As a city planner, you receive complaints from local residents about the safety

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 9030 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.

Max

Min

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

Explanation / Answer

let,

speed v=55 mi/hr=24.6 m/sec

mass of the bugs, m1=1250 lb=567 kg

mass of the truck, m2=9030 lb=4096 kg

coefficient of friction between a rolling wheel and asphalt range is,

0.842 - 0.941

coefficient of friction between a skidding wheel and asphalt range is,

0.550 - 0.754

maximum allowed distance d=155 ft =47.24 m

now,

in case of buggs:

K.E of buggs, K1=1/2*m1*v^2

K1=1/2*567*24.6^2=171562.86 J

worst case, frictional force, f1=u*m*g

f1=0.55*567*9.8=3056.13 N

and

W=f1*d

171562.86=3056.13*d

===> d=56.14 m

stopping distance at worst case, d=56.14 m <---------------answer

frictional force at best case, f1=0.941*567*9.8

f1=5228.76 N

and

stopping distance at best case, d=w/f1 =171562.86/5228.75

=====> d=32.81 m <-----------answer

=================================================================

in cae of truck:

K.E of truck, K2=1/2*m2*v^2

K1=1/2*4096*24.6^2=1239367.68 J

worst case, frictional force, f2=u*m2*g

f2=0.55*4096*9.8=22077.44 N

and

W=f2*d

1239367.68=22077.44*d

===> d=56.14 m

stopping distance at worst case, d=56.14 m <--------------

and

frictional force at best case, f2=0.941*4096*9.8

f2=37772.5 N

and

stopping distance at best case, d=w/f1 =1239367.68/37772.5

=====> d=32.81 m <--------------

===================================================

now,

maximum allowed distance, d=47.24 m

at worst case,

work done for the bug , w=f1*d

w=3056.13*47.24

w=144371.6 J

1/2*m*v^2=144371.6

1/2*567*v^2=144371.6

====> v=22.57 m/sec

maximum speed limit, vmax=22.57 m/sec =50.5 mi/hr --------->answer

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