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Anna35 [415]
2 years ago
12

You are observing traffic in a single lane of a highway at a specific location. You measure the average headway and average spac

ing of passing vehicles as 3 seconds and 150 feet, respectively. Calculate the flow, speed and density of the traffic stream in this lane.
Physics
1 answer:
HACTEHA [7]2 years ago
6 0

Answers:

1) flow of traffic =   1198.8 veh/h

2) average speed = 34.09 mi/h

3)density of traffic = 34.34 veh/mi

Explanation:

1) to find flow of traffic

we use the relation

q=1/h

where q is the traffic flow and h is average time headway.

h= 3s       (given)

insert the value

q=1/3=0.333veh/s=1198.8veh/h

2) to find average traffic speed

use the relation

u=S/h

where u is average speed S is average spacing and h average time

S= 150 ft   (given)

so inserting the values

u= 150/3*3600/5280=34.09 mi/h

3) density of traffic

K=q/u

where K is density of traffic q is flow of traffic and u is average speed

inserting values from above solved parts

K=1198.8/34.09=34.34 veh/mi

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Answer:

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Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.  

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

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Assuming the arm is just being stretched and there is no rotation going on,

                        T_{Deltoid} = 0

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       ⇒           T_{Deltoid} = T_{arm}

                  r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}

Where,

r_{d} is radius of the deltoid

F_{d} is the force of the deltiod

\alpha_{d} is the angle of the deltiod

r_{a} is the radius of the arm

F_{a} is the force of the arm , F_{a} = mg  which is the mass of the arm and acceleration due to gravity

\alpha_{a} is the angle of the arm

The force of the deltoid muscle is,

                                 F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}

but F_{a} = mg ,

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Explanation:

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