Question

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 0.536 and 0.599, whereas the effective coefficient of friction between a skidding (locked) wheel and asphalt ranges between 0.350 and 0.480. Vehicles of all types travel on the road, from small VW bugs weighing 1110 lb to large trucks weighing 8220 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.

Which of the following affect the soundness of your decision?

1. Reaction time of the drivers is not taken into account?

2. Newton\'s second law does not apply to this situation.

3. Precipitation from the fog can lower the coefficients of friction

4. Drivers cannot be expected to obey the posted speed limit.

Answer 1

We know that

1mph =1.46667ft/s then Residents complain that the speed limit in the area (55 mph) =80.7ft/s

Now to 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 is given by

From the kinetic energy of trucks

KE =(1/2)mv² =0.5*(8220 lb/32.2 ft/s²) * (80.7ft/s)²=831252.6056ft.lb

The worst case friction: Ffw = µmg = 0.536 * 8220lb = 4405.92 lb

then the stopping distance is given by d = 831252.6056ft.lb / 4405.92 lb = 188.667 ft

Now the best case friction: Ffb = 0.599 * 8220lb = 4923.78 lb

then the stopping distance is given by d = 831252.6056ft.lb / 4923.78lb = 168.82 ft

For the bugs now the kinetic energy is given by
Ek = 0.5(1110lb / 32.2ft/s²) * (80.7ft/s)² = 112249.4394 ft·lb

The worst case friction for the bug is : Ffw = 0.536 * 1110lb = 594.96 lb

then the stopping distance d = 188.66 ft
best case friction: Ffb = 0.599 * 1110lb = 664.89lb

then the stopping distance d =112249.4394 ft·lb/664.89lb = 168.824 ft

From the given problem the maximum allowable distance is 155 ft, we have to reduce the maximum allowable kinetic energy of the vehicles.

The worst case of friction for the bug over 155ft is =594.96lb*155ft =92218.2ft.lb

Now the corresponding kinetic energy for the vechiles is given by

                       KE =(1/2)mv²===> 92218.8ft.lb =0.5*(1110lb/32.2ft/s²)v² ====>v² =5350.352

Then speed v =73.146ft/s=49.87mph