As a city planner, you receive complaints from local residents about the safety

Question

As a city planner, you receive complaints from local residents about the safety of nearby rods and streets. One complaint concerns as stop sign at the corner of pine street and 1st street. Residence complain that the speed limit in the area 89km/h is too high to all vehicles to stop in time. Under normal conditions this is not a problem, but when fog rolls in visibility can be reduced to only 47 meters. Since fog is a common occurrence in this region, you decided to investigate. The state highway department states that the effective coefficient of friction between a rolling wheel and asphalt ranges between 0.842 and 0.941, whereas the effective coefficient of friction between a skidding (locked) wheel and asphalt ranges between 0.550 and 0.754. Vehicles of all types travel on the road, from small vw bugs weighing 475kg to large trucks weighing 4071 kg. Considering that some drivers will brakes properly when slowing and others will skid to stop calculate the minimum and maximum breaking distance need to ensure that all vehicles traveling at the posted speed limit can stop before reaching the intersection. Given that the goal to allow all vehicles to come safely to a stop before reaching the intersection, calculate the maximum desired speed limit Which of the following effects the soundness of your decision? - Newton's second law does not apply to this situation - Reaction time of the drivers is not taken into account -Drivers cannot be expected to obey the posted speed limit -Precipitation from the fog can lower the coefficients of friction

Explanation / Answer

Speed limit = 89 km/h = 24.72 m/s

distance = d = 47 m

We know that the minimum stopping distance is given by:

d = v^2/2 u g

for the drivers who skids: (for 0.842 and 0.941)

d1 = 24.72^2/2 x 0.842 x 9.81 = 37 m

d2 = 24.72^2/2 x 0.941 x 9.81 = 33.1

So the minimum and maximum distances are d(min) = 33.1 m and d(max) = 37 m

solving for v we get

v = sqrt (2 u g d)

for 0.842 and 0.941

for 0.55 and 0.754

v = sqrt (2 x 0.55 x 9.81 x 47) = 22.52 m/s = 81.07 km/h

v' = sqrt (2 x 0.754 x 9.81 x 47) = 26.37 m/s = 94.93 km/h

minimum and maximum speeds are v(min) = 22.5 m/s and v(max) = 26.37 m/s

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