The drag force on a boat varies jointly as a wetted area and the square of the velocity of a boat.If a boat going 6.5 mph experiences a drag force of 86 N when the wetted surface area is 41.2 ft\xb2,how fast must a boat with 28.5 ft\xb2 of wetted surface area goes in order to experience a drag force of 94 N?

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The drag force on a boat varies jointly as a wetted area and the square of the velocity of a boat.If a boat going 6.5 mph experiences a drag force of 86 N when the wetted surface area is 41.2 ft²,how fast must a boat with 28.5 ft² of wetted surface area goes in order to experience a drag force of 94 N?​

Problem:

The drag force on a boat varies jointly as a wetted area and the square of the velocity of a boat. If a boat going 6.5 mph experiences a drag force of 86 N when the wetted surface area is 41.2 ft², how fast must a boat with 28.5 ft² of wetted surface area goes in order to experience a drag force of 94 N?​

Solution:

F = force

A = Area of wetted surface

V = velocity of boat

k = constant of variation

F = 86N ; V = 6.5 mph ; A = 41.2 ft²

F = kAV²

86 = k(41.2)(6.5)²

\\begin{array}{l}F = kA{V^2}\\\\\frac{F}{{kA}} = \frac{{kA{V^2}}}{{kA}}\\\\{V^2} = \frac{F}{{kA}}\end{array}\

\V = \sqrt {\frac{F}{{kA}}} \

if

A = 28.5 ft²

F = 94 N

k = 0.049405411616

V = ?

\\begin{array}{l}V = \sqrt {\frac{F}{{kA}}} \\\\V = \sqrt {\frac{{94}}{{0.0494(28.5)}}} \\\\V = \sqrt {\frac{{94}}{{{\rm{1}}{\rm{.408054231056}}}}} \\\\V = \sqrt {{\rm{66}}{\rm{.758792}}} \\\\V = {\rm{8}}{\rm{.1706}}mph\end{array}\

Answer:

V = 8.17 mph

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