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(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. ...