Question

A car without ABS (antilock brake system) was moving at $15.0\mathrm{m}/\mathrm{s}$ when the driver hit the brake 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 acceleration of the car during braking is -4.22 m/s², and the stopping distance is approximately 26.89 m.

Step-by-step solution

Calculate the kinetic friction force

First, we need to find the kinetic friction force acting on the car when the driver hits the brake. The formula for kinetic friction force ($F_k$) is given by: $F_k={\mu }_k{F}_N$ where ${\mu }_k$ is the coefficient of kinetic friction and $F_N$ is the normal force, which is equal to the car's weight in this case. Since the car is on a horizontal road, the normal force ($F_N$) is equal to the gravitational force, which can be calculated as: $F_N=mg$ where $m$ is the mass of the car and $g$ is the acceleration due to gravity, approximately 9.81 $\mathrm{m}/{\mathrm{s}}^2$. We are not given the mass of the car, but we can express the friction force in terms of mass and then find the acceleration.

Calculate the acceleration during braking

Using Newton's second law of motion, we have: $F=ma$ where $F$ is the net force acting on the car, $m$ is the mass of the car, and $a$ is the acceleration. Since the net force acting on the car is the kinetic friction force, we have: $ma=F_k$ Substituting the expression for kinetic friction force from Step 1, we get: $ma={\mu }_{k}mg$ We can then solve for the acceleration ($a$): $a={\mu }_{k}g$ Insert the given values for the coefficient of kinetic friction and the acceleration due to gravity: $a=0.430\times 9.81$ $a=-4.22\mathrm{m}/{\mathrm{s}}^2$ (The negative sign indicates the acceleration is opposite to the initial velocity, as it's a decelerating motion) The acceleration of the car during the interval between braking and stopping is $-4.22\mathrm{m}/{\mathrm{s}}^2$.

Calculate the stopping distance

To find the distance the car traveled before stopping, we can use the following kinematic equation: $V_f^2=V_i^2+2as$ Here, $V_f$ is the final velocity, which is 0 since the car stopped, $V_i$ is the initial velocity, which is $15.0\mathrm{m}/\mathrm{s}$, $a$ is the acceleration we found in Step 2, which is $-4.22\mathrm{m}/{\mathrm{s}}^2$, and $s$ is the stopping distance we need to find. Plug in the given values and solve for $s$: $0=(15.0{)}^2+2(-4.22)s$ Solve for $s$: $s=\frac{(15.0{)}^2}{2\times 4.22}$ $s\approx 26.89\mathrm{m}$ The car traveled approximately $26.89\mathrm{m}$ before it stopped.

Key concepts

Kinetic Friction

Kinetic friction is a type of friction that acts against the movement of two surfaces sliding past each other. This force is crucial when calculating the stopping behavior of vehicles, such as a car without ABS coming to a halt after sudden braking. When the driver hits the brakes, kinetic friction opposes the car's motion, causing it to decelerate. The force of kinetic friction ($F_k$) is calculated using the equation:

• $F_k={\mu }_k{F}_N$

The coefficient of kinetic friction (${\mu }_k$) is a dimensionless number representing the friction level between surfaces—in this case, between the car tires and the road. The normal force ($F_N$) is the force exerted by a surface perpendicular to that surface. For a car on a flat road, the normal force is simply its weight ($mg$). Therefore, kinetic friction's magnitude also depends on the car's mass and gravity's acceleration. Understanding kinetic friction helps us determine the net force acting on the car, which is essential in using Newton's laws to find other values like acceleration.

Newton's Second Law

Newton's Second Law of motion is a fundamental principle used in physics to describe how forces act on objects and change their motion. The law is encapsulated by the equation:

• $F=ma$

Here, $F$ represents the net force acting on an object, $m$ is the mass of the object, and $a$ is the acceleration of the object. In scenarios like a car stopping, after the driver has braked, the only horizontal force considered is the kinetic friction force, which pushes against the direction of movement. By rearranging the formula, we can solve for acceleration:

• $a=\frac{F_k}{m}$

Substituting the expression for kinetic friction from earlier, we derive:

• $a={\mu }_k\cdot g$

The acceleration resulting from this calculation indicates the rate of deceleration of the car. For instance, in this exercise with a coefficient of kinetic friction of 0.430 and $g$ as 9.81 $\mathrm{m}/{\mathrm{s}}^2$, the car's acceleration is found to be $-4.22\mathrm{m}/{\mathrm{s}}^2$.

Kinematic Equations

Kinematic equations are used to analyze motion in physics, particularly when dealing with uniform acceleration. They relate initial and final velocities, acceleration, time, and displacement. To find out how far the car travels before stopping, we use one of the key kinematic equations:

• $V_f^2=V_i^2+2as$

In this context:- $V_f$ is the final velocity (0 $\mathrm{m}/\mathrm{s}$ since the car stops).- $V_i$ is the initial velocity of the car (15.0 $\mathrm{m}/\mathrm{s}$).- $a$ is the acceleration ($-4.22\mathrm{m}/{\mathrm{s}}^2$).- $s$ represents the stopping distance, which we need to calculate.By inputting these values, you rearrange and solve the equation to find the stopping distance:

• $s=\frac{(15.0{)}^2}{2\times 4.22}$
• $s\approx 26.89\mathrm{m}$

Thus, by understanding and applying the kinematic equations, one can precisely determine how far the car will travel until it fully stops, given an initial speed and known deceleration.