Answer: D = 23.09 pc/mi/h
LOS = C
Explanation:
we will begin by solving this with a step by step process for easy understanding;
given that the freeway has two lanes = 11 ft wide
and has a width of 5 ft right shoulder.
ramp density = 1 per mile in 3m
substituting the free flow speed value gives us;
Free flow speed = 75 - 1.9 - 0.6 - (3.22 × 1∧0.84) = 69.28 mph
we have that the length of the road = (1980 ft) × (1 mile / 5280 ft)
L = 0.375 mile
The next thing we will do is to calculate the proportion of bus and the truck
Pt = 300 + 300 / (2400 + 300 + 300)
Pt = 600/3000 = 20%
following up, we will consider the length and percentage of the buses and the trucks
Et = 2.0, Pr = 0, Er = 0
to calculate the percentage of truck
Ft = 1 / 1 + Pt (Et -1 ) + Pr (Er -1)
Ft = 1 / 1+0.2 (2-1) + 0 = 0.833
Truck percentage = 0.833
To the determine the traffic volume,
Vt = V / PHF × N × Ft × Fp
Vt = 2400/ (0.9×2×0.833×1) = 1600 pc/lane/hour
But the Density of the freeway is given thus;
D = Vp / FFS .............(1)
but to get the FFS, we will consider the graph of flow rate vs speed and show the level of service
FFS = 69.28 mi/h
From the above expression in (1) we have that
D = 1600/69.28 = 23.09 pc/mi/h
D = 23.09 pc/mi/h
now we have that the the density of the freeway segment is 23.09 pc/mi/h, we can thus safely say that the level of service (los) = C
cheers i hope this helps
