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SOVA2 [1]
3 years ago
8

A six-lane divided multilane highway (three lanes in each direction) has a measured free-flow speed of 50 mi/h. It is on mountai

nous terrain with a traffic stream consisting of 7% large trucks and buses and 3% recreational vehicles. The driver population adjustment in 0.92. One direction of the highway currently operates at maximum LOS C conditions and it is known that the highway has PHF = 0.90.
Required:
How many vehicles can be added to this highway before capacity is reached, assuming the proportion of vehicle types remains the same but the peak-hour factor increases to 0.95?
Engineering
1 answer:
Ilia_Sergeevich [38]3 years ago
4 0

Answer:

2901 vehicles

Explanation:

We are given;

Percentage of large trucks & buses; p_t = 7% = 0.07

Percentage of recreational vehicles; p_r = 3% = 0.03

PHF = 0.90

Driver population adjustment; f_p = 0.92

First of all, let's Calculate the heavy vehicle factor from the formula;

f_hv = 1/[1 + p_t(e_t - 1) + p_r(e_r - 1)]

Where;

e_t = passenger car equivalents for trucks and buses

e_r = passenger car equivalents for recreational vehicles

From the table attached, for a mountainous terrain, e_t = 6 and e_r = 4. Thus;

f_hv = 1/[1 + 0.07(6 - 1) + 0.03(4 - 1)]

f_hv = 1.44

Let's now calculate the initial hourly volume from the formula;

v_p = V1/(PHF × N × f_hv × f_p)

Where;

v_p = 15-minute passenger-car equivalent flow rate

V1 = hourly volume

N = number of lanes in each direction

From online tables of LOS criteria for multilane freeway segments, v_p = 1300 pc/hr/ln

Thus;

1300 = V1/(0.9 × 3 × 1.44 × 0.92)

V1 = 1300 × (0.9 × 3 × 1.44 × 0.92)

V1 = 4650 veh/hr

Now, let's Calculate the final hourly volume;

From online sources, the maximum capacity of a 6 lane highway with free-flow speed of 50 mi/h is 2000 pc/hr/ln.

We are told the online peak-hour factor increases to 0.95 and so PHF = 0.95.

Thus;

2000 = V2/(0.95 × 3 × 1.44 × 0.92)

V2 = 2000(0.95 × 3 × 1.44 × 0.92)

V2 = 7551 veh/hr

Number of vehicles added to the highway = V2 - V1 = 7551 - 4650 = 2901 vehicles

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The distance below the top of the cliff that the two balls cross paths is 7.53 meters.

<u>Given the following data:</u>

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<u>Scientific data:</u>

  • Acceleration due to gravity (a) = 9.8 m/s^2.

To determine how far (distance) below the top of the cliff that the two balls cross paths, we would apply the third equation of motion.

<h3>How to calculate the velocity.</h3>

Mathematically, the third equation of motion is given by this formula:

V^2 = U^2 +2aS

<u>Where:</u>

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Substituting the parameters into the formula, we have;

V^2 = 0^2 +2(9.8) \times 30\\\\V^2 = 588\\\\V=\sqrt{588}

V = 24.25 m/s.

<u>Note:</u> The final velocity of the first ball becomes the initial velocity of the second ball.

The time at which the two balls meet is calculated as:

Time = \frac{S}{U} \\\\Time = \frac{30}{24.25}

Time = 1.24 seconds.

The position of the ball when it is dropped from the cliff is calculated as:

y_1 = h-\frac{1}{2} at^2\\\\y_1 = 30-\frac{1}{2} \times 9.8 \times 1.24^2\\\\y_1 = 30-7.53\\\\y_1=22.47\;meters

Lastly, the distance below the top of the cliff is calculated as:

Distance = 30-22.47

Distance = 7.53 meters.

Read more on distance here: brainly.com/question/10545161

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The resultant force is directed along the positive x axis and has a magnitude of 1330 N.
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Answer:

the magnitude of F_A is 752 N

the direction theta of F_A is 57.9°

Explanations:

Given that,

Resultant force = 1330 N in x direction

∑Fx = R

from the diagram of the question which i uploaded along with this answer

FB = 800 N

FAsin∅ + FBcos30 = 1330 N

FAsin∅ = 1330 - (800 × cos30)

FA = 637.18 / sin∅

Now ∑Fx = 0

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(637.18 / sin∅)cos∅ = 800 × sin30

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