Stopping Distance of a Car on a Wet Surface: A Physics Problem Solving Guide

Stopping Distance of a Car on a Wet Surface: A Physics Problem Solving Guide

In this article, we will explore the physics behind the stopping distance of a car traveling on a wet surface. Using the given example, we will apply kinematic equations and Newton's laws to solve the problem step by step. Let's dive into the details.

Understanding the Given Information

The car in question has a mass of 1520 kg and is traveling at a speed of 21 m/s. The driver applies the brakes to stop the car on a wet surface, where the coefficient of friction is 0.40. Key points to consider:

Mass of the car: 1520 kg Initial speed (u): 21 m/s Coefficient of friction (μ): 0.40 Final speed (v): 0 m/s

Preliminary Calculations

First, let's calculate the momentum and the force exerted by the brakes.

Momentum Calculation

Momentum is given by the product of mass and velocity:

Momentum (p) mass (m) × velocity (u)

Momentum 1520 kg × 21 m/s 31920 kg-m/s

Force Exerted by the Brakes

The force exerted by the brakes can be calculated using the normal force and the coefficient of friction. The normal force (N) is the weight of the car, which is given by the mass times gravity (g ~ 9.8 m/s2).

N mass × g 1520 kg × 9.8 m/s2 14896 N

Using Newton's second law (F μN), the force exerted by the brakes is:

Force (F) 0.40 × 14896 N 5958.4 N

Applying Kinematic Equations

Now that we have the force, we can determine the acceleration caused by the braking force. Using Newton's second law (F ma), we can find the deceleration:

Acceleration (a) F/m 5958.4 N / 1520 kg

Acceleration -3.926 m/s2 (negative sign indicates deceleration)

We will use the kinematic equation to find the stopping distance (s). The equation that does not involve time is:

v2 u2 2as

Plugging in the values:

0 (21 m/s)2 2(-3.926 m/s2)s

0 441 - 7.852s

7.852s 441

s 441 / 7.852

s ≈ 56.14 m

Review and Practice

To solve similar problems, it's essential to be familiar with the four kinematic equations and to have them readily accessible. For this problem, the kinematic equation that does not involve time is particularly useful:

v2 u2 2as

Memorizing these equations can significantly simplify the problem-solving process. Additionally, having these equations on a 3x5 card or in a digital form can help you quickly access them during exams or real-world applications.

Conclusion

Understanding the physics behind the stopping distance of a car is crucial for both practical purposes and academic knowledge. By applying the principles of kinematics and Newton’s laws, you can calculate the stopping distance accurately. Remember, regular practice and familiarizing yourself with these fundamental equations can make problem-solving more efficient.

Stay curious and keep learning!