--- title: "Time and space mean speed from 14 travel times" subject: "Transportation Engineering II" chapter: 1 key: "semester_7/transportation_engineering_ii:time-space-mean-speed-14-vehicles" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/time-space-mean-speed-14-vehicles site: BCT Engineering --- # Time and space mean speed from 14 travel times Transportation Engineering II · Chapter 1: Traffic Engineering - Asked in: 2080 Baisakh, 4 marks - Video: [Space and Time Mean Speed,Traffic Flow and Density](https://www.youtube.com/watch?v=_lbHgPqiybY) — DAC ENGINEERING TUTORIALS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/time-space-mean-speed-14-vehicles - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1 ## Question Travel time data (in seconds) of 14 vehicles traversing a two kilometer road segment are: 80, 90, 96, 84.7, 72, 79.1, 120, 97.3, 85.7, 78.3, 80.9, 93.5, 76.6, and 86.7. Estimate (i) the time mean speed and (ii) the space mean speed (iii) What will be the average density of the above traffic stream if the mean headway is 8 sec? ## Short answer Compute each speed as 2 km divided by travel time; time mean speed is the arithmetic mean of speeds, 83.86 km/h, and space mean speed is distance divided by mean travel time, 82.57 km/h. With q = 3600/8 = 450 veh/h, density = q/v_s = 5.45 veh/km. ## Answer Speed of each vehicle $v = L/t$ with $L = 2$ km $= 2000$ m. | Vehicle | Travel time (s) | Speed (km/h) | |---|---|---| | 1 | 80 | 90.00 | | 2 | 90 | 80.00 | | 3 | 96 | 75.00 | | 4 | 84.7 | 85.01 | | 5 | 72 | 100.00 | | 6 | 79.1 | 91.02 | | 7 | 120 | 60.00 | | 8 | 97.3 | 74.00 | | 9 | 85.7 | 84.01 | | 10 | 78.3 | 91.95 | | 11 | 80.9 | 89.00 | | 12 | 93.5 | 77.01 | | 13 | 76.6 | 93.99 | | 14 | 86.7 | 83.04 | #### (i) Time mean speed $$ \bar v_t = \frac{\sum v_i}{n} = \frac{1174.04}{14} = 83.86\ \text{km/h} $$ #### (ii) Space mean speed Mean travel time $\bar t = 1220.8/14 = 87.200$ s. $$ \bar v_s = \frac{L}{\bar t} = \frac{2000}{87.200} = 22.936\ \text{m/s} = 82.57\ \text{km/h} $$ #### (iii) Average density Mean headway 8 s gives flow $q = 3600/8 = 450$ veh/h. Using $q = k\bar v_s$: $$ k = \frac{q}{\bar v_s} = \frac{450}{82.57} = 5.45\ \text{veh/km} $$ **Answer: time mean speed = 83.86 km/h; space mean speed = 82.57 km/h; density = 5.45 veh/km.** (The space mean speed is lower than the time mean speed, as always.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Time mean speed is calculated as: A. The arithmetic mean of individual vehicle speeds, sum(v)/n B. Length divided by the mean travel time C. The harmonic mean of travel times D. The median speed Answer: A. Time mean speed is the arithmetic mean of spot or individual speeds: 1174.04/14 = 83.86 km/h. (Key point: Time mean speed) 2. Space mean speed over a fixed length L is found as: A. Average of the individual speeds B. L divided by the average travel time C. L times the average travel time D. Sum of travel times divided by L Answer: B. v_s = L / mean t = 2000/87.2 = 22.94 m/s = 82.57 km/h. (Key point: Space mean speed) 3. Which statement is correct for the results 83.86 and 82.57 km/h? A. Space mean speed is always higher than time mean speed B. Space mean speed is lower than time mean speed C. They are always equal D. Space mean is higher only in congestion Answer: B. Time mean speed is always at least as high as space mean speed. (Key point: Relation of the two means) 4. How is density found from a mean headway of 8 s? A. k = 8 x 82.57 B. q = 3600/8 = 450 veh/h, then k = q / space mean speed = 5.45 veh/km C. k = 3600/8 = 450 veh/km D. k = time mean speed / 450 Answer: B. Using q = k v_s with q = 450 veh/h gives k = 450/82.57 = 5.45 veh/km. 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