As a city planner, you receive complaints from local residen

Step 1
Using newton’s second law,
The force of friction,
${\displaystyle F_f=\mu N}$
${\displaystyle =\mu mg}$
Step 2
Work done by friction force,
${\displaystyle W=\mu mg\mathrm{△}x}$
Step 3
The kinetic energy,
${\displaystyle K.E=\frac{1}{2}m{v}^2}$
Step 4
Using the law of conservation of energy,
${\displaystyle K.E=W}$
${\displaystyle \frac{1}{2}m{v}^2=\mu mg\mathrm{△}x}$
Step 5
The braking distance,
${\displaystyle \mathrm{△}x=\frac{v^2}{2\mu g}}$
Step 6
When the coefficient is minimum, the braking distance is maximum and vice-versa.
Step 7
The maximum braking distance,
${\displaystyle \mathrm{△}{x}_{max}=\frac{{\left(89k\frac{m}{h}\right)}^2}{2\left(0.350\right)\left(9.8\frac{m}{s^2}\right)}{\left(\frac{1000m}{1km}\right)}^2{\left(\frac{1h}{3600s}\right)}^2}$
${\displaystyle =89.10m}$
Step 8
The minimum braking distance,
${\displaystyle \mathrm{△}{x}_{min}=\frac{{\left(89k\frac{m}{h}\right)}^2}{2\left(0.599\right)\left(9.8\frac{m}{s^2}\right)}{\left(\frac{1000m}{1km}\right)}^2{\left(\frac{1h}{3600s}\right)}^2}$
${\displaystyle =52.10m}$
Step 9
The maximum desired speed limit,
${\displaystyle v_{max}=\sqrt{2\mu g\mathrm{△}x}}$
${\displaystyle =\sqrt{2\left(0.350\right)\left(9.8\frac{m}{s^2}\right)\left(47m\right)}}$
${\displaystyle =17.96\frac{m}{s}}$
${\displaystyle \approx 18.0\frac{m}{s}}$

For maximum acceleration, use maximum coefficient for rolling
${\displaystyle a_{max}=0.599\cdot 9.8=5.8702\frac{m}{s^2}}$
so, minimum distance covered
${\displaystyle d=\frac{v^2}{2}a}$
${\displaystyle v=89k\frac{m}{h}=24.722\frac{m}{s}}$
so, minimum distance covered
${\displaystyle d=52.058m}$
Now, for minimum acceleration, use minimum coefficient for locked wheels
${\displaystyle a_{min}=0.350\cdot 9.8=3.43\frac{m}{s^2}}$
so maximum distance covered ${\displaystyle =89.1m}$
Maximum desired speed limit
${\displaystyle v=\sqrt{2\cdot a_{min}\cdot 47}}$
${\displaystyle v=17.95\frac{m}{s}}$
or ${\displaystyle v=64.6k\frac{m}{h}}$

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 $=0.5\cdot m\cdot v^2$ $(W=mg\Rightarrow m=W/g=1030lb/32.2ft/s^2=31.99lb)$ $k.e=0.5(1030/32.2)(80.67{)}^{2}ftlb=104081.96ftlb$ Now the frictional force in bad condition, is $F1=mue\cdot mg=0.350\cdot 1030lb=360.5lb$ and the distance of stopping is,(k.e change $=\text{work done}$) work done by the frictional force $=\text{change in kinetic energy of the vehicle}$ $F1\cdot s1=0.5\cdot m\cdot (v{2}^2-v{1}^2)$ here $v2=0ft/s$ $F1\cdot s1=0.5\cdot m\cdot (-v{1}^2)$ $s1=k.e/F1$ $s1=104081.96/360.5ft$ $s1=288.72ft$ (maximum distance ) the frictional force in good condition, is $F2=mue\cdot mg=0.599\cdot 1030lb=616.97lb$ and the distance of stopping is,(k.e change = work done) work done by the frictional force = change in kinetic energy of the vehicle $F2\cdot s2=0.5\cdot m\cdot (v{2}^2-v{1}^2)$ here $v2=0ft/s$ $F2\cdot s2=0.5\cdot m\cdot (-v{1}^2)$ $s2=k.e/F2$ $s2=104081.96/616.97ft$ $s2=168.7ft$ ( 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 $=0.5\cdot m(v{2}^2-v{1}^2)=0.5(8580/32.2)(80.67{)}^2=867012.85ftlb$ bad condition the frictional force in bad condition, is $F1=mue\cdot mg=0.350\cdot 8580lb=3003lb$ $F1\cdot s1=0.5\cdot m\cdot (v{2}^2-v{1}^2)$ here $v2=0ft/s$ $F1\cdot s1=0.5\cdot m\cdot (-v{1}^2)$ $s1=k.e/F1$ $s1=867012.85/3003ft$ $s1=288.71ft$ (maximum distance) and at good condition frictional force is $F2=mue\cdot mg=0.599\cdot 8580lb=5139.32lb$ $F2\cdot s2=0.5\cdot m\cdot (-v{1}^2)$ $s2=k.e/F2$ $s2=104081.96/5139.32ft$ $s2=20.25ftft$ ( 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 $W=360.5\cdot 155ft.lb=55877.5ft.lb$ the kinetic energy is $k.e=$ work done by the friction force $0.5\cdot m\cdot v^2=W_f$ $0.5(1030/32.2)(v^2)=55877.5$ solving for v