Question

A car without ABS (antilock brake system) was moving at \(15.0 \mathrm{~m} / \mathrm{s}\) when the driver slammed on the brakes to make a sudden stop. The coefficients of static and kinetic friction between the tires and the road are 0.550 and \(0.430,\) respectively. a) What was the acceleration of the car during the interval between braking and stopping? b) How far did the car travel before it stopped?

Short answer

Answer: The car travels 26.65 meters before stopping.

Step-by-step solution

Part A: Finding the Acceleration

First, we need to find the force of kinetic friction acting on the car when the driver slams on the brakes. The force of kinetic friction is given by the formula: $$ F_k = μ_k F_n $$ where \(F_k\) is the force of kinetic friction, \(μ_k\) is the coefficient of kinetic friction, and \(F_n\) is the normal force (which is equal to the weight of the car in this case). Since the car is on a flat road, the normal force \(F_n\) is equal to the force due to gravity: \(mg\), where \(m\) is the mass of the car and \(g\) is the acceleration due to gravity (approximately \(9.81 \mathrm{~m/s^2}\)). We can plug in the given coefficient of kinetic friction, \(μ_k = 0.430\), to find the force of kinetic friction acting on the car: $$ F_k = 0.430 \cdot mg $$ Next, we can use Newton's second law of motion, \(F = ma\), where \(F\) is the net force acting on the car, \(m\) is its mass, and \(a\) is its acceleration. In this case, the force of kinetic friction is the only force acting on the car, so: $$ F_k = ma $$ Substituting the expression for \(F_k\) into this equation, we get: $$ 0.430 \cdot mg = ma $$ Notice that the mass \(m\) cancels out, leaving us with the acceleration of the car during braking: $$ a = 0.430 \cdot g $$ Now we can find the acceleration of the car by plugging in the value for \(g = 9.81 \mathrm{~m/s^2}\): $$ a = 0.430 \cdot 9.81 \mathrm{~m/s^2} = -4.22 \mathrm{~m/s^2} $$ The negative sign indicates that the acceleration is opposing the initial motion of the car (decelerating it). So, the acceleration during braking is \(-4.22 \mathrm{~m/s^2}\).

Part B: Finding the Distance Traveled

Now that we have found the acceleration of the car during braking, we can use the equations of motion with constant acceleration to find the distance it traveled before stopping. Since the car comes to a stop, the final velocity \(v_f\) is \(0\,\mathrm{m/s}\). And the initial velocity \(v_i\) is \(15.0\,\mathrm{m/s}\), as given in the exercise. We also already know the acceleration \(a = -4.22 \mathrm{~m/s^2}\). We can use the equation: $$ v_f^2 = v_i^2 + 2a \Delta x $$ where \(\Delta x\) is the distance traveled. We need to solve this equation for \(\Delta x\). Since \(v_f = 0\), the equation simplifies to: $$ 0 = v_i^2 + 2a \Delta x $$ Solving for \(\Delta x\), we get: $$ \Delta x = -\frac{v_i^2}{2a} $$ Now we can substitute in the values for \(v_i\) and \(a\): $$ \Delta x = -\frac{(15.0\,\mathrm{m/s})^2}{2 \cdot (-4.22\,\mathrm{m/s^2})} = 26.65\,\mathrm{m} $$ The car traveled a distance of \(26.65\,\mathrm{m}\) before stopping.

Key concepts

Newton's Second Law of Motion

Newton's second law of motion is central to understanding how objects move and interact with forces in our world. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (\(F = ma\)). This law illustrates that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. In other words, a larger force leads to greater acceleration, while a greater mass leads to smaller acceleration, assuming the force is constant.

For example, when a car brakes suddenly, the force exerted by the brakes through friction creates a negative acceleration, or deceleration, causing the car to slow down. If there were no other forces acting on the car—like air resistance or additional friction from the road—the problem would involve only the force of kinetic friction and the weight of the car. This simplicity allows us to apply Newton's second law directly, where the force of kinetic friction is the singular force acting in the opposite direction of the car's initial movement, thus slowing it down.

Equations of Motion

Equations of motion are formulas that describe the relationship between displacement, time, velocity, and acceleration when an object is moving with uniform acceleration. They are often employed to solve problems related to kinematics in physics.

One of the key equations is the one connecting initial velocity (\(v_i\)), final velocity (\(v_f\)), acceleration (\(a\)), and displacement (\( \text{Δ} x\)) : \[ v_f^2 = v_i^2 + 2a \text{Δ} x \$$. This equation reveals how a moving object's velocity changes with respect to its acceleration and the distance covered. In practical terms, if you know any three of these quantities, you can determine the fourth. For instance, in the absence of final velocity, such as when a car comes to a stop after braking, this equation becomes instrumental in calculating the distance traveled during that stopping phase, as illustrated in the provided solution.

Coefficient of Kinetic Friction

The coefficient of kinetic friction (\(μ_k\)) is a dimensionless quantity that characterizes the frictional resistance between two surfaces in relative motion. It is specific to the materials of the contacting surfaces and their conditions, such as smoothness or roughness, and whether they are wet or dry.

The force of kinetic friction is hence calculated by multiplying the coefficient of kinetic friction by the normal force (\(F_n\)) acting perpendicular to the surfaces in contact: \[ F_k = μ_k F_n \$$. The normal force usually equals the weight of the object when it is on a horizontal surface, making it equal to the product of mass (\(m\)) and gravitational acceleration (\(g\)).

Understanding the coefficient of kinetic friction is essential in solving problems that involve objects sliding across surfaces, such as the car in our exercise example. It helps us determine how much force is required to keep an object moving or to bring it to a stop, playing a crucial role in designing safety features, like brakes on vehicles.