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 (89 km/h) 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 47 meters. Because 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 cars with a mass of 571 kg to large trucks with a mass of 4239 kg. Considering that some drivers will brake properly when slowing down and others will skid to stop, calculate the minimum and maximum braking distance needed to ensure that all vehicles traveling at the posted speed limit can stop before reaching the intersection. maximum: minimum: Im 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. maximum speed limit: km/h

Answer 1

For maximum acceleration, use maximum coefficient for rolling

a_max = 0.599 * 9.8 = 5.8702 m/s²

so,

minimum distance covered

d = v² / 2a

v = 89 km/h = 24.722 m/s

so,

minimum distance covered

d = 52.058 m

Now,

for minimum acceleration, use minimum coefficient for locked wheels

a_min = 0.350 * 9.8 = 3.43 m/s²

so

maximum distance covered = 89.1 m

----------------------------------------------------------------

Maximum desired speed limit

v = sqrt ( 2 * a_min * 47)

v = 17.95 m/s

or

v = 64.6 km/h

Answer 2

Given data  speed limit is  55 mph = 80.67 ft/s 

under fog conditions visibility can reduce to  155 ft

coefficient of friction between a rolling wheel and asphalt ranges between 0.536 and 0.599 given by the state highway department

and the effective coefficient of friction between a skidding (locked) wheel and asphalt ranges between 0.350 and 0.480

here one thing we should remember, regarding the coefficient of friction , if the coefficient of friction is more so that more work will be done by the friction so that the vehicle can come to rest in short time and distance ( called best case most suitable ) and in reverse less work come to rest in long time or making much displacement ( called worst case or bad conditions)

so first we can calculate the kinetic energies of small VW bugs and large trucks weighing 1030 lb, 8580 lb respectively

For small VW bugs 

kinetic energy is  k.e= 0.5*m*v^2 

(W = mg ==>m = W/g = 1030 lb/32.2 ft/s^2 = 31.99 lb)

k.e = 0.5(1030/32.2)(80.67)^2 ft lb = 104081.96 ft lb

Now the frictional force in bad condition, is  F1 = mue*mg = 0.350*1030 lb = 360.5 lb

and the distance of stopping is,(k.e change  = work done)

work done by the frictional force  = change in kinetic energy of the vehicle 

F1*s1 = 0.5*m*(v2^2-v1^2) 

here v2= 0 ft/s

F1*s1 = 0.5*m*(-v1^2)

s1 = k.e/F1 

s1 = 104081.96/360.5 ft  

s1 =  288.72 ft  (maximum distance )

the frictional force in good condition, is  F2 = mue*mg = 0.599*1030 lb = 616.97 lb

and the distance of stopping is,(k.e change  = work done)

work done by the frictional force  = change in kinetic energy of the vehicle 

F2*s2 = 0.5*m*(v2^2-v1^2) 

here v2= 0 ft/s

F2*s2 = 0.5*m*(-v1^2)

s2 = k.e/F2 

s2 = 104081.96 /616.97 ft

s2 = 168.7 ft ( minimum distance )

similarly for the large trucks we can calculate the s1 and s2 values 

 large truck of weight 8580 lb 

kinetic energy change is  k.e = 0.5*m(v2^2-v1^2) = 0.5(8580/32.2)(80.67)^2  = 867012.85 ft lb

bad condition

the frictional force in bad condition, is  F1 = mue*mg = 0.350*8580 lb = 3003 lb

F1*s1 = 0.5*m*(v2^2-v1^2) 

here v2= 0 ft/s

F1*s1 = 0.5*m*(-v1^2)

s1 = k.e/F1 

s1 = 867012.85/3003 ft  

s1 =  288.71 ft   (maximum distance)

and at good condition 

frictional force is  F2 = mue*mg = 0.599*8580 lb = 5139.32 lb

F2*s2 = 0.5*m*(-v1^2)

s2 = k.e/F2 

s2 = 104081.96 /5139.32 ft

s2 = 20.25 ft ft ( minimum distance)

----

2. to allow all vehicles to come safely to a stop before reaching the intersection, desired maximum speed limit is  v_l =?

it is more appropriate to take the case of small VW bugs, bad condition so that the other cases the speed is less than this case so 

given limit of distance is  155 ft and the work done W = 360.5*155 ft.lb = 55877.5 ft.lb

the kinetic energy is  k.e = work done by the friction force

0.5*m*v^2 = W_f 

0.5(1030/32.2)(v^2) = 55877.5 

solving for  v 

v = 59.11 ft/s = 40.30 mph