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 0.842 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 1190 lb to large trucks weighing 8730 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.
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.
Thank you!
2 Answers

55 mph ≈ 80.7 ft/s
trucks:
Ek = ½mv² = ½ * (8730lb / 32.2 ft/s²) * (80.7ft/s)² = 882 100 ft·lb
worst case friction: Ffw = µmg = 0.55 * 8730lb = 4802 lb
→ stopping distance d = Ek / Ffw = 184 ft
best case friction: Ffb = 0.941 * 8730lb = 8215 lb
→ stopping distance d = Ek / Ffb = 107 ft
bugs:
Ek = ½ * (1190lb / 32.2ft/s²) * (80.7ft/s)² = 120 240 ft·lb
worst case friction: Ffw = 0.55 * 1190lb = 655 lb
→ stopping distance d = 184 ft
best case friction: Ffb = 0.941 * 1190lb = 1120 lb
→ stopping distance d = 107 ft
Given that the maximum allowable distance is 155 ft, we’ve got to reduce the maximum allowable Ek of the vehicles, and it appears not to matter which one we analyze.
worst case friction for bug over 155 ft entails Work = 655lb * 155ft = 101 525 ft·lb
This corresponds to Ek = 101 525 ft·lb = ½ * (1190lb / 32.2ft/s²) * v²
→ v ≈ 74 ft/s ≈ 50 mph ← maximum desired speed limit

Newton’s 2d regulation states that: Resultant stress= Mass(kg) x Acceleration(m/s^2) yet this could additionally be written as: F=ma a million) convert g into kg: mass= 5.2×1000= 5200kg F=ma = 5200/ 308=sixteen.883 =sixteen.9 N