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A section of highway has the following flowdensity relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the capacity of the highway section, the speed at capacity, and the density when the highway is at one-quarter of its capacity?

Answers

Answer 1

Answer:

(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:

$q = 50k - 0.156 k^2$

At max. flow density,

$(dd)/(dk) =0$

$((dq)/(dt) ) = 50-0.321k =0$

(a)

→ $k = ((50)/(0.312) )$

$= 160.3 \ or \ 160 \ veh/mi$

By substituting the value,

→ $q = 50k-0.156k^2$

$= 50* 160.3-0.156* (160.3)^2$

$= 4006.4 \ veh/h$

(b)

The speed will be:

→ $U = (q)/(k)$

$= (4006.4)/(160.3)$

$= 25 \ mph$

(c)

The density be:

→ $1001.6 = 50k-0.156k^2$

$0.156k^2-50k+1001.6 =0$

$k = 21.5 \ veh/mi \ or \ 299 \ veh/mi$

Thus the responses above are correct.

Find out more information about density here:

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Answer 2

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!!!

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