Particle Motion: Position, Velocity, & Acceleration At T=3s

by Alex Johnson 60 views

Understanding the motion of particles is a fundamental concept in physics. This article breaks down a classic problem: determining the position, velocity, and acceleration of a particle moving along the x-axis, given its equation of motion. We'll walk through the steps to solve this problem, focusing on clarity and providing a comprehensive explanation for each step.

Understanding the Problem: Kinematics in Action

The core of this problem lies in the realm of kinematics, which is the branch of physics that describes the motion of objects without considering the forces that cause the motion. We are given the particle's position as a function of time, x(t) = 2.00 + 3.00t - 1.00t^2, where x is in meters and t is in seconds. Our goal is to find the particle's position, velocity, and acceleration at a specific time, t = 3.00 s. This involves understanding the relationships between position, velocity, and acceleration, particularly how they are connected through calculus. Velocity, in essence, is the rate of change of position with respect to time, and acceleration is the rate of change of velocity with respect to time. These relationships are crucial for solving a wide range of physics problems, from simple projectile motion to more complex dynamics. By mastering these concepts, you gain a powerful toolset for analyzing and predicting the motion of objects in various scenarios. The application of calculus, specifically differentiation, allows us to move from the position function to velocity and then to acceleration. This step-by-step process reveals the intricate details of the particle's motion at any given instant. Furthermore, understanding these relationships helps in visualizing the motion – whether the particle is speeding up, slowing down, or changing direction. This problem serves as an excellent foundation for tackling more advanced topics in mechanics and dynamics. Let's dive into the step-by-step solution.

1. Determining the Position at t = 3.00 s

The particle's position is described by the equation x(t) = 2.00 + 3.00t - 1.00t^2. To find the position at t = 3.00 s, we simply substitute t = 3.00 into the equation. This is a direct application of the given information and requires careful attention to arithmetic. Replacing 't' with '3.00' in the equation, we get x(3.00) = 2.00 + 3.00(3.00) - 1.00(3.00)^2. Now, we perform the calculations step by step: First, calculate 3.00 * 3.00, which equals 9.00. Then, calculate (3.00)^2, which also equals 9.00. Next, multiply 1.00 by 9.00, resulting in 9.00. Now, the equation becomes x(3.00) = 2.00 + 9.00 - 9.00. Simplifying further, we add 2.00 and 9.00, which gives 11.00. Finally, subtract 9.00 from 11.00, which equals 2.00. Therefore, the position of the particle at t = 3.00 s is 2.00 meters. This means that at the specified time, the particle is located 2 meters along the x-axis from the origin. This straightforward substitution and calculation demonstrate a fundamental principle in physics: using equations of motion to predict an object's position at a given time. The result provides a snapshot of the particle's location at a specific moment, which is a crucial piece of information for understanding its overall motion.

2. Calculating the Velocity at t = 3.00 s

To determine the velocity, we need to find the derivative of the position function with respect to time. The velocity, denoted as v(t), is the rate of change of position, and in calculus terms, it's the first derivative of the position function x(t). Given x(t) = 2.00 + 3.00t - 1.00t^2, we apply the power rule of differentiation. The derivative of a constant (like 2.00) is zero. The derivative of 3.00t is 3.00 (since the power of t is 1, and we multiply by the coefficient and reduce the power by 1). For the term -1.00t^2, we multiply by the power (2) and reduce the power by 1, resulting in -2.00t. Therefore, the velocity function v(t) is the derivative of x(t), which is v(t) = 0 + 3.00 - 2.00t, simplifying to v(t) = 3.00 - 2.00t. Now that we have the velocity function, we can find the velocity at t = 3.00 s by substituting t = 3.00 into the velocity equation. So, v(3.00) = 3.00 - 2.00(3.00). Calculating this, we get v(3.00) = 3.00 - 6.00, which equals -3.00 m/s. The negative sign indicates that the particle is moving in the negative x-direction at this time. The velocity of -3.00 m/s provides critical information about the particle's motion at t = 3.00 s, indicating not only its speed but also its direction of movement. This step highlights the power of calculus in physics, allowing us to derive velocity from the position function.

3. Finding the Acceleration at t = 3.00 s

Now, let's determine the acceleration of the particle. Acceleration, denoted as a(t), is the rate of change of velocity with respect to time. In calculus terms, it's the first derivative of the velocity function v(t) or the second derivative of the position function x(t). We already found the velocity function to be v(t) = 3.00 - 2.00t. To find the acceleration, we differentiate v(t) with respect to time. The derivative of a constant (like 3.00) is zero. The derivative of -2.00t is -2.00 (since the power of t is 1). Therefore, the acceleration function a(t) is a(t) = -2.00 m/s². Notice that the acceleration is constant and does not depend on time. This means the particle is undergoing uniform acceleration. Since the acceleration is constant, its value at t = 3.00 s is the same as at any other time. Therefore, the acceleration at t = 3.00 s is -2.00 m/s². The negative sign indicates that the acceleration is in the negative x-direction, which means the particle's velocity is decreasing (or becoming more negative) over time. This constant acceleration provides a clear picture of how the particle's velocity changes uniformly. This step underscores the significance of understanding derivatives in physics, as they allow us to move from velocity to acceleration and gain a deeper insight into the motion of the particle. The constant acceleration simplifies the analysis in this case, but in many real-world scenarios, acceleration can also be time-dependent, requiring more advanced mathematical techniques to solve.

Summary of Results

At t = 3.00 s, we have determined the following:

  • Position: The particle's position is x = 2.00 meters.
  • Velocity: The particle's velocity is v = -3.00 m/s.
  • Acceleration: The particle's acceleration is a = -2.00 m/s².

These results provide a complete picture of the particle's motion at the specified time. The particle is located 2 meters along the x-axis, moving in the negative x-direction at a speed of 3.00 m/s, and its velocity is decreasing (or becoming more negative) at a rate of 2.00 m/s² due to the negative acceleration. This comprehensive analysis demonstrates the power of kinematics in describing the motion of objects. By knowing the equation of motion, we can precisely determine the position, velocity, and acceleration at any given time. These quantities are fundamental in understanding and predicting the behavior of physical systems. Moreover, this problem highlights the interconnectedness of these kinematic variables. The position function dictates the velocity, and the velocity function dictates the acceleration. This chain of relationships is crucial for solving a wide range of physics problems, from projectile motion to oscillations and waves. Understanding these concepts is essential for students and professionals alike in physics and engineering. The ability to analyze and interpret motion is a cornerstone of many scientific and technological applications. This problem serves as an excellent illustration of how theoretical concepts in physics can be applied to practical scenarios.

Conclusion

In conclusion, by applying the principles of kinematics and calculus, we have successfully determined the position, velocity, and acceleration of a particle at a specific time. This problem showcases the importance of understanding the relationships between position, velocity, and acceleration, as well as the mathematical tools used to analyze them. These concepts are essential for further studies in physics and related fields.

For further exploration of kinematics and particle motion, visit Khan Academy's Physics Section.