Answer:
a) capacity of the highway section = 4006.4 veh/h
b) The speed at capacity = 25 mph
c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi
Explanation:
q = 50k - 0.156k²
with q in veh/h and k in veh/mi
a) capacity of the highway section
To obtain the capacity of the highway section, we first find the k thay corresponds to the maximum q.
q = 50k - 0.156k²
At maximum flow density, (dq/dk) = 0
(dq/dt) = 50 - 0.312k = 0
k = (50/0.312) = 160.3 ≈ 160 veh/mi
q = 50k - 0.156k²
q = 50(160.3) - 0.156(160.3)²
q = 4006.4 veh/h
b) The speed at the capacity
U = (q/k) = (4006.4/160.3) = 25 mph
c) the density when the highway is at one-quarter of its capacity?
Capacity = 4006.4
One-quarter of the capacity = 1001.6 veh/h
1001.6 = 50k - 0.156k²
0.156k² - 50k + 1001.6 = 0
Solving the quadratic equation
k = 21.5 veh/mi or 299 veh/mi
Hope this Helps!!!
