Question

A car driving on level ground at 20.0 m/s slams on its brakes and skids to a halt. If the coefficient of kinetic
friction between the car’s tires and the road is 0.750, how far does the car skid before stopping? How far would
the car have skidded if it had been moving at 40.0 m/s?

Answer 1

To solve the problem, we can use the principles of physics related to motion and friction.

1. **Determine the deceleration caused by friction:**

The force of kinetic friction (\( F_k \)) can be calculated using the equation:

\[
F_k = \mu_k \cdot N
\]

where:
- \( \mu_k = 0.750 \) (the coefficient of kinetic friction),
- \( N = mg \) is the normal force (\( m \) is the mass of the car and \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity).

Therefore, the frictional force is:

\[
F_k = \mu_k \cdot mg = 0.750 \cdot mg
\]

The deceleration (\( a \)) of the car can be found using Newton's second law (\( F = ma \)), where the only horizontal force acting on the car when it brakes is the frictional force, which acts opposite to the direction of motion:

\[
ma = -F_k \implies a = -\mu_k g
\]

Substituting the known values:

\[
a = -0.750 \cdot 9.81 \approx -7.36 \, \text{m/s}^2
\]

2. **Calculate the distance skidded using the kinematic equation:**

We can use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance:

\[
v^2 = u^2 + 2as
\]

where:
- \( v = 0 \, \text{m/s} \) (final velocity),
- \( u = 20.0 \, \text{m/s} \) (initial velocity when braking),
- \( a = -7.36 \, \text{m/s}^2 \) (deceleration),
- \( s \) is the distance skidded.

Rearranging the equation to solve for \( s \):

\[
0 = (20.0)^2 + 2(-7.36)s
\]

\[
0 = 400 - 14.72s
\]

\[
14.72s = 400 \implies s = \frac{400}{14.72} \approx 27.17 \, \text{m}
\]

So, for the initial speed of 20.0 m/s, the car skids approximately **27.17 meters** before coming to a stop.

3. **Repeat the calculation for an initial speed of 40.0 m/s:**

Using the same kinematic equation with \( u = 40.0 \, \text{m/s} \):

\[
0 = (40.0)^2 + 2(-7.36)s
\]

\[
0 = 1600 - 14.72s
\]

\[
14.72s = 1600 \implies s = \frac{1600}{14.72} \approx 108.92 \, \text{m}
\]

Therefore, if the car had been moving at 40.0 m/s, it would have skidded approximately **108.92 meters** before stopping.

### Summary:
- Distance skidded at 20.0 m/s: **27.17 meters**
- Distance skidded at 40.0 m/s: **108.92 meters**