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Time and space mean speed from 20 spot speeds

Short answer

Time mean speed = 49.50 km/h (arithmetic mean of 20 observations). Space mean speed = 46.95 km/h (harmonic mean). For the 0.5 km segment, mean travel time 38.34 s gives space mean speed 0.5/0.01065 h = 46.95 km/h, matching the harmonic mean.

In a field survey of spot speed measurement, the following twenty observations were taken. Find time mean speed, and space mean speed.

50, 40, 60, 54, 45, 31, 72, 58, 43, 52, 46, 56, 43, 65, 33, 69, 34, 51, 47, 41.

Also, assuming these vehicle speeds are fixed over a half km segment, calculate the corresponding travel times and show that the space mean speed calculated using travel times is equal to the point estimate.

Time mean speed

The time mean speed is the arithmetic mean of the spot speeds:

vˉt=∑vin=99020=49.50 km/h\bar v_t = \frac{\sum v_i}{n} = \frac{990}{20} = 49.50\ \text{km/h}

Space mean speed

The space mean speed is the harmonic mean of the spot speeds:

vˉs=n∑(1/vi)=200.42600=46.95 km/h\bar v_s = \frac{n}{\sum (1/v_i)} = \frac{20}{0.42600} = 46.95\ \text{km/h}

Travel times over 0.5 km

ti=0.5/vi×3600t_i = 0.5/v_i \times 3600 s:

VehicleSpeed (km/h)Travel time (s)
15036.00
24045.00
36030.00
45433.33
54540.00
63158.06
77225.00
85831.03
94341.86
105234.62
114639.13
125632.14
134341.86
146527.69
153354.55
166926.09
173452.94
185135.29
194738.30
204143.90

Mean travel time tˉ=766.80/20=38.340\bar t = 766.80/20 = 38.340 s =0.010650= 0.010650 h.

vˉs=Ltˉ=0.50.010650=46.95 km/h\bar v_s = \frac{L}{\bar t} = \frac{0.5}{0.010650} = 46.95\ \text{km/h}

This equals the harmonic mean of the spot speeds (46.95 km/h), because L/tˉ=L/(1n∑L/vi)=n/∑(1/vi)L/\bar t = L/\left(\frac{1}{n}\sum L/v_i\right) = n/\sum(1/v_i).

Answer: time mean speed = 49.50 km/h; space mean speed = 46.95 km/h (same from travel times). The space mean speed is lower than the time mean speed.

✓ Answer checked by Claude Opus 5.5 (high effort)

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