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jeka94
3 years ago
6

Two routes connect an origin and a destination. Routes 1 and 2 have performance functions t1 = 2 + X1 and t2 = 1 + X2, where the

t's are in minutes and the x's are in thousands of vehicles per hour. The travel times on the routes are known to be in user equilibrium. If an observation for route 1 finds that the gaps between 30% of the vehicles are less than 6 seconds. Estimate the volume and average travel times for the two routes
Engineering
1 answer:
Musya8 [376]3 years ago
7 0

Solution :

Given

$t_1=2+x_1$

$t_2=1+x_2$

Now,

$P(h

$0.4=1-P(h \geq5)$

$0.6=P(h \geq5)$

$0.6= e^{\frac{-x_1 5}{3600}}$

Therefore,   $x_1=368 \ veh/h$

                        $=\frac{368}{1000} = 0.368$

Given,   $t_1=2+x_1$

                 = 2 + 0.368

                 = 2.368 min

At user equilibrium, $t_2=t_1$

∴  $t_2$ = 2.368 min

$t_2=1+x_2$

$2.368=1+x_2$

$x_2 = 1.368$

$x_2 = 1.368 \times 1000$

    = 1368 veh/h

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Answer:

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Explanation:

We know that for two point charges of magnitude q_{1},q_{2} the magnitude of force between them is given by

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Initially when the charges are separated by 2.4 meters the force can be calculated as

F_{1}=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\10.8=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\\therefore k_{e}\cdot q_{1}q_{2}=10.8\times 2.4^{2}=62.208

Now when the separation is reduced to 0.7 meters the force is similarly calculated as

F_{2}=\frac{k_{e}\cdot q_{1}q_{2}}{0.7^{2}}

Applying value of the constant we get

F_{1}=\frac{62.208}{0.7^{2}}

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The force that does work on the electron is the one due to the electric field E, and by definition of electric field, can be expressed as follows:

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maw [93]

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