Question

The coefficient of kinetic friction between the tires of a car and a horizontal road surface is 0.52. If the car is traveling at an initial speed of 25 m/s, and then slams on the breaks so the car skids straight ahead to a stop, how far does the car skid?

Answer 1

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 ($a$), measured in meters per square second, is determined by the following expression:

$a = \mu_{k}\cdot g$ (1)

Where:

$\mu_{k}$ - Coefficient of kinetic friction, no unit.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $\mu_{k} = 0.52$ and $g = -9.807\,\frac{m}{s^{2}}$, then the net deceleration of the vehicle is:

$a = 0.52\cdot \left(-9.807\,\frac{m}{s^{2}} \right)$

$a = -5.1\,\frac{m}{s^{2}}$

The distance covered by the car is finally calculated by this kinematic expression:

$\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$ (2)

Where:

$v_{o}$, $v$ - Initial and final speed, measured in meters per second.

$a$ - Net deceleration, measured in meters per square second.

If we know that $v_{o} = 25\,\frac{m}{s}$, $v = 0\,\frac{m}{s}$ and $a = -5.1\,\frac{m}{s^{2}}$, then the distance covered by the car is:

$\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(25\,\frac{m}{s} \right)^{2}}{2\cdot \left(-5.1\,\frac{m}{s^{2}} \right)}$

$\Delta s = 61.275\,m$

The car skids in a distance of 61.275 meters.