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

The car skids in a distance of 61.275 meters.

Step-by-step 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\,(m)/(s^(2))`, then the net deceleration of the vehicle is:

`a = 0.52\cdot \left(-9.807\,(m)/(s^(2)) \right)`

`a = -5.1\,(m)/(s^(2))`

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

`\Delta s = (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\,(m)/(s)`,
`v = 0\,(m)/(s)` and
`a = -5.1\,(m)/(s^(2))`, then the distance covered by the car is:

`\Delta s = (\left(0\,(m)/(s) \right)^(2)-\left(25\,(m)/(s) \right)^(2))/(2\cdot \left(-5.1\,(m)/(s^(2)) \right))`

`\Delta s = 61.275\,m`

The car skids in a distance of 61.275 meters.