UFPA Physics Problem Solving Acceleration In Uniformly Varied Motion
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Introduction
In the realm of physics, understanding uniformly varied motion is crucial for analyzing how objects move under constant acceleration. This article delves into a classic problem from the Federal University of Pará (UFPA) that illustrates the principles of uniformly varied motion. We will dissect the problem, apply the relevant equations, and arrive at the solution in a clear, step-by-step manner. This approach not only helps in solving the specific problem but also reinforces the fundamental concepts applicable to a wide range of physics scenarios. By the end of this article, you will have a solid understanding of how to calculate acceleration given the initial conditions, displacement, and final velocity in a uniformly varied motion problem.
Problem Statement
The problem we're tackling today involves a point mass starting from rest and traversing a distance of 12 meters with uniformly varied motion. At the end of this journey, the point mass attains a velocity of 6 meters per second. Our task is to determine the acceleration of the point mass during this motion. This is a typical problem that appears in introductory physics courses and is essential for grasping the dynamics of moving objects. The problem statement provides us with key information: the initial state (at rest), the distance covered, and the final velocity. By carefully analyzing these details, we can select the appropriate kinematic equation to solve for the unknown acceleration. The ability to break down a problem into its fundamental components and identify the relevant formulas is a crucial skill in physics, and this problem provides an excellent opportunity to practice that skill.
Identifying Key Concepts
Before diving into the solution, let's identify the key concepts involved in this problem. First and foremost, we are dealing with uniformly varied motion, which means the acceleration is constant throughout the motion. This allows us to use a set of kinematic equations that relate displacement, initial velocity, final velocity, acceleration, and time. In this specific problem, we are given the initial velocity (0 m/s since the point starts from rest), the displacement (12 m), and the final velocity (6 m/s). We are asked to find the acceleration. Notice that the time taken for the motion is not provided, which suggests we should use a kinematic equation that does not involve time directly. Among the various kinematic equations, there is one that perfectly fits our needs: the equation that relates final velocity, initial velocity, acceleration, and displacement. Understanding this relationship is crucial for solving this problem efficiently. Additionally, recognizing that the motion starts from rest simplifies the equation further, making the calculation more straightforward. By mastering these key concepts, you'll be well-equipped to tackle similar problems involving uniformly varied motion.
Relevant Equations
The core of solving any physics problem lies in identifying and applying the relevant equations. In this case, we are dealing with uniformly varied motion, which means the acceleration is constant. The kinematic equations for uniformly varied motion provide the mathematical framework for relating displacement, initial velocity, final velocity, acceleration, and time. However, since the problem does not provide the time taken for the motion, we need to select an equation that does not include time as a variable. The appropriate equation for this scenario is: v² = u² + 2as, where: v represents the final velocity, u represents the initial velocity, a represents the acceleration, and s represents the displacement. This equation is derived from the fundamental definitions of velocity and acceleration under constant acceleration conditions. It is a powerful tool for solving problems where time is not a known quantity. Understanding the origin and application of this equation is essential for mastering kinematics. By correctly identifying and applying this equation, we can directly calculate the acceleration using the given values for final velocity, initial velocity, and displacement. This equation is a cornerstone of kinematics and is frequently used in various physics problems involving constant acceleration.
Step-by-Step Solution
Now, let's walk through the step-by-step solution to the problem. We begin by restating the given information: The point mass starts from rest, so the initial velocity (u) is 0 m/s. The displacement (s) is 12 meters. The final velocity (v) is 6 m/s. Our goal is to find the acceleration (a). We have already identified the relevant equation: v² = u² + 2as. The next step is to substitute the given values into the equation: (6 m/s)² = (0 m/s)² + 2 * a * (12 m). This simplifies to 36 m²/s² = 0 + 24a m. Now, we isolate the acceleration (a) by dividing both sides of the equation by 24 m: a = (36 m²/s²) / (24 m). Performing the division, we get: a = 1.5 m/s². Therefore, the acceleration of the point mass is 1.5 meters per second squared. This step-by-step approach ensures clarity and accuracy in the solution process. By carefully substituting the values and performing the algebraic manipulation, we arrive at the correct answer. This methodical approach is crucial for solving any physics problem, as it minimizes the chances of errors and helps in understanding the underlying principles.
Detailed Calculation
To provide an even clearer understanding, let's delve into the detailed calculation of the acceleration. We have the equation: v² = u² + 2as. Substituting the known values: (6 m/s)² = (0 m/s)² + 2 * a * (12 m). This simplifies to: 36 m²/s² = 0 + 24a m. To isolate the acceleration (a), we divide both sides by 24 m: a = (36 m²/s²) / (24 m). Now, let's break down the division: a = 36 / 24 * (m²/s²) / m. Simplifying the fraction 36/24, we get 1.5. The units also simplify: (m²/s²) / m = m/s². Therefore, the acceleration is: a = 1.5 m/s². This detailed breakdown illustrates how the units are handled during the calculation, ensuring that the final answer has the correct units for acceleration (meters per second squared). Understanding the unit analysis is crucial in physics, as it helps verify the correctness of the calculations and the consistency of the physical quantities involved. This thorough calculation reinforces the step-by-step solution and provides a deeper understanding of the mathematical process involved.
Final Answer
After performing the calculations, we arrive at the final answer: The acceleration of the point mass is 1.5 m/s². This means that the velocity of the point mass is increasing at a rate of 1.5 meters per second every second. The positive value indicates that the acceleration is in the same direction as the motion, causing the point mass to speed up. This answer is consistent with the given information: the point mass starts from rest and reaches a final velocity of 6 m/s over a distance of 12 meters. Understanding the physical meaning of the answer is crucial. The acceleration of 1.5 m/s² tells us how the velocity changes over time, which is a fundamental concept in kinematics. This final answer provides a quantitative solution to the problem and solidifies our understanding of uniformly varied motion. By clearly stating the final answer, we complete the problem-solving process and provide a concise and meaningful result.
Conclusion
In conclusion, we have successfully solved the UFPA physics problem by applying the principles of uniformly varied motion. We identified the key concepts, selected the relevant equation, and performed a step-by-step calculation to determine the acceleration of the point mass. The problem highlighted the importance of understanding the relationships between displacement, initial velocity, final velocity, and acceleration in constant acceleration scenarios. The final answer, 1.5 m/s², provides a quantitative measure of the rate at which the velocity of the point mass changes. This exercise reinforces the fundamental concepts of kinematics and demonstrates how to apply them to solve real-world problems. Furthermore, the problem-solving approach used in this article can be applied to a wide range of physics problems involving motion. By mastering these concepts and techniques, students can build a strong foundation in physics and excel in their studies. The ability to break down complex problems into manageable steps and apply the appropriate equations is a valuable skill that extends beyond the realm of physics and is applicable to various problem-solving scenarios.
Keywords
Uniformly Varied Motion, Acceleration, Kinematics, UFPA Physics Problem, Physics Equations, Step-by-Step Solution, Problem Solving, Initial Velocity, Final Velocity, Displacement, Constant Acceleration, Physics Education, Mechanics, Dynamics
