Question

(a) If the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you can stop a car by locking the brakes when the car is traveling at 28.7 m/s (about 65 mi/h)? (b) On wet pavement the coefficient of kinetic friction may be only 0.25. How fast should you drive on wet pavement to be able to stop in the same distance as in part (a)? (\(Note\): Locking the brakes is \(not\) the safest way to stop.)

Short answer

(a) 52.7 meters. (b) 16.1 m/s.

Step-by-step solution

Identify Given Information for Dry Pavement

The coefficient of kinetic friction on dry pavement is given as \( \mu_k = 0.80 \), and the initial velocity of the car is \( v_i = 28.7 \text{ m/s} \). We need to find the stopping distance \( d \).

Use the Kinetic Friction Formula

The force of kinetic friction \( f_k \) can be expressed as \( f_k = \mu_k \cdot N \), where \( N \) is the normal force. For a car on a horizontal road, \( N = mg \). Thus, \( f_k = \mu_k mg \), where \( m \) is the mass of the car, and \( g \) is the acceleration due to gravity (\( 9.8 \text{ m/s}^2 \)).

Apply Newton's Second Law to Find Deceleration

According to Newton's Second Law, \( f_k = ma \). Substituting \( f_k \) from Step 2, we have \( \mu_k mg = ma \). This simplifies to \( a = \mu_k g \). For dry pavement, \( a = 0.80 \times 9.8 = 7.84 \text{ m/s}^2 \). The acceleration works as a deceleration in this context.

Use Kinematic Equation to Find Stopping Distance

The kinematic equation \( v_f^2 = v_i^2 + 2ad \) can be used, setting the final velocity \( v_f = 0 \) (since the car stops). Rearranging gives \( d = \frac{-v_i^2}{2a} \). Substituting the known values: \( d = \frac{-(28.7)^2}{2(-7.84)} = 52.7 \text{ meters} \).

Determine Conditions for Wet Pavement

For wet pavement, the coefficient of kinetic friction is \( \mu_k = 0.25 \). We need to find the initial velocity \( v_i \) to stop at the same distance \( d = 52.7 \text{ meters} \).

Calculate Deceleration on Wet Pavement

The deceleration on wet pavement is \( a = \mu_k g = 0.25 \times 9.8 = 2.45 \text{ m/s}^2 \).

Use Kinematic Equation for Wet Pavement

Use the equation \( v_f^2 = v_i^2 + 2ad \) again with \( v_f = 0 \) and solve for \( v_i \) to find: \( v_i = \sqrt{-2ad} \). Substitute \( a = -2.45 \) and \( d = 52.7 \): \( v_i = \sqrt{2 \times 2.45 \times 52.7} \approx 16.1 \text{ m/s} \).

Key concepts

Coefficient of Kinetic Friction

The coefficient of kinetic friction, denoted as \( \mu_k \), is a crucial factor in understanding how different surfaces interact with each other when they are in motion. This coefficient is a dimensionless value, meaning it has no units. It essentially quantifies the amount of frictional force that acts between two surfaces sliding against one another.

For instance, when driving on different pavements, like dry versus wet, \( \mu_k \) can greatly affect the stopping distance of a vehicle. On dry pavement, the \( \mu_k \) is higher, such as 0.80, meaning there is more frictional force helping the car come to a stop. Conversely, on wet pavement, the \( \mu_k \) drops to around 0.25, indicating less friction and a longer stopping distance if the same speed is maintained.

• Higher \( \mu_k \): more friction, shorter stopping distance.
• Lower \( \mu_k \): less friction, longer stopping distance.

Understanding \( \mu_k \) can aid in mindful driving decisions based on road conditions to ensure safety.

Newton's Second Law

Newton's Second Law of Motion is a fundamental principle that describes the relationship between the force applied to an object, its mass, and its acceleration. This law is succinctly expressed as:\[ F = ma \] where \( F \) is the force applied to the object, \( m \) is its mass, and \( a \) is the acceleration produced. In the context of stopping a car, this becomes particularly relevant. When frictional force acts on the car, Newton's Second Law helps us relate this force to the deceleration experienced by the vehicle.

For a car on a horizontal plane, the normal force, \( N \), equals the weight of the car \( mg \) (mass times gravity), and the frictional force \( f_k \) can thus be expressed as \( \mu_k \cdot N = \mu_k mg \).

• Newton's Second Law can help calculate the needed deceleration to stop.
• It connects the friction force, mass of the car, and its acceleration directly.

By using Newton's law, we can calculate how quickly a car can stop and make better predictions about safe driving speeds under different friction conditions.

Kinematic Equations

Kinematic equations are essential tools in physics that describe the motion of objects. These equations help us solve problems involving the distances traveled, time intervals, and velocities of moving objects. For the given problem, the relevant kinematic equation is:\[ v_f^2 = v_i^2 + 2ad \]This formula connects the initial velocity \( v_i \), final velocity \( v_f \), acceleration \( a \), and distance \( d \). It is used under constant acceleration conditions—crucial for calculating stopping distances for vehicles.

In this exercise, \( v_f \) is set to zero because we want to find the stopping distance (\( d \)) when the car comes to a full stop. By knowing the initial speed and deceleration (calculated from friction), we rearrange to find the stopping distance:

• When decelerating, \( d = \frac{-v_i^2}{2a} \)

Using these kinematic equations, drivers and engineers can determine how different variables, like friction and speed, impact stopping distances and ensure safety measures are adequate for various driving conditions.

Deceleration

Deceleration is a specific form of acceleration where the object's speed decreases over time. In the context of a vehicle, deceleration occurs when it slows down and comes to a stop due to friction between its tires and the road surface. This is essentially negative acceleration.

The formula for deceleration derived from the force of friction is:\[a = \mu_k \cdot g\] where \( a \) is the deceleration, \( \mu_k \) is the coefficient of kinetic friction, and \( g \) is the acceleration due to gravity (approx. \( 9.8 \text{ m/s}^2 \)). This equation helps to quantify how quickly a car can slow down based on the friction available.

• Higher friction coefficient results in greater deceleration.
• Deceleration ensures vehicles stop over a safe distance.

Understanding deceleration allows calculating necessary stopping distances under various road conditions, helping to manage driving safety and prevent accidents.