Question

The driver of a car traveling up a 2% grade at an initial speed V0 applied the brakes abruptly and the vehicle slid to a complete stop at an average deceleration of 8 ft/s^2.
Was the pavement wet or dry?

Answer 1

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)

$\Sigma F_{x'} = -\mu_{k}\cdot N-m\cdot g \cdot \sin \theta = m\cdot a\\\Sigma F_{y'} = N-m\cdot g \cos \theta = 0$

After some algebraic handling, the following expression is constructed:

$-\mu_{k}\cdot m \cdot g \cdot \cos \theta - m \cdot g \cdot \sin \theta = m \cdot a$

$-g\cdot (\mu_{k} \cdot \cos \theta +\sin \theta) = a$

$\mu_{k} \cdot \cos \theta = -\frac{a}{g} - \sin \theta$

$\mu_{k} = -\frac{1}{\cos \theta}\cdot \left(\frac{a}{g} +\sin \theta \right)$

The angle of the incline is:

$\theta = \tan^{-1} 0.02$

$\theta \approx 1.146^{\textdegree}$

Now, the kinetic coefficient of friction is:

$\mu_{k} = -\frac{1}{\cos 1.146^{\textdegree}}\cdot \left(-\frac{8\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }+\sin 1.146^{\textdegree}\right)$

$\mu_{k}\approx 0.796$

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.