# 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

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

Density

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

Time Mean Speed

${\displaystyle {\overline {v_{t}}}={\frac {1}{N}}\sum \limits _{n=1}^{N}{v_{n}}={\frac {1}{4}}\left({72+90+80+88}\right)=82.5km/hr\,\!}$

Space Mean Speed

${\displaystyle {\begin{array}{l}{\overline {v_{s}}}={\frac {N}{\sum \limits _{n=1}^{N}{\frac {1}{v_{i}}}}}={\frac {4}{{\frac {1}{72}}+{\frac {1}{90}}+{\frac {1}{80}}+{\frac {1}{88}}}}=81.86\\t_{i}=L/v_{i}\\t_{A}=L/v_{A}=0.28/88=0.00318hr\\t_{B}=L/v_{B}=0.28/80=0.00350hr\\t_{C}=L/v_{C}=0.28/90=0.00311hr\\t_{D}=L/v_{D}=0.28/72=0.00389hr\\{\overline {v_{s}}}={\frac {NL}{\sum \limits _{n=1}^{N}{t_{n}}}}={\frac {4*0.28}{\left({0.00318+0.00350+0.00311+0.00389}\right)}}=81.87km/hr\\\end{array}}\,\!}$