Answer :
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!!!
(a) The capacity will be "4006.4 veh/h".
(b) The speed at capacity be "25 mph".
(c) The density will be "299 veh/mi".
Given:
- [tex]q = 50k - 0.156 k^2[/tex]
At max. flow density,
- [tex]\frac{dd}{dk} =0[/tex]
- [tex](\frac{dq}{dt} ) = 50-0.321k =0[/tex]
(a)
→ [tex]k = (\frac{50}{0.312} )[/tex]
[tex]= 160.3 \ or \ 160 \ veh/mi[/tex]
By substituting the value,
→ [tex]q = 50k-0.156k^2[/tex]
[tex]= 50\times 160.3-0.156\times (160.3)^2[/tex]
[tex]= 4006.4 \ veh/h[/tex]
(b)
The speed will be:
→ [tex]U = \frac{q}{k}[/tex]
[tex]= \frac{4006.4}{160.3}[/tex]
[tex]= 25 \ mph[/tex]
(c)
The density be:
→ [tex]1001.6 = 50k-0.156k^2[/tex]
[tex]0.156k^2-50k+1001.6 =0[/tex]
[tex]k = 21.5 \ veh/mi \ or \ 299 \ veh/mi[/tex]
Thus the responses above are correct.
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