The drag force F on a boat varies jointly with the wet surface area A of the boat and the square of the speed s of the boat. A boat with a wet surface area of 83 ft 2 traveling at 20 mph experiences a drag force of 996 N . Find the wet surface area of a boat traveling 18 mph and experiencing a drag force of 1215 N .

Wet surfaces areaA=+25.3ft^2

Explanation:

Using F= K×A× S^2

Where F= drag force

A= surface area

S= speed

Given : F=996N S=20mph A= 83ft^2

K = F/AS^2=996/(83×20^2)

K= 996/33200 = 0.03

1215= (0.03)× A × 18^2

1215=9.7A

A=1215/9.7=125.3ft^2

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s