Solve the problem.
-A formula used by an engineer to determine the safe radius of a curve, R, when designing a particular road is: $\mathrm { R } = \frac { \mathrm { V } ^ { 2 } } { g ( \mathrm { f } + \tan \alpha ) }$, where $\alpha$ is the superelevation of the road and $\mathrm { V }$ is the velocity (in feet per second) for which the curve is designed. If $\alpha = 2.1 ^ { \circ } , \mathrm { f } = 0.1 , \mathrm {~g} = 30$, and $\mathrm { R } = 1195.11 \mathrm { ft }$, find V. Round to the nearest foot per second.
A) $V = 68 \mathrm { ft }$ per sec
B) $V = 73 \mathrm { ft }$ per sec
C) $\mathrm { V } = 70 \mathrm { ft }$ per sec
D) $V = 66 \mathrm { ft }$ per sec

Question

Quizplus · Accepted Answer

To find $V$, we rearrange the formula to solve for $V$: $V = \sqrt{R \cdot g \cdot (f + \tan(\alpha))}$. Substituting the given values: $\alpha = 2.1^\circ$, $f = 0.1$, $g = 30$, and $R = 1195.11$, we calculate $V = \sqrt{1195.11 \cdot 30 \cdot (0.1 + \tan(2.1^\circ))}$. This calculation gives $V \approx 70$ ft/sec when rounded to the nearest foot per second.

Solve the Problem R=V2g(f+tan⁡α)\mathrm { R } = \frac { \mathrm { V } ^ { 2 } } { g ( \mathrm { f } + \tan \alpha ) }R=g(f+tanα)V2​

Solve the Problem $\mathrm { R } = \frac { \mathrm { V } ^ { 2 } } { g ( \mathrm { f } + \tan \alpha ) }$