四辆汽车以恒定速度在 X 和 Y（相距 280 米）之间行驶，在某一时刻观察到它们的方位和速度。在 X 点的观察者观察到四辆汽车在 15 秒内通过 X 点。车辆的速度分别测量为 88、80、90 和 72 公里/小时。计算车辆的流量、密度、时间平均速度和空间平均速度。
流量
q = N ( 3600 t m e a s u r e d ) = 4 ( 3600 15 ) = 960 v e h / h r {\displaystyle q=N\left({\frac {3600}{t_{measured}}}\right)=4\left({\frac {3600}{15}}\right)=960veh/hr\,\!}
密度
k = N L = 4 ∗ 1000 280 = 14.2 v e h / k m {\displaystyle k={\frac {N}{L}}={\frac {4*1000}{280}}=14.2veh/km\,\!}
时间平均速度
v t ¯ = 1 N ∑ n = 1 N v n = 1 4 ( 72 + 90 + 80 + 88 ) = 82.5 k m / h r {\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\,\!}
空间平均速度
v s ¯ = N ∑ n = 1 N 1 v i = 4 1 72 + 1 90 + 1 80 + 1 88 = 81.86 t i = L / v i t A = L / v A = 0.28 / 88 = 0.00318 h r t B = L / v B = 0.28 / 80 = 0.00350 h r t C = L / v C = 0.28 / 90 = 0.00311 h r t D = L / v D = 0.28 / 72 = 0.00389 h r v s ¯ = N L ∑ n = 1 N t n = 4 ∗ 0.28 ( 0.00318 + 0.00350 + 0.00311 + 0.00389 ) = 81.87 k m / h r {\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}}\,\!}
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