Let's solve this step-by-step.

Problem Summary:

The drag force $F$ on a boat varies jointly as the wetted surface area $A$ and the square of the velocity $v$ of the boat. This can be modeled by the formula:

$F = k \cdot A \cdot v^2$

where $k$ is a constant of proportionality.

Given:

1. $F = 86 \, \text{N}$, $A = 41.2 \, \text{ft}^2$, $v = 6.5 \, \text{mph}$
2. We need to find the velocity $v$ when $F = 94 \, \text{N}$ and $A = 28.5 \, \text{ft}^2$.

Step 1: Find the constant $k$

Using the values from the first scenario:

$86 = k \cdot 41.2 \cdot (6.5)^2$

Solving for $k$:

1. Calculate $6.5^2$: $6.5^2 = 42.25$

2. Substitute and solve for $k$: $86 = k \cdot 41.2 \cdot 42.25$ $k = \frac{86}{41.2 \times 42.25}$

   Calculating this gives: $k \approx 0.0493$

Step 2: Use $k$ to find the unknown velocity $v$ in the second scenario

Now, using $F = 94$, $A = 28.5$, and $k \approx 0.0493$:

$94 = 0.0493 \cdot 28.5 \cdot v^2$

1. Solve for $v^2$: $v^2 = \frac{94}{0.0493 \times 28.5}$

2. Calculating this value: $v^2 \approx 66.33$

3. Take the square root of both sides to find $v$: $v \approx \sqrt{66.33} \approx 8.14 \, \text{mph}$

Answer:

The boat must travel at approximately 8.14 mph to experience a drag force of 94 N with a wetted surface area of 28.5 ft².

Would you like further details on any part of this calculation, or do you have any questions?

Here are some additional questions for exploration:

1. How would the drag force change if the velocity were doubled?
2. What would happen to the drag force if the surface area were halved?
3. How sensitive is the drag force to changes in velocity compared to surface area?
4. How would a different constant $k$ affect the velocity required for the same drag force?
5. Can we use this formula for different units, such as SI units for area and velocity?

Tip: In joint variation problems, identifying the constant first simplifies finding unknown values for other scenarios.