Calculating Acceleration And Distance A Car's Uniform Acceleration Example

Introduction

In the realm of physics, understanding motion is fundamental, and one of the most common types of motion we encounter is uniform acceleration. This article delves into the concept of uniform acceleration by exploring a practical example: a car accelerating from 30 km/h to 80 km/h in 5 minutes. We will meticulously calculate the car's acceleration and the total distance it covers during this period. This discussion is crucial for anyone studying introductory physics or simply interested in the mechanics of motion. By the end of this article, you will have a clear understanding of how to apply kinematic equations to solve problems involving uniform acceleration. The principles discussed here are applicable in various real-world scenarios, from understanding the motion of vehicles to analyzing the trajectories of projectiles.

Problem Statement: A Car's Acceleration

To begin, let's restate the problem clearly. A car initially moves at a speed of 30 kilometers per hour (km/h), and it accelerates uniformly to a final speed of 80 km/h. This change in speed occurs over a time interval of 5 minutes. Our primary goal is to determine two key quantities: the acceleration of the car and the total distance the car covers during this acceleration period. This problem is a classic example of a uniformly accelerated motion problem, and it requires us to apply the fundamental equations of kinematics to find the solution. The concepts of initial velocity, final velocity, time, acceleration, and distance are all intertwined in this scenario. Understanding how these quantities relate to each other is essential for solving problems related to motion in physics. We will break down the problem step by step, ensuring that each stage of the solution is clearly explained and justified. This approach will not only help in solving this specific problem but also provide a framework for tackling similar problems in the future.

Step 1: Converting Units

Before we can proceed with any calculations, it is essential to ensure that all our units are consistent. In this problem, we have speeds given in kilometers per hour (km/h) and time given in minutes. To work with the standard units used in physics, which are meters for distance, seconds for time, and meters per second (m/s) for speed, we need to convert the given values. First, let's convert the initial speed of 30 km/h to m/s. To do this, we multiply by the conversion factor ${ \frac{1000 \text{ m}}{1 \text{ km}} }$ to convert kilometers to meters and by ${ \frac{1 \text{ h}}{3600 \text{ s}} }$ to convert hours to seconds. This gives us:

${ 30 \frac{\text{km}}{\text{h}} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 8.33 \text{ m/s} }$

Next, we convert the final speed of 80 km/h to m/s using the same conversion factors:

${ 80 \frac{\text{km}}{\text{h}} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 22.22 \text{ m/s} }$

Finally, we need to convert the time interval of 5 minutes to seconds. Since there are 60 seconds in a minute, we have:

${ 5 \text{ minutes} \times 60 \frac{\text{seconds}}{\text{minute}} = 300 \text{ seconds} }$

Now that we have the initial speed (8.33 m/s), the final speed (22.22 m/s), and the time interval (300 seconds) all in consistent units, we are ready to proceed with the calculation of the acceleration.

Step 2: Calculating Acceleration

Acceleration is defined as the rate of change of velocity over time. In the case of uniform acceleration, this rate is constant. The formula to calculate acceleration (${a}$) is given by:

${ a = \frac{v_f - v_i}{t} }$

where ${v_f}$ is the final velocity, ${v_i}$ is the initial velocity, and ${t}$ is the time interval. We have already converted these values into consistent units in the previous step. The initial velocity ${v_i}$ is 8.33 m/s, the final velocity ${v_f}$ is 22.22 m/s, and the time interval ${t}$ is 300 seconds. Plugging these values into the formula, we get:

${ a = \frac{22.22 \text{ m/s} - 8.33 \text{ m/s}}{300 \text{ s}} }$

${ a = \frac{13.89 \text{ m/s}}{300 \text{ s}} }$

${ a ≈ 0.0463 \text{ m/s}^2 }$

Therefore, the acceleration of the car is approximately 0.0463 meters per second squared. This value represents the constant rate at which the car's velocity is increasing. A positive acceleration indicates that the car is speeding up in the direction of its motion. Understanding this calculation is crucial as acceleration is a fundamental concept in classical mechanics and plays a significant role in understanding the motion of objects under the influence of forces. The calculated acceleration will now be used to determine the distance covered by the car during this acceleration period.

Step 3: Calculating the Distance Covered

Now that we have calculated the acceleration, the next step is to determine the distance the car covers while accelerating. For uniformly accelerated motion, we can use the following kinematic equation to find the distance (${d}$):

${ d = v_i t + \frac{1}{2} a t^2 }$

where ${d}$ is the distance, ${v_i}$ is the initial velocity, ${t}$ is the time, and ${a}$ is the acceleration. We have all these values: the initial velocity ${v_i}$ is 8.33 m/s, the time ${t}$ is 300 seconds, and the acceleration ${a}$ is 0.0463 m/s². Plugging these values into the equation, we get:

${ d = (8.33 \text{ m/s})(300 \text{ s}) + \frac{1}{2} (0.0463 \text{ m/s}^2)(300 \text{ s})^2 }$

First, let's calculate the two terms separately:

${ (8.33 \text{ m/s})(300 \text{ s}) = 2499 \text{ m} }$

and

${ \frac{1}{2} (0.0463 \text{ m/s}^2)(300 \text{ s})^2 = \frac{1}{2} (0.0463 \text{ m/s}^2)(90000 \text{ s}^2) = 2083.5 \text{ m} }$

Now, adding these two terms together, we get:

${ d = 2499 \text{ m} + 2083.5 \text{ m} }$

${ d ≈ 4582.5 \text{ m} }$

Therefore, the car covers approximately 4582.5 meters while accelerating from 30 km/h to 80 km/h in 5 minutes. This distance calculation is crucial in understanding the dynamics of the car's motion during the acceleration phase. It provides a complete picture of how far the car travels while its speed is increasing uniformly. The combination of the acceleration and distance calculations gives a comprehensive understanding of the car's motion, which is a fundamental aspect of physics.

Conclusion

In summary, we have successfully calculated the acceleration and the distance covered by a car accelerating uniformly from 30 km/h to 80 km/h in 5 minutes. By converting the initial and final speeds to meters per second and the time to seconds, we were able to use the kinematic equations for uniform acceleration. We found that the acceleration of the car is approximately 0.0463 m/s², and the distance covered during this acceleration period is approximately 4582.5 meters. These calculations demonstrate the practical application of physics principles in understanding the motion of objects in the real world. The process involved unit conversions, the application of the acceleration formula, and the use of a kinematic equation to find the distance. This problem is a valuable exercise in understanding uniform acceleration and reinforces the importance of consistent units in physics calculations. The concepts and methods used here can be applied to a wide range of similar problems involving motion, making this a foundational topic in physics education.

This detailed analysis not only provides the solution to the problem but also explains the underlying physics concepts, making it a comprehensive learning resource. Understanding uniform acceleration is crucial for students and anyone interested in the dynamics of motion, and this article serves as a valuable guide to mastering this fundamental concept.