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TProblem 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.
Example Solution:
Flow
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\,\!}
Density
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\,\!}
Time Mean Speed
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\,\!}
Space Mean Speed
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}}\,\!}
