Answer:
83.1946504051 m
Explanation:
u = Initial velocity = [tex]60\ mph=\dfrac{60\times 1609.34}{3600}=26.82233\ m/s[/tex]
s = Displacement = [tex]123\ ft=\dfrac{123}{3.281}=37.4885705578\ m[/tex]
[tex]\theta[/tex] = Angle = [tex]26^{\circ}[/tex]
[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-26.82233^2}{2\times 37.4885705578}\\\Rightarrow a=-9.5954230306\ m/s^2[/tex]
Coefficient of friction
[tex]\mu=-\dfrac{a}{g}\\\Rightarrow \mu=\dfrac{9.5954230306}{9.81}\\\Rightarrow \mu=0.978126710561[/tex]
[tex]mg sin\theta - u mg cos\theta = ma\\\Rightarrow a=9.81(sin26-0.978126710561cos26)\\\Rightarrow a=-4.32382\ m/s^2[/tex]
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-26.82233^2}{2\times -4.32382}\\\Rightarrow s=83.1946504051\ m[/tex]
The stopping distance is 83.1946504051 m
The stopping distance on a roadway sloping downward at an angle of 26.0° is 83.1 m.
The acceleration of the Chevrolet Corvette is calculated by applying the third kinematic equation as follows;
[tex]a = \frac{v^2 - u^2}{2s}[/tex]
where;
[tex]a = \frac{0 - (26.82)^2}{2(37.49)} \\\\a = - 9.6 \ m/s^2[/tex]
The coefficient of friction is calculated as follows;
[tex]\mu = \frac{a}{g} \\\\\mu = \frac{9.6}{9.8}\\\\ \mu = 0.98[/tex]
[tex]mg sin(\theta) - \mu mg cos(\theta) = ma\\\\g(sin\theta \ - \mu cos\theta) = a\\\\9.8(sin26- \ 0.98cos26) = a\\\\-4.33 \ m/s^2 = a[/tex]
The stopping distance is calculated as follows;
[tex]v^2 = u^2 - 2as\\\\s = \frac{v^2 -u^2}{2s} \\\\s = \frac{0 - (26.82)^2}{2(-4.33)} \\\\s = 83.1 \ m[/tex]
Thus, the stopping distance on a roadway sloping downward at an angle of 26.0° is 83.1 m.
Learn more about coefficient of friction here: https://brainly.com/question/20241845
