A vertical curve has a length of 5.5 stations. The entering grade is 5.2%, the exiting grade is 0.8%, and the PVI Station and Elevation is 73+40.00 and 254.16 feet respectively. a) Compute the station and elevation of the high point on the curve. b) Compute the elevation on the curve directly below the PVI? Is this the high point on the curve? Why or why not? c) Station the curve using 100 foot stations. d) Compute curve elevations for each station. Show your 1st and 2nd differences to check your work. e) Clearly show the high point on the elevation curve on your sketch. Will the grade curve ever have a slope of zero? Show your answer on the grade curve.

A vertical curve has g1 = -3% and g2 = +6 for a vertical curve length of 1200 ft. What is the elevation at the PVT if the PVI = 53.50? a. 89.50 ft b. 83.50 ft c. 56.50 ft d. 53.80 ft

A vertical curve has g1 = -3% and g2 = +6 for a vertical curve length of 1200 ft. How far is the low point located from the PVC? a. 400 ft b. 500 ft c. 666.67 ft d. 333.33 ft

A simple horizontal curve has a radius of 600 ft and central angle of 70o. Compute the Tangent distance. a. 420.12 ft b. 306.93 ft c. 230.20 ft d. 350.10 ft10 points   QUESTION 3A simple horizontal curve has a radius of 600 ft and central angle of 70o. Compute the long chord. a. 939.69 ft b. 965.93 ft c. 573.58 ft d. 688.29 ft10 points   QUESTION 4A simple horizontal curve has a radius of 600 ft and central angle of 70o. Compute the external distance. a. 132.46 ft b. 110.39 ft c. 118.84 ft d. 268.64 ft10 points   QUESTION 5A simple horizontal curve has a radius of 600 ft and central angle of 70o. Compute the middle ordinate. a. 90.42 ft b. 108.51 ft c. 103.32 ft d. 328.99 ft10 points   QUESTION 6A simple horizontal curve has a radius of 600 ft and central angle of 70o. Compute the degree of curve. a. 14.32o b. 36o c. 9.55o d. 11.45o10 points   QUESTION 7A curve has a degree of curve of 16o. This means that: a. the angle for a 100 ft length of curve is 4o b. the angle for a 100 ft length of curve is 8o c. the total (central) angle of the curve is 16o d. the angle for a 100 ft length of curve is 16o10 points   QUESTION 8A vertical curve has g1 = -3% and g2 = +6 for a vertical curve length of 1200 ft. What is the elevation at the PVI if PVC= 53.50 ft?  a. 38.50 ft b. 68.50 ft c. 53.35 ft d. 35.50 ft10 points   QUESTION 9A vertical curve has g1 = -3% and g2 = +6 for a vertical curve length of 1200 ft. What is the elevation at the PVT if the PVI = 53.50? a. 89.50 ft b. 83.50 ft c. 56.50 ft d. 53.80 ft10 points   QUESTION 10A vertical curve has g1 = -3% and g2 = +6 for a vertical curve length of 1200 ft. How far is the low point located from the PVC? a. 400 ft b. 500 ft c. 666.67 ft d. 333.33 ft10 points   QUESTION 11A vertical curve has g1 = -3% and g2 = +6 for a vertical curve length of 1200 ft. What is the elevation at the low point if the low point is 400 ft from the PVC?  PVC elevation is 53.50 ft.  a. 59.50 ft b. 47.50 ft c. 53.50 ft d. 50.00 ft10 points   QUESTION 12Compute the volume of fill between station 5+00 (Area = 250 ft2) and 6+00 (Area= 500 ft2). a. 375 CY b. 1388.89 CY c. 750 CY d. 500 CY

The first point on the line coincides with the _________ rate:the rate of increase in elevation of _______ feet per hour.The graph starts at the _____________, because the hikersstarted at base camp.The last point on the line indicates the total elevation increasefor the duration of the hike.

The height of an EDM set up at station A is 1.415 m. The height of the reflector set up at station B is 1.124 m. The height of the theodolite at station A used to measure the vertical angle is 1.623 m. The height of the target at station B on which the vertical angle sight is taken is 1.175 m. The measured vertical angle is 5° 32' 45" (upward). The slope distance, after meteorological corrections, is 1537.248 m. Determine the horizontal distance between stations A and B.