A highway must traverse a 6% followed by a -2% grade. The length of the crest vertical curve is 2040 m. The highest point of the curve (Y max) is?

An equal-tangent vertical curve is designed for the year 2030 for a design speed of 70 mi/h to connect grades G1 = 1.2% and G2 = - 2.1%. In 2030, vehicles have become smaller so that the driver's eye height is assumed to be 3.0 ft above the pavement and roadway objects are assumed to be 1.0 ft above the pavement surface. Compute the design curve length for the year 2030. A. 1178.02 В. 1032.09 C. 974.12 D. 980.25

1) An equal-tangent crest vertical curve has a 50 mi/h design speed. The initial grade is +3%. The high point is at station 33+37.43 and the PVT is at station 37+18.26. What is the elevation difference between the high point and the PVT?   2) A 1400-ft-long sag vertical curve (equal tangent) has a PVC at stateion 115+00 and elevation 750ft.  The intial grade is -3.5% and the final grade is +6.5%.  Determine the elevation and stationing of the low point, PVI, and PVT.   If you could show as many steps taken as possible that would be helpful in me actually understanding how to do the problem.  Thanks

An equal-tangent crest vertical curve is 1100 ft long and connects a +2.5% and a -1.5% grade. If the design speed of the roadway is 85 mi/h, determine the length of curve, does this curve have adequate passing sight distance?