Observers stationed at two sections XX and YY, 500 ft apart on a highway, record the arrival times of five vehicles as shown in the accompanying table. If the total time of observation at XX was 15 sec, determine (a) the time mean speed, (b) the space mean speed, and (c) the flow at section XX.
What will be an ideal response?
First, calculate speeds of individual vehicles:
u1 = L/t = 500 ft / (7.6 sec – 0 sec) = 65.79 ft/sec
u2 = L/t = 500 ft / (9.9 sec – 3.4 sec) = 76.92 ft/sec
u3 = L/t = 500 ft / (14.6 sec – 7.9 sec) = 74.63 ft/sec
u4 = L/t = 500 ft / (20.4 sec – 12.0 sec) = 59.52 ft/sec
u5 = L/t = 500 ft / (21.7 sec – 14.9 sec) = 73.53 ft/sec
a) Time mean speed (TMS), using Equation 6.2, 
ut = (65.79 + 76.92 + 74.63 + 59.52 + 73.53) / 5
ut = 70.08 ft/sec
ut = 70.08/1.47 = 47.7 mi/h
b) Space mean speed (SMS), using Equation 6.3, 
us = (500 ft)(5) / (7.6 + 6.5 + 6.7 + 8.4 + 6.8) sec
us = 69.44 ft/sec
us = 47.2 mi/h
c) Flow at XX, using Equation 6.1, q = n (3600) / T
q = 5 (3600 sec/h) / 15 sec
q = 1200 veh/h
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