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

Charles Kingsley

Charles Kingsley

Answered

2022-01-16

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 ft. 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 1310 lb to large trucks weighing 8.40×103lb.
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.
Minimum braking distance: ? ft
Maximum braking distance: ? ftask
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: ? mph
Which factors affect the soundness of your decision?
Reaction time of the drivers is not taken into account.

Answer & Explanation

enlacamig

enlacamig

Expert

2022-01-17Added 30 answers

Step 1
Using newton’s second law,
The force of friction,
Ff=μN
=μmg
Step 2
Work done by friction force,
W=μmgx
Step 3
The kinetic energy,
K.E=12mv2
Step 4
Using the law of conservation of energy,
K.E=W
12mv2=μmgx
Step 5
The braking distance,
x=v22μg
Step 6
When the coefficient is minimum, the braking distance is maximum and vice-versa.
Step 7
The maximum braking distance,
xmax=(89kmh)22(0.350)(9.8ms2)(1000m1km)2(1h3600s)2
=89.10m
Step 8
The minimum braking distance,
xmin=(89kmh)22(0.599)(9.8ms2)(1000m1km)2(1h3600s)2
=52.10m
Step 9
The maximum desired speed limit,
vmax=2μgx
=2(0.350)(9.8ms2)(47m)
=17.96ms
18.0ms
Alex Sheppard

Alex Sheppard

Expert

2022-01-18Added 36 answers

For maximum acceleration, use maximum coefficient for rolling
amax=0.5999.8=5.8702ms2
so, minimum distance covered
d=v22a
v=89kmh=24.722ms
so, minimum distance covered
d=52.058m
Now, for minimum acceleration, use minimum coefficient for locked wheels
amin=0.3509.8=3.43ms2
so maximum distance covered =89.1m
Maximum desired speed limit
v=2amin47
v=17.95ms
or v=64.6kmh
alenahelenash

alenahelenash

Expert

2022-01-23Added 366 answers

Given data speed limit is 55mph=80.67ft/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 =0.5mv2 (W=mgm=W/g=1030lb/32.2ft/s2=31.99lb) k.e=0.5(1030/32.2)(80.67)2ft lb=104081.96ft lb Now the frictional force in bad condition, is F1=muemg=0.3501030lb=360.5lb and the distance of stopping is,(k.e change =work done) work done by the frictional force =change in kinetic energy of the vehicle F1s1=0.5m(v22v12) here v2=0ft/s F1s1=0.5m(v12) s1=k.e/F1 s1=104081.96/360.5ft s1=288.72ft (maximum distance ) the frictional force in good condition, is F2=muemg=0.5991030lb=616.97lb and the distance of stopping is,(k.e change = work done) work done by the frictional force = change in kinetic energy of the vehicle F2s2=0.5m(v22v12) here v2=0ft/s F2s2=0.5m(v12) s2=k.e/F2 s2=104081.96/616.97ft s2=168.7ft ( 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 =0.5m(v22v12)=0.5(8580/32.2)(80.67)2=867012.85ft lb bad condition the frictional force in bad condition, is F1=muemg=0.3508580lb=3003lb F1s1=0.5m(v22v12) here v2=0ft/s F1s1=0.5m(v12) s1=k.e/F1 s1=867012.85/3003ft s1=288.71ft (maximum distance) and at good condition frictional force is F2=muemg=0.5998580lb=5139.32lb F2s2=0.5m(v12) s2=k.e/F2 s2=104081.96/5139.32ft s2=20.25ft ft ( minimum distance) 2. to allow all vehicles to come safely to a stop before reaching the intersection, desired maximum speed limit is vl=? 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=360.5155ft.lb=55877.5ft.lb the kinetic energy is k.e= work done by the friction force 0.5mv2=Wf 0.5(1030/32.2)(v2)=55877.5 solving for v

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