PE Civil (NCEES Principles and Practice of Engineering) — All Questions

Level of service F on a facility corresponds to conditions where:

• a.demand exceeds capacity (v/c > 1) and flow breaks down with queuing✓
• b.the road is empty
• c.only trucks are present
• d.traffic flows freely at high speed

LOS F represents breakdown: arriving demand exceeds capacity, queues form and grow, and speeds collapse. It is the failing condition that capacity improvements aim to avoid.

A bottleneck serves 800 veh/hr while 1,000 veh/hr arrive for 1 hour (deterministic queuing). What is the maximum queue length at the end of that hour?

• a.125 veh
• b.1,800 veh
• c.200 veh✓
• d.1,000 veh

For deterministic queuing, queue growth = (arrival rate - service rate) x time = (1,000 - 800) x 1 = 200 vehicles. The queue accumulates at the difference between demand and capacity.

A parabolic vertical curve 400 ft long begins (BVC) at elevation 100.00 ft on a +2% grade and ends on a -2% grade. What is the curve elevation at 200 ft from the BVC?

• a.100.0 ft
• b.104.0 ft
• c.102.0 ft✓
• d.98.0 ft

Elevation = BVC + g1·x + [(g2 - g1)/(2L)]·x^2 = 100 + 0.02(200) + [(-0.04)/(800)](200^2) = 100 + 4 - 2 = 102.0 ft. The parabolic offset (-2 ft) pulls the point below the tangent elevation of 104 ft.

A vertical curve 400 ft (4 stations) long joins a +2% grade to a -2% grade. What is the rate of grade change per station?

• a.-4 %/station
• b.-0.01 %/station
• c.-1 %/station✓
• d.+1 %/station

Rate of change r = (g2 - g1)/L = (-2 - 2)/4 stations = -1 percent per station. The negative sign indicates a crest curve; the magnitude sets how quickly the grade transitions.

A reverse curve consists of:

• a.a single curve with a spiral
• b.two curves turning the same direction with different radii
• c.two circular curves of opposite direction (turning opposite ways) joined at a common tangent point✓
• d.a straight tangent only

A reverse curve has two arcs curving in opposite directions, one right and one left, meeting at a common point. A compound curve, by contrast, has two same-direction arcs of different radii.

The primary purpose of a spiral (transition) curve between a tangent and a circular curve is to:

• a.gradually introduce curvature and superelevation for a smooth, comfortable transition✓
• b.eliminate the need for superelevation
• c.shorten the overall alignment
• d.increase the design speed automatically

A spiral provides a continuously changing radius so lateral acceleration and superelevation build up gradually rather than abruptly at the PC. This improves driver comfort, safety, and the natural steering path.

On a horizontal curve, the superelevation runoff length is the distance required to:

• a.provide passing sight distance
• b.change the horizontal radius to zero
• c.stop a vehicle from the design speed
• d.transition the pavement cross-slope from zero (adverse crown removed) to full superelevation✓

Superelevation runoff is the length over which the outside lane rotates from a level (or crown-removed) cross-section to the full banked cross-slope of the curve. It is set to keep the edge-of-pavement grade change gradual.

When selecting a design vehicle for intersection turning templates, the largest turning radius is required by:

• a.a passenger car (P)
• b.a bicycle
• c.a large tractor-semitrailer (e.g., WB-67)✓
• d.a single-unit truck (SU)

Larger, longer design vehicles like the WB-67 tractor-semitrailer track wide and need the largest turning paths. Designing curb returns and turning lanes for the largest expected vehicle ensures smaller vehicles are accommodated too.

A vehicle accelerates uniformly from rest to 60 mph (88 ft/s) in 10 seconds. What is its acceleration?

• a.8.8 ft/s^2✓
• b.6.0 ft/s^2
• c.88 ft/s^2
• d.0.88 ft/s^2

a = (v - v0)/t = (88 - 0)/10 = 8.8 ft/s^2 (60 mph = 88 ft/s). Uniform acceleration is the change in velocity divided by the time interval.

A vehicle traveling 60 mph (88 ft/s) decelerates at 15 ft/s^2. What braking distance is needed to stop?

• a.516 ft
• b.129 ft
• c.88 ft
• d.258 ft✓

Using v^2 = v0^2 - 2a·d with final v = 0: d = v0^2/(2a) = 88^2/(2 x 15) = 7,744/30 = 258 ft. Braking distance grows with the square of speed.

The standard AASHTO perception-reaction time used for stopping sight distance design is:

• a.1.0 second
• b.2.5 seconds✓
• c.5.0 seconds
• d.0.5 second

AASHTO uses a 2.5-second brake-reaction (perception-reaction) time for stopping sight distance, which accommodates most drivers under most conditions. It combines the time to perceive, identify, decide, and initiate braking.

Skid marks 100 ft long are measured on a surface with a friction factor of 0.70. What was the approximate initial speed? (Use v = sqrt(30·f·d), v in mph, d in ft.)

• a.2,100 mph
• b.30 mph
• c.45.8 mph✓
• d.21 mph

v = sqrt(30·f·d) = sqrt(30 x 0.70 x 100) = sqrt(2,100) = 45.8 mph. This accident-reconstruction formula estimates speed from skid length and pavement friction.

For a fixed design vehicle and roadway, increasing the design speed generally requires the maximum allowable grade to be:

• a.lower (flatter)✓
• b.unchanged
• c.exactly zero
• d.higher (steeper)

Higher design speeds call for flatter maximum grades so that heavy vehicles can maintain speed on upgrades and stopping distances remain adequate on downgrades. AASHTO tables reduce allowable grade as design speed rises.

A roadway direction must carry a directional design hour volume of 1,800 veh/hr, and each lane provides a service flow of 900 veh/hr at the target LOS. How many lanes are needed in that direction?

• a.1 lane
• b.2 lanes✓
• c.4 lanes
• d.3 lanes

Number of lanes = DDHV / per-lane service flow = 1,800/900 = 2 lanes. Dividing the directional demand by the acceptable per-lane service volume gives the lanes required to hold the target level of service.

An approach has a design speed of 30 mph (44 ft/s), a driver perception-reaction time of 1.0 s, and a comfortable deceleration of 10 ft/s^2 on a level grade. What is the minimum yellow (change) interval? (Use y = t + V/[2a].)

• a.1.0 s
• b.2.2 s
• c.4.4 s
• d.3.2 s✓

Yellow change interval y = t + V/(2a) = 1.0 + 44/(2 x 10) = 1.0 + 2.2 = 3.2 s. It gives a driver time to react and either stop or clear the stop line, avoiding the dilemma zone.