Particle Motion Analysis Determining Velocity At A Specific Displacement
Introduction
In this article, we will delve into the fascinating world of particle motion and analyze the movement of a particle along a straight line. We will explore how its displacement, velocity, and acceleration change over time. Specifically, we will examine a scenario where the particle's displacement, denoted by s meters, from a fixed point O is given by the equation s = 5 - (15 / (2t + 3)), where t represents the time in seconds after the commencement of motion. This equation provides a mathematical model of the particle's position as a function of time, allowing us to investigate its dynamic behavior. Our primary objective is to determine the velocity of the particle at a specific instant when its displacement reaches a particular value, namely 3 meters from the fixed point O. To achieve this, we will employ concepts from calculus, including differentiation, to derive the velocity function from the displacement function. The velocity function will then enable us to calculate the particle's velocity at any given time. Understanding the motion of particles is fundamental in physics and engineering, with applications ranging from projectile motion to the design of mechanical systems. By analyzing this particular example, we can gain valuable insights into the principles governing particle dynamics and develop our problem-solving skills in this area. The ability to model and analyze motion is crucial in many fields, allowing us to predict and control the behavior of objects in various systems. This article will provide a step-by-step approach to solving the problem, making it accessible to readers with a basic understanding of calculus and physics. Furthermore, we will discuss the physical interpretation of the results, enhancing our comprehension of the particle's motion. By the end of this analysis, you will have a deeper appreciation for the interplay between mathematics and physics in describing the motion of objects.
Problem Statement
A particle moves in a straight line, and its displacement s (in meters) from a fixed point O is described by the equation:
s = 5 - (15 / (2t + 3))
where t is the time in seconds after the start of motion. Our goal is to find the velocity of the particle when it is 3 meters away from the fixed point O. This problem requires us to understand the relationship between displacement, velocity, and time, and how to use calculus to solve it. The displacement equation provides a complete description of the particle's position at any time t. However, to find the velocity, which is the rate of change of displacement with respect to time, we need to differentiate the displacement equation. This is a fundamental concept in calculus and is widely used in physics and engineering to analyze motion. Once we have the velocity function, we can determine the velocity at any given time. However, in this problem, we are not given the time directly but rather the displacement. Therefore, we first need to find the time at which the particle is 3 meters away from the fixed point O. This can be done by setting s = 3 in the displacement equation and solving for t. Once we have the time, we can substitute it into the velocity function to find the velocity at that instant. This problem highlights the importance of understanding the physical meaning of mathematical equations and how to manipulate them to solve real-world problems. It also emphasizes the connection between displacement, velocity, and acceleration, which are fundamental concepts in kinematics. By solving this problem, we will gain a better understanding of how to analyze and interpret the motion of objects using mathematical tools.
Solution
1. Find the time when s = 3 meters
To begin, we need to determine the time (t) at which the particle's displacement (s) is equal to 3 meters. We can achieve this by substituting s = 3 into the given displacement equation:
3 = 5 - (15 / (2t + 3))
Now, let's solve for t:
First, subtract 5 from both sides:
-2 = - (15 / (2t + 3))
Next, multiply both sides by -1:
2 = 15 / (2t + 3)
Now, multiply both sides by (2t + 3):
2(2t + 3) = 15
Expand the left side:
4t + 6 = 15
Subtract 6 from both sides:
4t = 9
Finally, divide both sides by 4:
t = 9 / 4 = 2.25 seconds
Therefore, the particle is 3 meters away from the fixed point O at t = 2.25 seconds. This step is crucial because it allows us to connect the displacement information with the time variable, which is necessary for finding the velocity. The algebraic manipulations involved in solving for t are fundamental skills in mathematics and are essential for solving many physics problems. By carefully following each step, we can ensure that we arrive at the correct solution. This value of time will be used in the subsequent steps to calculate the velocity of the particle. The ability to solve equations like this is a key skill in physics, as it allows us to relate different physical quantities and make predictions about the behavior of systems. In this case, we have used the displacement equation to find the time at which the particle is at a specific location. This is a common type of problem in kinematics, and the techniques used here can be applied to a wide range of scenarios.
2. Find the velocity function
The velocity of the particle, denoted by v, is the rate of change of displacement (s) with respect to time (t). Mathematically, this is represented as the derivative of s with respect to t: v = ds/dt. To find the velocity function, we need to differentiate the displacement equation:
s = 5 - (15 / (2t + 3))
with respect to t. It's often helpful to rewrite the equation to make differentiation easier. We can rewrite the term 15 / (2t + 3) as 15(2t + 3)^-1. Now, we can apply the chain rule of differentiation:
ds/dt = d/dt [5 - 15(2t + 3)^-1]
The derivative of the constant 5 is 0. For the second term, we use the chain rule:
d/dt [-15(2t + 3)^-1] = -15 * (-1) * (2t + 3)^-2 * 2
Simplifying, we get:
v = ds/dt = 30 / (2t + 3)^2
This is the velocity function, which gives us the velocity of the particle at any time t. The process of differentiation is a core concept in calculus and is essential for understanding rates of change. In this case, we have used differentiation to find the velocity function from the displacement function. The chain rule is a particularly important technique for differentiating composite functions, such as the one we encountered in this problem. By carefully applying the chain rule, we can correctly find the derivative and obtain the velocity function. This function is crucial for answering the original question, as it allows us to calculate the velocity at any given time. The velocity function provides a complete description of how the particle's velocity changes over time. It is a powerful tool for analyzing the motion of the particle and making predictions about its future behavior.
3. Calculate the velocity at t = 2.25 seconds
Now that we have the velocity function, v = 30 / (2t + 3)^2, we can calculate the velocity of the particle at t = 2.25 seconds. To do this, we simply substitute t = 2.25 into the velocity function:
v = 30 / (2 * 2.25 + 3)^2
First, calculate the expression inside the parentheses:
2 * 2.25 + 3 = 4.5 + 3 = 7.5
Now, substitute this back into the velocity equation:
v = 30 / (7.5)^2
Calculate the square:
(7.5)^2 = 56.25
Finally, divide to find the velocity:
v = 30 / 56.25 ≈ 0.533 meters per second
Therefore, the velocity of the particle when it is 3 meters away from the fixed point O is approximately 0.533 meters per second. This step is the culmination of our analysis, where we use the velocity function and the previously calculated time to find the particle's velocity at the specified location. The calculation involves simple arithmetic operations, but it is important to perform them carefully to avoid errors. The final result gives us a quantitative measure of how fast the particle is moving at the moment it is 3 meters away from the origin. The units of velocity are meters per second, which is consistent with the units of displacement (meters) and time (seconds) used in the problem. This result provides a complete answer to the original question and demonstrates the power of using calculus to analyze motion. By combining the concepts of displacement, velocity, and differentiation, we have been able to determine the particle's velocity at a specific point in its trajectory.
Conclusion
In this analysis, we successfully determined the velocity of a particle moving in a straight line when its displacement from a fixed point O was 3 meters. We began by finding the time at which the particle was at this displacement by solving the displacement equation for t. This involved algebraic manipulation and careful attention to detail. Next, we derived the velocity function by differentiating the displacement equation with respect to time. This step utilized the chain rule of differentiation, a fundamental concept in calculus. Finally, we substituted the calculated time value into the velocity function to find the velocity of the particle at that instant. The result showed that the particle's velocity was approximately 0.533 meters per second when it was 3 meters away from the fixed point. This problem demonstrates the power of calculus in analyzing motion and the importance of understanding the relationships between displacement, velocity, and time. The ability to solve problems like this is essential in physics and engineering, where understanding motion is crucial for designing and analyzing systems. The techniques used in this analysis can be applied to a wide range of problems involving particle motion and other dynamic systems. By mastering these concepts and techniques, we can gain a deeper understanding of the physical world and our ability to model and predict its behavior. The connection between mathematics and physics is clearly illustrated in this problem, where calculus provides the tools to describe and analyze the motion of a particle. The solution process involved multiple steps, each requiring careful attention to detail and a solid understanding of the underlying principles. The final result provides a quantitative answer to the original question and demonstrates the effectiveness of the mathematical approach.
