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 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 7710 lb. 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: ft 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.

(A) Applying Newton's second lawalong x-axis \Rightarrow F_{r}=m aalong y-axis \Rightarrow \quad N=m gAs we knowF_{r}=\mu N(1) [fr \rightarrow frictional force]Equating equation (D, (iI) and (III)\begin{array}{l}\mu m g=m a \\\Rightarrow a=f i g \\\end{array}whire g=9.81 \mathrm{~m} / \mathrm{s}^{2}(00) g=32 \mathrm{ft} / \mathrm{s}^{2}for minimum \mu,\mu=0.350, a_{1}=11.2 \mathrm{ft} / \mathrm{s}^{2} \quad\left[a_{1} \Rightarrow \text { minimum value }\right]For maximum \mu\begin{array}{l}\text { For maximum } \mu \\\qquad \mu=0.599, \quad a_{2}=19.168 \mathrm{ft} / \mathrm{s}^{2}\left[a_{2} \Rightarrow \text { maximumvalue }\right] \\v_{0}=55 \mathrm{mph} \times\left(\frac{5280 \mathrm{ft}}{1 \text { mile }}\right) \times\left(\frac{2 \mathrm{~h}}{3600 \mathrm{~s}}\right) \\v_{0}=80.667 \mathrm{ft} / \mathrm{s}^{2}\end{array}(1) Maximum braking distance \left(x_{1}\right)\begin{array}{l}x_{1}=\frac{v_{0}^{2}}{2} \times a_{1}=\frac{(80.667)^{2}}{2} \times: 11.2 \\x_{1}=290.5 \mathrm{ft}\end{array}(11) Minimum breaking distance \left(x_{2}\right)\begin{array}{l}x_{2}=\frac{v_{0}^{2}}{2} \times a_{2}=\frac{(80.667)^{2}}{2} \times 19.168 \\x_{2}=169.74 \mathrm{ft}\end{array}(B) Maximum desired speed limit ( v_{\text {max }} )V_{0}=\sqrt{2 \times a \times x}\left[a=11.2 \mathrm{ft} / \mathrm{s}^{2} \text { and } 19.1 .68 \mathrm{ft} / \mathrm{s}^{2}\right][x=155 f t]V_{0}=58.92 \mathrm{ft} / \mathrm{s} \text { and } 77.08 \mathrm{ft} / \mathrm{s}There fore maximum desired speed limitv_{\text {max }}=58.92 \mathrm{ft} / \mathrm{s} \text { which is with }minimum acceleration. ...