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Homework 2 – Basic Traffic Flow: Worked Solutions

Here are all five problems worked through with the underlying relationships shown, so you can transcribe them onto engineering paper with the steps intact.

Key relationships used throughout:

  • Space headway (ft) = speed (ft/s) × time headway (s), where speed (ft/s) = speed (mph) × 1.467
  • Density k (vpm) = 5280 / space headway (ft)
  • Gap (ft) = space headway (ft) − vehicle length (ft)
  • Occupancy (%) = [vehicle length (ft) / space headway (ft)] × 100
  • Flow q (vph) = 3600 / time headway (s) = density (vpm) × speed (mph)

Question 1

Condition a: 50 mph, 4 sec headway, L = 15 ft

  • Speed = 50 × 1.467 = 73.33 ft/s
  • A. Headway (ft) = 73.33 × 4 = 293.3 ft
  • B. Density = 5280 / 293.3 = 18.0 vpm
  • C. Gap = 293.3 − 15 = 278.3 ft
  • D. Occupancy = 15/293.3 × 100 = 5.1%
  • E. Flow = 3600/4 = 900 vph

Condition b: 30 mph, 3 sec headway, L = 15 ft

  • Speed = 30 × 1.467 = 44.0 ft/s
  • A. Headway (ft) = 44.0 × 3 = 132.0 ft
  • B. Density = 5280 / 132.0 = 40.0 vpm
  • C. Gap = 132.0 − 15 = 117.0 ft
  • D. Occupancy = 15/132.0 × 100 = 11.4%
  • E. Flow = 3600/3 = 1200 vph

Question 2

Condition a: 20 vpm, 15 mph, L = 15 ft

  • A. Headway (ft) = 5280/20 = 264.0 ft
  • Speed = 15 × 1.467 = 22.0 ft/s
  • B. Headway (sec) = 264.0/22.0 = 12.0 sec
  • C. Gap = 264.0 − 15 = 249.0 ft
  • D. Occupancy = 15/264.0 × 100 = 5.7%
  • E. Flow = 20 × 15 = 300 vph (check: 3600/12 = 300 ✓)

Condition b: 15 vpm, 75 mph, L = 15 ft

  • A. Headway (ft) = 5280/15 = 352.0 ft
  • Speed = 75 × 1.467 = 110.0 ft/s
  • B. Headway (sec) = 352.0/110.0 = 3.2 sec
  • C. Gap = 352.0 − 15 = 337.0 ft
  • D. Occupancy = 15/352.0 × 100 = 4.3%
  • E. Flow = 15 × 75 = 1125 vph (check: 3600/3.2 = 1125 ✓)

Question 3 (Fricker 2.3)

15-min counts: 264, 204, 357, 305

A. Flow rate for each period (vph) — multiply each 15-min count by 4 (since 15 min = ¼ hr):

  • Period 1: 264 × 4 = 1056 vph
  • Period 2: 204 × 4 = 816 vph
  • Period 3: 357 × 4 = 1428 vph
  • Period 4: 305 × 4 = 1220 vph

B. Peak hour factor:
PHF = Hourly volume / (4 × peak 15-min count)

  • Hourly volume = 264+204+357+305 = 1130 veh
  • Peak 15-min count = 357
  • PHF = 1130 / (4 × 357) = 1130/1428 = 0.79

Question 4 (Fricker 2.16)

Given: 21 other vehicles counted over 3 EB lanes, 0.78 mi ahead, speed ≈ 52 mph.

A. Density per lane:

  • Total density (all lanes) = 21 veh / 0.78 mi = 26.92 vpm
  • Per lane: 26.92 / 3 lanes = 8.97 vpm/lane

B. Flow rate:

  • Flow = density × speed = 26.92 vpm × 52 mph = ≈1400 vph (total, 3 lanes)
  • (Per lane, that's 8.97 × 52 ≈ 467 vph/lane)

Question 5 (Fricker 2.17)

Truck length = 31.8 ft, detector effective length = 9 ft, speed = 46.9 mph.

The detector stays activated while the truck travels a distance equal to its own length plus the detector's effective length:

  • Distance = 31.8 + 9 = 40.8 ft
  • Speed = 46.9 × 1.467 = 68.79 ft/s
  • Time = 40.8 / 68.79 = ≈0.59 sec