"A northbound freeway segment is on a 4% upgrade from station 430+20 to 450+00 and has two 11-ft wide lanes, a 5-ft right shoulder, and has a ramp density of 1 per mile in the 3 miles before and after station 440+10. The peak-hour factor is 0.9. Northbound traffic during the peak hour is 2550 cars, 300 STs, and 300 TTs. Determine the density and LOS of the freeway segment."

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

answer:

try looking it up that is always

explanation:

answer:

ur mom

explanation: