Solve the problem.
-A formula used by an engineer to determine the safe radius of a curve, $R$, when designing a particular road is: $\mathrm { R } = \frac { \mathrm { V } ^ { 2 } } { \mathrm {~g} ( \mathrm { f } + \tan \alpha ) }$, 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 $\mathrm { R } = 1229.5 \mathrm { ft }$, find V. Round to the nearest foot per second.
A) $\mathrm { V } = 74 \mathrm { ft }$ per sec
B) $V = 69 \mathrm { ft }$ per sec
C) $V = 71 \mathrm { ft }$ per sec
D) $V = 67 \mathrm { ft }$ per sec

Question

Quizplus · Accepted Answer

To find $V$, rearrange the formula: $V = \sqrt{R \cdot g \cdot (f + \tan \alpha)}$. Plug in the values: $R = 1229.5$, $g = 30$, $f = 0.1$, and $\alpha = 2.1^\circ$. Calculate $\tan(2.1^\circ)$, then solve for $V$. The result is approximately 71 ft/sec.

Solve the Problem RRR , When Designing a Particular Road Is R=V2 g(f+tan⁡α)\mathrm { R } = \frac { \mathrm { V } ^ { 2 } } { \mathrm {~g} ( \mathrm { f } + \tan \alpha ) }R= g(f+tanα)V2​

Solve the Problem $R$ , When Designing a Particular Road Is $\mathrm { R } = \frac { \mathrm { V } ^ { 2 } } { \mathrm {~g} ( \mathrm { f } + \tan \alpha ) }$