civil engineering

highway curves.... proper banking angle, theta, for a car making a turn of radius, r feet at a velocity v feet per second is given by tan theta = v^2 /gr, where g is the acceleration due to gravity. assume g = 32 ft/s^2. an engineer is designing a curve with a radius of 1000 feet. if the speed limit on the curve will be 55mph, at what angle should the curve be banked?

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Originally Posted by mathAna1ys!5

highway curves.... proper banking angle, theta, for a car making a turn of radius, r feet at a velocity v feet per second is given by tan theta = v^2 /gr, where g is the acceleration due to gravity. assume g = 32 ft/s^2. an engineer is designing a curve with a radius of 1000 feet. if the speed limit on the curve will be 55mph, at what angle should the curve be banked?

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Substitute that infromation in to the equation (wonder if this equation is true :rolleyes: )
$\displaystyle \tan\theta=\frac{v^2}{gr}$
Thus,
$\displaystyle (g,r)=(32,1000)$
However,$\displaystyle v$ must be converted to feet per second,
$\displaystyle 55\frac{\mbox{miles}}{\mbox{hour}}=55\frac{\mbox{5 ,280 feet}}{\mbox{3600 seconds}}=80.\bar 6$
Thus,
$\displaystyle \tan\theta=\frac{(80.\bar 6)^2}{32\times 1000}\approx .203$
Thus, (use arctan)
$\displaystyle \theta \approx 11.5$

I had another problem like this and was able to solve it just fine. thanks for the help.

Welcome, I can see you do not like trigonometry that much.