Exercise 9 Calculate Distance And Displacement On 1/5 Circle

Introduction

In physics, understanding the concepts of distance and displacement is crucial for describing motion accurately. While both terms relate to how an object's position changes, they represent different aspects of that change. Distance is a scalar quantity that refers to the total length of the path traveled by an object, regardless of direction. Displacement, on the other hand, is a vector quantity that refers to the change in position of an object, considering both the magnitude and direction from the initial point to the final point. This article delves into exercise 9, which focuses on calculating the distance and displacement between the initial and final points of an object moving along a circular path, specifically $\frac{1}{5}$\ of a circle. We will explore the formulas and methodologies required to solve this problem, providing a comprehensive understanding of these fundamental concepts.

Defining Distance and Displacement

To effectively tackle exercise 9, we must first clearly define distance and displacement. Imagine a scenario where you walk from point A to point B, then to point C. The distance you've traveled is the sum of the lengths of the paths AB and BC. It's the total ground you've covered. Displacement, however, is simply the straight-line distance from point A to point C, along with the direction. It’s the shortest path and direction from the starting point to the ending point. Understanding this distinction is vital because, in many real-world scenarios, the distance traveled can be significantly greater than the magnitude of the displacement.

Circular Motion and Its Significance

Circular motion is a common phenomenon in physics, observed in various contexts, from the orbit of planets around stars to the movement of objects on a carousel. Analyzing circular motion provides valuable insights into concepts such as centripetal force, angular velocity, and periodic motion. When an object moves along a circular path, the distance it travels is the arc length along the circle, while the displacement is the straight-line distance between the initial and final positions on the circle. In our exercise, the object travels $\frac{1}{5}$\ of a circle, which means it covers a specific fraction of the circle's circumference. This fraction directly influences both the distance and displacement calculations. The circular path introduces a geometric element to the problem, requiring us to consider angles and radii in our calculations, thus enhancing our understanding of physics and geometry interplay.

Importance of Accurate Calculations

In physics, accurate calculations are paramount. A small error in calculation can lead to significant discrepancies in results, affecting the overall analysis and interpretation of physical phenomena. For distance and displacement calculations, precision is essential, especially in applications such as navigation, trajectory prediction, and engineering design. Understanding how to correctly apply formulas and consider units ensures that our calculations reflect the true physical situation. Furthermore, accurate calculations build a strong foundation for more advanced physics concepts and problem-solving skills. In the context of exercise 9, precise calculation of both distance and displacement will not only provide the correct answers but also reinforce the importance of meticulousness in physics problem-solving.

Methodology for Calculating Distance

To calculate the distance traveled when an object moves along a circular path, we need to determine the arc length corresponding to the given fraction of the circle. The formula for the circumference (C) of a circle is C = 2πr, where r is the radius of the circle. Since the object travels $\frac{1}{5}$\ of the circle, the distance (d) traveled can be calculated as d = $\frac{1}{5}$\ * C = $\frac{1}{5}$\ * 2πr. This formula gives us the length of the arc that the object has traversed. Understanding this approach is crucial because it connects geometric properties of the circle with the physical concept of distance traveled. Moreover, this method can be extended to scenarios where the object travels any fraction of the circular path, making it a versatile tool in problem-solving. Proper application of this formula ensures accurate determination of the distance, which is essential for further analysis of motion.

Step-by-Step Calculation of Distance

Let's break down the calculation of distance into manageable steps. First, identify the radius (r) of the circle. This value is typically provided in the problem statement. If not, additional information may be given that allows you to deduce the radius. Second, apply the formula d = $\frac{1}{5}$\ * 2πr. This involves multiplying the fraction $\frac{1}{5}$\ by 2π and the radius. Ensure you use the correct units for the radius (e.g., meters, centimeters) and that the final distance is expressed in the same unit. Third, perform the calculation carefully, paying attention to significant figures and rounding rules if applicable. For example, if the radius is 10 meters, the distance would be $\frac{1}{5}$\ * 2 * 3.14159 * 10 ≈ 12.57 meters. By following these steps, you can systematically calculate the distance traveled along the circular path, minimizing the chances of error and ensuring a clear understanding of the process. This structured approach not only helps in solving this particular problem but also enhances your overall problem-solving skills in physics.

Example Calculation

To solidify the method for calculating distance, let’s consider a concrete example. Suppose an object moves $\frac{1}{5}$\ of a circle with a radius of 5 meters. To find the distance traveled, we use the formula d = $\frac{1}{5}$\ * 2πr. Substituting the given values, we get d = $\frac{1}{5}$\ * 2 * π * 5 meters. Approximating π as 3.14159, the calculation becomes d = $\frac{1}{5}$\ * 2 * 3.14159 * 5 meters ≈ 6.283 meters. Therefore, the distance traveled by the object is approximately 6.283 meters. This example illustrates the direct application of the formula and the importance of accurate substitution and calculation. By working through such examples, you can build confidence in your ability to calculate distance in various circular motion scenarios. Moreover, these practical applications reinforce the theoretical concepts, making them more intuitive and easier to remember.

Methodology for Calculating Displacement

Calculating displacement involves finding the straight-line distance between the initial and final points of the object's motion. When an object travels $\frac{1}{5}$\ of a circle, the initial and final points form a chord of the circle. To find the length of this chord, we can use geometric principles. Imagine drawing radii from the center of the circle to both the initial and final points. These radii, along with the chord, form an isosceles triangle. The angle (θ) subtended at the center by the arc is $\frac{1}{5}$\ of the total angle in a circle, which is 360 degrees. Thus, θ = $\frac{1}{5}$\ * 360° = 72°. We can use the law of cosines to find the length of the chord (which is the displacement). If r is the radius of the circle, the displacement (Δx) is given by the formula Δx = √(r² + r² - 2r²cos(θ)) = r√(2 - 2cos(θ)). This formula allows us to calculate the displacement based on the radius and the central angle. Understanding this approach requires a good grasp of both geometry and trigonometry, highlighting the interdisciplinary nature of physics.

Step-by-Step Calculation of Displacement

The step-by-step calculation of displacement involves a systematic application of geometric and trigonometric principles. First, identify the radius (r) of the circle, which is typically provided in the problem or can be derived from given information. Second, calculate the central angle (θ) subtended by the arc. Since the object travels $\frac{1}{5}$\ of the circle, θ = $\frac{1}{5}$\ * 360° = 72°. Third, use the formula for displacement: Δx = r√(2 - 2cos(θ)). This involves substituting the values of r and θ into the formula. Fourth, evaluate the cosine of the angle and perform the necessary arithmetic operations. Ensure that your calculator is set to degree mode when calculating the cosine. For example, if r = 5 meters, then Δx = 5√(2 - 2cos(72°)) ≈ 5√(2 - 2 * 0.309) ≈ 5√(1.382) ≈ 5 * 1.175 ≈ 5.875 meters. By following these steps, you can accurately calculate the displacement. This structured approach not only simplifies the calculation but also enhances your understanding of the underlying concepts and their application.

Example Calculation

To illustrate the calculation of displacement, let's consider an example where an object moves $\frac{1}{5}$\ of a circle with a radius of 5 meters. We follow the steps outlined earlier. First, we identify the radius, r = 5 meters. Second, we calculate the central angle, θ = $\frac{1}{5}$\ * 360° = 72°. Third, we apply the formula for displacement: Δx = r√(2 - 2cos(θ)). Substituting the values, we get Δx = 5√(2 - 2cos(72°)). We evaluate the cosine of 72°, which is approximately 0.309. Thus, Δx ≈ 5√(2 - 2 * 0.309) ≈ 5√(2 - 0.618) ≈ 5√(1.382). Calculating the square root, we get Δx ≈ 5 * 1.175 ≈ 5.875 meters. Therefore, the displacement of the object is approximately 5.875 meters. This example provides a clear demonstration of how to apply the formula and perform the calculations to find the displacement. By practicing with such examples, you can gain confidence in your ability to solve similar problems and deepen your understanding of the relationship between geometric and physical quantities.

Comparing Distance and Displacement

Having calculated both distance and displacement, it is crucial to compare these values to understand their differences and significance. In our example, the distance traveled by the object moving $\frac{1}{5}$\ of a circle with a radius of 5 meters is approximately 6.283 meters, while the displacement is approximately 5.875 meters. The distance represents the actual path length along the circular arc, whereas the displacement represents the straight-line distance between the initial and final points. The fact that distance is greater than the magnitude of the displacement illustrates a fundamental concept in physics: the path taken by an object can be longer than the shortest distance between its starting and ending points. This difference highlights the importance of understanding both scalar and vector quantities in describing motion. Moreover, comparing distance and displacement in different scenarios helps in developing a more comprehensive understanding of kinematics, which is the study of motion. This comparative analysis enhances problem-solving skills and conceptual clarity.

Significance of the Difference

The difference between distance and displacement is not merely a mathematical curiosity; it has significant implications in various real-world applications and theoretical contexts. In navigation, for instance, understanding the distance traveled is crucial for calculating fuel consumption and travel time, especially in scenarios involving curved paths or complex routes. Displacement, on the other hand, is more relevant when considering the net change in position, such as in determining the effectiveness of a journey or the overall progress made. In physics, this distinction is vital in analyzing motion under different conditions, such as motion with constant velocity versus motion with acceleration. Distance and displacement play different roles in equations and calculations related to work, energy, and momentum. Moreover, the difference between these two quantities can provide insights into the nature of the path taken by an object, whether it is a straight line or a curved path. Therefore, recognizing and understanding the significance of the difference between distance and displacement is essential for a thorough understanding of physics.

Practical Implications

The practical implications of understanding the difference between distance and displacement are vast and varied, spanning across multiple disciplines. In sports, athletes and coaches use these concepts to analyze performance. For example, a runner's distance in a race measures the total ground covered, while displacement indicates the net progress towards the finish line. In robotics, the distance traveled by a robot arm can determine wear and tear on its motors, while displacement helps in assessing the accuracy of its movements. In civil engineering, calculating distance and displacement is critical in planning transportation routes and designing infrastructure. For example, the distance a vehicle travels affects road design, while displacement considerations impact the placement of bridges and tunnels. Furthermore, in fields like geographic information systems (GIS), distance and displacement are fundamental in spatial analysis and mapping applications. By recognizing and applying these concepts in practical contexts, professionals can make informed decisions and optimize outcomes in their respective fields. Thus, the distinction between distance and displacement is not just an academic exercise but a vital tool in real-world problem-solving.

Conclusion

In conclusion, exercise 9 provides a valuable opportunity to understand and apply the concepts of distance and displacement in the context of circular motion. By calculating the distance traveled along $\frac{1}{5}$\ of a circle and comparing it with the displacement between the initial and final points, we gain a deeper appreciation for the differences between these two fundamental quantities. The step-by-step methodologies outlined for calculating distance and displacement, along with the example calculations, offer a practical guide for solving similar problems. The significance of the difference between distance and displacement extends beyond theoretical physics, impacting fields such as navigation, sports, robotics, and engineering. Therefore, mastering these concepts is crucial for anyone seeking a comprehensive understanding of motion and its applications. By reinforcing our understanding of these principles, we enhance our problem-solving skills and pave the way for more advanced topics in physics.