Amultilane highway (two lanes in each direction) is on level terrain. the free-flow speed has been measured at 45 mi/h. the peak-hour directional traffic flow is 1300 vehicles with 6% large trucks and buses and 2% recreational vehicles (f_p = 0.95).if the peak-hour factor is 0.85, determine the highway's level of service.

no of vehicles added = 1667.2395 vehicles

Explanation:

from the question given, we have that;

amount of vehicles (V) = 1300

peak hour factor (Phf) = 0.85

the peak hour traffic of buses and trucks (Ptb) = 6% = 0.06

the peak hour traffic of recreational vehicles  (Prv) = 2% = 0.02

Fp = 0.95

N = no of lanes

from these values, we calculate the value of the heavy vehicle factor hvF given as;

hvF = 1 / (1 + Ptb(Et-1) + Prv(Er-1))

where Et = 1.5

and Er = 1.2

putting values of Et and Er we have,

hvF = 1 / (1 + 0.06(1.5-1) + 0.02(1.2-1))

this gives

hvF = 0.967

we already have V as 1300, applying the formula for Vp;

Vp = V / (phF * N * hvF * Fp) (1)

Vp = 1300 / (0.85 x 2 x 0.967 x 0.95)

Vp = 832.3 pc / min/in

given from the question that the free-flow speed is measured at 45 mi/h

we have that phF = 0.85, Fp = 0.95 and no of lanes N = 2,

Vp = 1900

from equation (1)

new vehicle added (Vn) = Vp * phF * N * hvF * Fp = 1900 x  0.85 x 2 x 0.967 x 0.95 = 2967.2395

∴ Vnew - Voriginal = 2967.2395 - 1300 = 1667.2395 vehicles

cheers, i hope this helps

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explanation: