Solve the problem.
-If an automobile is traveling at velocity $\mathrm { V }$ (in feet per second), the safe radius $\mathrm { R }$ for a curve with superelevation $\alpha$ is given by the formula $\mathrm { R } = \frac { \mathrm { V } ^ { 2 } } { \mathrm {~g} ( \mathrm { f } + \tan \alpha ) }$, where $\mathrm { f }$ and $g$ are constants. A road is being constructed for automobiles traveling at 50 miles per hour. If $\alpha = 4 ^ { \circ } , \mathrm { g } = 32.4$, and $\mathrm { f } = 0.16$, calculate R. Round to the nearest foot. (Hint: 1 mile $= 5280$ feet)
A) $R = 1172 \mathrm { ft }$
B) $R = 706 \mathrm { ft }$
C) $R = 722 \mathrm { ft }$
D) $\mathrm { R } = 742 \mathrm { ft }$

Question

Quizplus · Accepted Answer

First, convert the velocity from miles per hour to feet per second: $50 \text{ mph} = 50 \times \frac{5280 \text{ ft}}{3600 \text{ s}} = 73.33 \text{ ft/s}$. Then, calculate $R$ using the given formula with $\alpha = 4^\circ$, which means $\tan(\alpha) = \tan(4^\circ)$. Thus, $R = \frac{73.33^2}{32.4(0.16 + \tan(4^\circ))}$. Using a calculator, $\tan(4^\circ) \approx 0.070$, so $R \approx \frac{5378.89}{32.4(0.16 + 0.070)} = \frac{5378.89}{32.4 \times 0.23} \approx \frac{5378.89}{7.452} \approx 721.7$, which rounds to $722$ feet.

Solve the Problem V\mathrm { V }V (In Feet Per Second), the Safe Radius

Solve the Problem $\mathrm { V }$ (In Feet Per Second), the Safe Radius