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Akimi4 [234]
3 years ago
13

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 complet

e stop at an average deceleration of 8 ft/s^2.
Was the pavement wet or dry?
Engineering
1 answer:
mafiozo [28]3 years ago
3 0

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.

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