Answer:
Explanation:
From the given information;
Let assume that:
the wheel radius = 15 inches
the driveline slippage = 3%; &
the gear reduction ratio (overall) = 2.5 to 1
So; using the equation:
[tex]v_1= \dfrac{2 \pi r n_o (1 -i)}{\varepsilon_o}[/tex]
[tex]v_1= \dfrac{2 \times 3.14 \times \dfrac{15}{12} \times \dfrac{85}{100} (1 -0.03)}{2.5}[/tex]
[tex]v_1= \dfrac{2 \times 3.14 \times \dfrac{15}{12} \times \dfrac{85}{100} (0.97)}{2.5}[/tex]
[tex]v_1 = 126.92 \ fp^3[/tex]
[tex]frl = 0.01 ( 1+ \dfrac{v}{147}) \ if \ v \ is \ ft/sec[/tex]
[tex]frl = 0.01 \Bigg( 1+ \dfrac{\dfrac{126.92 +(20)1.47 }{2} }{147}\Bigg)[/tex]
[tex]S = \dfrac{v_b ( v_1^r-v_2^r)}{2g(n_b \mu + frl \pm sin \ y}[/tex]
where;
[tex]\mu = 0.6[/tex]
[tex]291 = \dfrac{1.64( 126.92^2-29.9^2)}{64.4(n_b \times 0.6 +0.01532 +0.03}[/tex]
[tex]n_b = 1.33 \to which \ is \ not \ possible[/tex]
However;
[tex]n_b \mu = 1.33(0.6) = 0.80[/tex]
[tex]\mu = 0.9 \to[/tex] if the car's anti-clock breaking system did not fail
Thus;
[tex]n_b (0.9) = 0.80[/tex]
[tex]n_b =\dfrac{ 0.80}{(0.9) }[/tex]
[tex]n_b = 0.89[/tex]
Hence, the distance is possible if the anti-clock breaking system did not fail.
