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Chapter 17: Problem 31

At what angle, in degrees, should a road be slanted to prevent an automobile traveling at \(25 \mathrm{~m} / \mathrm{s}\) from tending to slip? The radius of curvature is \(200 \mathrm{~m}\). a) 22 c) 18 b) 20 d) 16

Short Answer

Expert verified
Answer: 18 degrees

Step by step solution

01

Find the centripetal acceleration of the car

First, we need to find the centripetal acceleration of the car, which is given by the formula: \(a_c = \dfrac{v^2}{r}\) where \(a_c\) is the centripetal acceleration, \(v\) is the speed of the car (25 m/s), and \(r\) is the radius of curvature of the road (200 m). Plugging in the given values, we get: \(a_c = \dfrac{(25 \mathrm{~m/s})^2}{200 \mathrm{~m}}\) Calculating the above expression, we find that the centripetal acceleration of the car is: \(a_c = 3.125 \mathrm{~m/s^2}\)
02

Relate centripetal acceleration to the angle of banking

Next, we need to relate the centripetal acceleration to the angle of banking of the road. This relationship is given by the formula: \(a_c = g \cdot \tan{\theta}\) where \(a_c\) is the centripetal acceleration, \(g\) is the acceleration due to gravity (approximately \(9.81 \mathrm{~m/s^2}\)), and \(\theta\) is the angle of banking. We can solve for the angle \(\theta\) using the above formula. First, plug in the value of \(a_c\) that we found in Step 1, and re-arrange the formula to get: \(\tan{\theta} = \dfrac{a_c}{g} = \dfrac{3.125 \mathrm{~m/s^2}}{9.81 \mathrm{~m/s^2}}\)
03

Calculate the angle in degrees

Finally, we need to find the angle \(\theta\) in degrees by performing the inverse tangent operation and converting from radians to degrees. To do this, we can use the formula: \(\theta = \arctan{\dfrac{3.125 \mathrm{~m/s^2}}{9.81 \mathrm{~m/s^2}}} \cdot \dfrac{180}{\pi}\) Calculating the above expression, we find that the angle of banking required to prevent the car from tending to slip is approximately: \(\theta \approx 18^\circ\) Thus, the correct answer is: c) 18

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