Fundamentals of Transportation/Traffic Flow/Solution

Problem:

Four vehicles are traveling at constant speeds between sections X and Y (280 meters apart) with their positions and speeds observed at an instant in time. An observer at point X observes the four vehicles passing point X during a period of 15 seconds. The speeds of the vehicles are measured as 88, 80, 90, and 72 km/hr respectively. Calculate the flow, density, time mean speed, and space mean speed of the vehicles.

Solution:

Flow

$q = N\left( {\frac{{3600}}{{t_{measured} }}} \right) = 4\left( {\frac{{3600}}{{15}}} \right) = 960veh/hr \,\!$

Density

$k = \frac{N}{L} = \frac{4*1000}{{280}} = 14.2veh/km \,\!$

Time Mean Speed

$\overline {v_t } = \frac{1}{N}\sum\limits_{n = 1}^N {v_n } = \frac{1}{4}\left( {72 + 90 + 80 + 88} \right) = 82.5km/hr \,\!$

Space Mean Speed

$\begin{array}{l} \overline {v_s } = \frac{N}{{\sum\limits_{n = 1}^N {\frac{1}{{v_i }}} }} = \frac{4}{{\frac{1}{{72}} + \frac{1}{{90}} + \frac{1}{{80}} + \frac{1}{{88}}}} = 81.86 \\ t_i = L/v_i \\ t_A = L/v_A = 0.28/88 = 0.00318hr \\ t_B = L/v_B = 0.28/80 = 0.00350hr \\ t_C = L/v_C = 0.28/90 = 0.00311hr \\ t_D = L/v_D = 0.28/72 = 0.00389hr \\ \overline {v_s } = \frac{{NL}}{{\sum\limits_{n = 1}^N {t_n } }} = \frac{{4*0.28}}{{\left( {0.00318 + 0.00350 + 0.00311 + 0.00389} \right)}} = 81.87km/hr \\ \end{array} \,\!$