Answer:
The car skids in a distance of 61.275 meters.
Explanation:
Since the only force exerted on the car is the kinetic friction between the car and the horizontal road, deceleration of the vehicle ([tex]a[/tex]), measured in meters per square second, is determined by the following expression:
[tex]a = \mu_{k}\cdot g[/tex] (1)
Where:
[tex]\mu_{k}[/tex] - Coefficient of kinetic friction, no unit.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If we know that [tex]\mu_{k} = 0.52[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], then the net deceleration of the vehicle is:
[tex]a = 0.52\cdot \left(-9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]a = -5.1\,\frac{m}{s^{2}}[/tex]
The distance covered by the car is finally calculated by this kinematic expression:
[tex]\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}[/tex] (2)
Where:
[tex]v_{o}[/tex], [tex]v[/tex] - Initial and final speed, measured in meters per second.
[tex]a[/tex] - Net deceleration, measured in meters per square second.
If we know that [tex]v_{o} = 25\,\frac{m}{s}[/tex], [tex]v = 0\,\frac{m}{s}[/tex] and [tex]a = -5.1\,\frac{m}{s^{2}}[/tex], then the distance covered by the car is:
[tex]\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(25\,\frac{m}{s} \right)^{2}}{2\cdot \left(-5.1\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\Delta s = 61.275\,m[/tex]
The car skids in a distance of 61.275 meters.
