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A minor road intersects a major 4-lane divided road with a design speed of 50 mph and a median width of 12 ft. The intersection is controlled with a stop sign on the minor road. If the design vehicle is a passenger car, determine the minimum sight distance required on the major road that will allow a stopped vehicle on the minor road to safely turn left if the approach grade on the minor road is 5%.

Respuesta :

Answer:

minimum sight distance = 699 ft

Explanation:

given data

road lane = 4 divided road

median width = 12 ft

grade road = 5%

solution

we take here time gap factor for minor road vehicle when enter to major road  from table

time gap = 8.1  sec

and for median width of 12 ft

time gap = 8.2 + 0.7 ( 1 + [tex]\frac{12}{12}[/tex]  )  

time gap = 9.5 second

so minimum sight distance will be

minimum sight distance = 1.47 × design speed × time gap  

minimum sight distance = 1.47 × 50  × 9.5

minimum sight distance = 699 ft

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