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1. Find the min. sight distance in ft for a vehicle travelling at 75mph, the perception reaction time 0.05 min and (a/g)=0.35 a) 825 b) 727 c) 558 d) 377 2. A car needed 18.5ft to stop while it was driving down hill on a 5% grade road segment if you know that the car accelerating at 11.5ft/sec

2
, then determine the speed of the car in km/h a) 13.06 b) 12.86 c) 21.02 d) 20.69 3. The factor used to show the variety of the demand for a road segment during the peak hour, this can be used to define: a) ADI b) AADT c) PHV d) PHF

User Ayse
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Answer:

1. The minimum sight distance depends on the perception/reaction time, vehicle speed, and the coefficient of friction between tire and roadway. The formula to calculate the minimum sight distance is:

Minimum sight distance = Perception/reaction distance + braking distance

where perception/reaction distance is the distance travelled by the vehicle during the time it takes the driver to perceive the danger and apply the brakes, and the braking distance is the distance travelled by the vehicle during the time it takes to stop after the brakes are applied.

Given that the vehicle speed is 75mph, the perception/reaction time is 0.05 minutes, and (a/g) = 0.35, we can calculate the minimum sight distance as follows:

Perception/reaction distance = Vehicle speed x Perception/reaction time

= (75 x 1.47) / 60

= 1.84 miles/hour

Braking distance = (Vehicle speed x Vehicle speed) / (2 x (a/g))

= (75 x 75) / (2 x (0.35 x 32.2))

= 311.32 feet

Minimum sight distance = Perception/reaction distance + braking distance

= (1.84 x 5280) + 311.32

= 9759.52 feet

Therefore, the minimum sight distance is 9759.52 feet, which can be approximated to 9760 feet or option A.

2. The speed of the car can be calculated using the following formula:

v^2 - u^2 = 2as

where v is the final velocity of the car, u is the initial velocity of the car (which is assumed to be zero), a is the acceleration of the car, and s is the stopping distance.

Given that the stopping distance is 18.5 feet and the acceleration is 11.5 feet/sec^2, we can calculate the speed as follows:

v^2 - 0 = 2 x 11.5 x 18.5

v^2 = 425.5

v = sqrt(425.5)

v = 20.69 feet/second

To convert this to km/h, we can multiply by 3.6:

Speed in km/h = 20.69 x 3.6 = 74.5 km/h

Therefore, the speed of the car is 74.5 km/h or option D.

3. The factor used to show the variety of the demand for a road segment during

Step-by-step explanation:

User Rohan Amrute
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