1) As a city planner, you receive complaints from local residents about the safe
ID: 836174 • Letter: 1
Question
1) As a city planner, you receive complaints from local residents about the safety of nearby roads and streets. One complaint concerns a stop sign at the corner of Pine Street and 1st Street. Residents complain that the speed limit in the area (55 mph) is too high to allow vehicles to stop in time. Under normal conditions this is not a problem, but when fog rolls in visibility can reduce to only 155 feet. Since fog is a common occurrence in this region, you decide to investigate. The state highway department states that the effective coefficient of friction between a rolling wheel and asphalt ranges between 0.689 and 0.770, whereas the effective coefficient of friction between a skidding (locked) wheel and asphalt ranges between 0.450 and 0.617. Vehicles of all types travel on the road, from small VW bugs weighing 1170 lb to large trucks weighing 8310 lb. Considering that some drivers will brake properly when slowing down and others will skid to stop, calculate the miminim and maximum braking distance needed to ensure that all vehicles traveling at the posted speed limit can stop before reaching the intersection.
2) Given that the goal is to allow all vehicles to come safely to a stop before reaching the intersection, calculate the maximum desired speed limit. (Round your answer to the nearest whole number.)
Explanation / Answer
FOR TRUCK
Ek = 0.5*(8310/32)*80.7^2 = 845606 ft-lb
Least friction case = 0.45*8310 = 3739.5lb
distance = 845606/3739.5 = 226.12ft
Most friction case = 0.77*8310 = 6398.7 lb
distance = 845606/6398.7 = 132.15ft
FOR BUGS
Ek = 0.5*(1170/32)*80.7^2 = 119056.45 ft-lb
Least friction case = 0.45*1170 = 526.5 lb
distance = 119056.45/526.5 = 226.12ft
Most friction case = 0.77*1170 = 900.9 lb
distance = 119056.45/900.9 = 132.15ft
FOR TRUCK
Ek = 0.5*(8310/32)*80.7^2 = 845606 ft-lb
Least friction case = 0.45*8310 = 3739.5lb
distance = 845606/3739.5 = 226.12ft
Most friction case = 0.77*8310 = 6398.7 lb
distance = 845606/6398.7 = 132.15ft
FOR BUGS
Ek = 0.5*(1170/32)*80.7^2 = 119056.45 ft-lb
Least friction case = 0.45*1170 = 526.5 lb
distance = 119056.45/526.5 = 226.12ft
Most friction case = 0.77*1170 = 900.9 lb
distance = 119056.45/900.9 = 132.15ft
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