Solve.
-Wind resistance or atmospheric drag tends to slow down moving objects. Atmospheric drag varies jointly as an object's surface area A and velocity v. If a car traveling at a speed of 40 mph with a surface area of $37 \mathrm { ft } ^ { 2 }$ experiences a drag of $236.8 \mathrm {~N}$ (Newtons), how fast must a car with $48 \mathrm { ft } ^ { 2 }$ of surface area travel in order to
A)66 mph
B)69 mph
C)71 mph
D)74 mph

Question

Quizplus · Accepted Answer

The drag force $D$ is given by $D = kAv$, where $k$ is a constant, $A$ is the surface area, and $v$ is the velocity. Given $D_1 = 236.8$ N for $A_1 = 37 \mathrm{ft}^2$ and $v_1 = 40$ mph, we can find $k$ and then solve for $v_2$ given $A_2 = 48 \mathrm{ft}^2$. Since $D$ varies jointly with $A$ and $v$, we have $\frac{D_1}{A_1v_1} = \frac{D_2}{A_2v_2}$. Substituting the given values and solving for $v_2$ yields approximately 69 mph.

Solve -Wind Resistance or Atmospheric Drag Tends to Slow Down 37ft237 \mathrm { ft } ^ { 2 }37ft2

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-Wind Resistance or Atmospheric Drag Tends to Slow Down $37 \mathrm { ft } ^ { 2 }$