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 0.842 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 1250 lb to large trucks weighing 7560 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.

Answer 1

Consider the case when the car is rolling to a stop :

acceleration of the car due to friction is given as

a = - $\mu$_r g                                              where $\mu$_r = coefficient of friction force for rolling

so a_min = - $\mu$_rmin g = - (0.842) (9.8) = - 8.2516 m/s²

a_max = - $\mu$_rmax g = - (0.941) (9.8) = - 9.2218 m/s²

v_f = final velocity = 0 m/s

v_i = initial velocity = 55 mph = 55 mph (0.447 m/s/1 mph) = 24.58 m/s

d = stopping distance

using the equation

v_f² = v_i² + 2 a d

0² = 24.58² + 2 a d

d = - 24.58² /(2a)

Since stopping distance is inversly related to the acceleration , the stopping distance is minimum when the acceleration due to friction is maximum

hence d_min = minimum stopping distance = - 24.58² /(2a_max) = - 24.58² /(2(- 9.2218)) = 32.76 m = 32.76 m (3.28 ft/1 m) = 107.5 ft

d_max = maximum stopping distance = - 24.58² /(2a_min) = - 24.58² /(2(- 8.2516)) = 36.6 m = 36.6 m (3.28 ft/1 m) = 120.1 ft

Consider the case when the car is skidding to a stop :

acceleration of the car due to friction is given as

a = - $\mu$_s g                                              where $\mu$_s = coefficient of friction force for skidding

so a_min = - $\mu$_smin g = - (0.550) (9.8) = - 5.39 m/s²

a_max = - $\mu$_smax g = - (0.754) (9.8) = - 7.39 m/s²

v_f = final velocity = 0 m/s

v_i = initial velocity = 55 mph = 55 mph (0.447 m/s/1 mph) = 24.58 m/s

d = stopping distance

using the equation

v_f² = v_i² + 2 a d

0² = 24.58² + 2 a d

d = - 24.58² /(2a)

Since stopping distance is inversly related to the acceleration , the stopping distance is minimum when the acceleration due to friction is maximum

hence d_min = minimum stopping distance = - 24.58² /(2a_max) = - 24.58² /(2(- 7.39)) = 40.88 m = 40.88 m (3.28 ft/1 m) = 134.1 ft

d_max = maximum stopping distance = - 24.58² /(2a_min) = - 24.58² /(2(- 5.39)) = 56.05 m = 56.05 m (3.28 ft/1 m) = 183.8 ft