A 2200 kg car is travelling at 45 km per hour when its brakes are applied and it skids to a stop. If the coefficient of friction between the road and the tires is 0.70 how far does the car go before stopping?

11.3m i think

To find the distance the car travels before stopping, we can use the equation:

\[ F_{friction} = \mu \cdot m \cdot g \]

where:
- \( F_{friction} \) is the frictional force between the car and the road,
- \( \mu \) is the coefficient of friction (0.70 in this case),
- \( m \) is the mass of the car (2200 kg),
- \( g \) is the acceleration due to gravity (approx. 9.8 m/s²).

The frictional force can be calculated by:

\[ F_{friction} = m \cdot a \]

where:
- \( a \) is the deceleration of the car.

Since the car is skidding to a stop, we can assume that the acceleration is negative and equal to:

\[ a = \frac{{v_f - v_i}}{{t}} \]

where:
- \( v_f \) is the final velocity (0 m/s),
- \( v_i \) is the initial velocity (45 km/h converted to m/s),
- \( t \) is the time taken to stop (unknown).

Now let's solve for time:

\[ t = \frac{{v_f - v_i}}{{a}} \]

Substituting the given values into the equation:

\[ t = \frac{{0 \, \text{m/s} - (45 \, \text{km/h} \cdot \frac{{1000 \, \text{m}}}{{3600 \, \text{s}}})}}{{a}} \]

Next, let's find the deceleration:

\[ a = \mu \cdot g \]

Substituting the given values:

\[ a = 0.70 \cdot 9.8 \, \text{m/s²} \]

Finally, we can calculate the time \( t \):

\[ t = \frac{{-45 \, \text{km/h} \cdot \frac{{1000 \, \text{m}}}{{3600 \, \text{s}}}}}{{0.70 \cdot 9.8 \, \text{m/s²}}} \]

Simplifying the equation further:

\[ t = \frac{{-45 \, \text{km/h} \cdot \frac{{1000 \, \text{m}}}{{3600 \, \text{s}}}}}{{0.70 \cdot 9.8 \, \text{m/s²}}} \approx -13.2 \, \text{s} \]

We have a negative time because we assumed that the car is decelerating.

To find the distance \( d \) traveled by the car, we can use the formula:

\[ d = v_i \cdot t + \frac{1}{2} \cdot a \cdot t^2 \]

Substituting the known values:

\[ d = (45 \, \text{km/h} \cdot \frac{{1000 \, \text{m}}}{{3600 \, \text{s}}}) \cdot (-13.2 \, \text{s}) + \frac{1}{2} \cdot (0.70 \cdot 9.8 \, \text{m/s²}) \cdot (-13.2 \, \text{s})^2 \]

Evaluating the equation:

\[ d \approx -63.5 \, \text{m} \]

The negative sign indicates that the car is moving in the opposite direction. However, since we are interested in the magnitude of the distance, we can take the absolute value:

\[ |d| \approx 63.5 \, \text{m} \]

Therefore, the car skids approximately 63.5 meters before coming to a stop.

To determine how far the car skids before coming to a stop, we can use the equation of motion that relates the distance traveled, the initial velocity, the final velocity, and the coefficient of friction.

The equation we'll use is:

\( v_f^2 = v_i^2 + 2ad \),

where:
\( v_f \) = final velocity (which is 0 m/s since the car comes to a stop)
\( v_i \) = initial velocity
\( a \) = acceleration
\( d \) = distance covered

First, let's convert the initial velocity from km/h to m/s. Since 1 km is equal to 1000 m and 1 hour is equal to 3600 seconds, we can calculate the initial velocity as:

\( v_i = 45 \times \frac{1000}{3600} \) m/s.

\( v_i = 12.5 \) m/s.

Next, we need to calculate the acceleration. The acceleration of the car is given by the equation:

\( a = \mu \times g \),

where:
\( \mu \) = coefficient of friction
\( g \) = acceleration due to gravity (approximately 9.8 m/s^2)

\( a = 0.70 \times 9.8 \) m/s^2.

\( a = 6.86 \) m/s^2.

Now, we can use the equation of motion to calculate the distance the car skids:

\( 0^2 = (12.5)^2 + 2 \times 6.86 \times d \).

Simplifying the equation:

\( 0 = 156.25 + 13.72 \times d \).

Re-arranging the equation:

\( 13.72 \times d = -156.25 \).

Finally, we can solve for distance \( d \):

\( d = \frac{-156.25}{13.72} \).

\( d \approx -11.38 \) meters.

Since distance cannot be negative in this scenario, we disregard the negative sign. Thus, the car skids approximately 11.38 meters before coming to a stop.

Vf^2=Vi^2+2ad where a= -F/mass=-mu*mg/m=-mu*g

solve for d.