 1 = \frac{7}{2}$. Let's plug this into the equation to find $a$:

$\frac{7}{2} = a(4 - 3)(4 - 17)$ $\frac{7}{2} = a(1)(-13)$ $\frac{7}{2} = -13a$ $a = -\frac{7}{26}$

Final Equation:

$y = -\frac{7}{26}(x - 3)(x - 17)$

This is the equation modeling the diver's vertical position $y$ with respect to time $x$.

Would you like more details or have any other questions?

Related Questions:

1. How do you derive the equation of a parabola from two points and a slope?
2. How does the rate of descent influence the shape of the parabolic curve?
3. What does the vertex of the parabola represent in this context?
4. Can you calculate the maximum depth the diver reaches?
5. How do changes in the descent rate affect the coefficients in the equation?

Tip:

Always check the units and signs of your constants to ensure they match the physical context of the problem, especially in motion problems!