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Time and space mean speed from 14 travel times

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.

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?

Speed of each vehicle v=L/tv = L/t with L=2L = 2 km =2000= 2000 m.

VehicleTravel time (s)Speed (km/h)
18090.00
29080.00
39675.00
484.785.01
572100.00
679.191.02
712060.00
897.374.00
985.784.01
1078.391.95
1180.989.00
1293.577.01
1376.693.99
1486.783.04

(i) Time mean speed

vˉt=∑vin=1174.0414=83.86 km/h\bar v_t = \frac{\sum v_i}{n} = \frac{1174.04}{14} = 83.86\ \text{km/h}

(ii) Space mean speed

Mean travel time tˉ=1220.8/14=87.200\bar t = 1220.8/14 = 87.200 s.

vˉs=Ltˉ=200087.200=22.936 m/s=82.57 km/h\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=450q = 3600/8 = 450 veh/h. Using q=kvˉsq = k\bar v_s:

k=qvˉs=45082.57=5.45 veh/kmk = \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.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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