Final answer:
Using the relationship of joint variation of drag force with the surface area and the square of speed, and the given information from the first boat, we can calculate and find that the speed of the second boat experiencing a drag force of 405 N with a wet surface area of 36 ft is 15 mph.
Step-by-step explanation:
To solve for the speed of a boat experiencing a drag force of 405 N with a wet surface area of 36 ft, we use the concept that the drag force varies jointly with the surface area of the boat and the square of its speed. We start by using the given information of the first boat: a drag force of 210 N, a wet surface area of 42 ft, and a speed of 10 mph.
Let's denote the constant of variation by k. So, the initial relationship is given by F = k * A * V2, where F is the drag force, A is the surface area, and V is the speed. From the given information for the first boat, we can find k by rearranging the formula: k = F / (A * V2) and substituting the known values to obtain k.
With k determined, we can find the speed of the second boat by using the formula again with the new known values of F (405 N) and A (36 ft) and solving for V. This results in a quadratic equation which we solve to find the speed of the second boat. The correct speed corresponding to the drag force of 405 N and a surface area of 36 ft is 15 mph, which is an option (b).