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A manhole is located in a street that has a 4-inch parabolic crown. It is 16.16 feet from the centerline, and the street is 42 feet wide. What is the elevation of the center of the manhole if the elevation at the street centerline is 277.18 feet?

a. 277.06 ft
b. 276.74 ft
c. 276.85 ft
d. 276.98 ft
e. 276.61 ft

Answer :

Final answer:

To determine the manhole elevation on a parabolically crowned street, the parabolic shape must be considered to calculate the difference in elevation from the street centerline. The exact answer requires the formula for the street's cross-section, which is not provided. However, we know the manhole elevation will be less than 277.18 ft.

Explanation:

The elevations of the manhole relative to the street centerline can be determined by understanding the parabolic shape of the street's surface. Given that you are 16.16 ft from the centerline and the street has a 4-inch (0.333 ft) parabolic crown at the center, we need to find out how much of that crown's height is lost over the distance to the manhole. This is a basic problem of applying the properties of a parabola.

The width of the street does not directly affect the calculation, but it helps us understand that the highest point of the crown is at the center of the street. Since the full width of the street is 42 ft, the manhole is located less than halfway to the edge, which means that a simple ratio of the distances can be used to calculate the reduction in height due to the parabolic shape.

Using this approach, we can calculate the expected drop in elevation over the 16.16 ft distance. However, without the specific formula for the street's cross-section, an exact answer cannot be provided. Nevertheless, it's clear that the elevation at 16.16 ft from the centerline will be less than 277.18 ft due to the curvature of the street surface.

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