Answer:
The pavement was dry.
Explanation:
The car is modelled by means of the equations of equilibrium: (x' is for the axis parallel to the incline, y' is for the axis perpendicular to the incline)
[tex]\Sigma F_{x'} = -\mu_{k}\cdot N-m\cdot g \cdot \sin \theta = m\cdot a\\\Sigma F_{y'} = N-m\cdot g \cos \theta = 0[/tex]
After some algebraic handling, the following expression is constructed:
[tex]-\mu_{k}\cdot m \cdot g \cdot \cos \theta - m \cdot g \cdot \sin \theta = m \cdot a[/tex]
[tex]-g\cdot (\mu_{k} \cdot \cos \theta +\sin \theta) = a[/tex]
[tex]\mu_{k} \cdot \cos \theta = -\frac{a}{g} - \sin \theta[/tex]
[tex]\mu_{k} = -\frac{1}{\cos \theta}\cdot \left(\frac{a}{g} +\sin \theta \right)[/tex]
The angle of the incline is:
[tex]\theta = \tan^{-1} 0.02[/tex]
[tex]\theta \approx 1.146^{\textdegree}[/tex]
Now, the kinetic coefficient of friction is:
[tex]\mu_{k} = -\frac{1}{\cos 1.146^{\textdegree}}\cdot \left(-\frac{8\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }+\sin 1.146^{\textdegree}\right)[/tex]
[tex]\mu_{k}\approx 0.796[/tex]
A typical kinetic coeffcient of friction between a car tire and asphalt is about 0.6, if pavement would be wet, such indicator would be significantly lower. Therefore, the deceleration occurs on dry pavement.
