A minor road intersects a major 4-lane divided road with a design speed of 55mph and a median width of 8 feet. 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 3%.
1 answer:
Answer:
stopping sight diatnce is =158 meters
Explanation:
The Design Speed =55 MPH(Miles per Hour)
1 miles per hour =0.447 meter pe second
then 55MPH=55*0.447=24.58 meter per second
Consider recation time =2.5 second
Consider Co efficinet friction is 0.35
As per the Stopping sight distance based on the Breaking distance+ Lag distance
SSD=Vt+(V^2/(2gf))
SSD=24.58+(24.58^2/(2*9.81*0.35))
The total stopping sight distance is =150 meters
if we consider the Approaching garde on the minor road is 3%
SSD=Vt+(V^2/(2g(f-n/100))
SSD=24.58+(24.58^2/(2*9.81*(0.35-3/100))
then stopping sight diatnce is =158 meters
See attachment for workings
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Answer:
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