Particle Motion Velocity And Position Analysis

In the realm of physics and calculus, understanding the motion of particles is a fundamental concept. This article delves into the relationship between position, velocity, and time, using a specific example to illustrate these principles. We will explore how to determine the velocity of a particle at any given time, and then calculate its velocity at a specific time, $t = 3$ seconds. This exploration will provide a clear understanding of how calculus is applied in analyzing motion. The position function, $s(t)=7t^3+4t+9$, serves as our starting point, allowing us to dissect the intricacies of particle movement and its mathematical representation. Let's embark on this journey to uncover the dynamics of motion.

(a) Finding the Velocity at Time t

To find the velocity at time t, we need to understand the fundamental relationship between position and velocity. In calculus, velocity is defined as the rate of change of position with respect to time. Mathematically, this means that the velocity function, v(t), is the derivative of the position function, s(t). Given the position function $s(t) = 7t^3 + 4t + 9$, we will apply the rules of differentiation to find its derivative. The power rule, which states that the derivative of $t^n$ is $nt^{n-1}$, will be particularly useful here. We will also use the constant multiple rule, which allows us to differentiate terms multiplied by constants separately.

Applying these rules, we differentiate each term in the position function. The derivative of $7t^3$ is $7 imes 3t^{3-1} = 21t^2$. The derivative of $4t$ is simply 4, since the derivative of $t$ is 1. The derivative of the constant term 9 is 0, as the derivative of any constant is zero. Combining these results, the velocity function v(t) is the sum of these individual derivatives. Therefore, $v(t) = 21t^2 + 4$. This equation gives us the velocity of the particle at any time t. The units of velocity are meters per second (m/s), assuming the position is measured in meters and time in seconds. Understanding this relationship between position and velocity is crucial in analyzing the motion of objects. The derivative provides a powerful tool for uncovering the instantaneous rate of change, offering insights into how the particle's speed and direction evolve over time. This concept forms the bedrock of kinematics, the study of motion, and is indispensable in various scientific and engineering applications.

The velocity function, $v(t) = 21t^2 + 4$, is a quadratic function, indicating that the particle's velocity changes non-linearly with time. The $21t^2$ term dominates the behavior of the velocity at larger values of t, suggesting that the particle accelerates significantly as time increases. The constant term, 4, represents the initial velocity component, which is the velocity at t = 0. This means that the particle already has a velocity of 4 m/s before any acceleration due to the $21t^2$ term takes effect. Analyzing the velocity function further, we can see that the velocity is always positive since $t^2$ is always non-negative, and the constant term 4 ensures that the velocity never becomes zero or negative. This implies that the particle is always moving in the positive direction. The rate at which the velocity changes is given by the acceleration, which is the derivative of the velocity function. In this case, the acceleration function would be the derivative of $21t^2 + 4$, which is $42t$. This shows that the acceleration is linearly increasing with time, meaning the particle's velocity is increasing at an increasing rate. This detailed analysis of the velocity function provides a comprehensive understanding of the particle's motion, revealing how its speed and direction evolve over time.

(b) Finding the Velocity at Time t = 3 Seconds

Now that we have the velocity function, $v(t) = 21t^2 + 4$, we can easily find the velocity at any specific time. In this case, we want to find the velocity at t = 3 seconds. To do this, we simply substitute t = 3 into the velocity function. This process involves replacing the variable t in the equation with the value 3 and then performing the arithmetic operations to find the corresponding velocity value. This is a direct application of the function evaluation, a fundamental concept in mathematics. The velocity at a specific time provides a snapshot of the particle's motion at that instant, giving us a precise measure of its speed and direction at that moment. Understanding how to evaluate functions at specific points is essential in analyzing physical systems, as it allows us to predict and understand the behavior of objects at particular times or locations.

Substituting t = 3 into the velocity function, we get $v(3) = 21(3)^2 + 4$. First, we calculate $3^2$, which is 9. Then, we multiply 21 by 9, which equals 189. Finally, we add 4 to 189, resulting in 193. Therefore, the velocity at t = 3 seconds is 193 m/s. This calculation demonstrates how the velocity function can be used to determine the instantaneous velocity of the particle at any given time. The result, 193 m/s, indicates that the particle is moving at a significant speed at 3 seconds. This value is much higher than the initial velocity of 4 m/s, which we found earlier, confirming that the particle is accelerating over time. The large velocity at t = 3 seconds is a direct consequence of the $21t^2$ term in the velocity function, which causes the velocity to increase quadratically with time. This quantitative result provides a concrete understanding of the particle's motion, illustrating how its velocity changes and how to calculate it at a specific point in time. Such calculations are crucial in various applications, such as predicting the trajectory of projectiles or analyzing the motion of vehicles.

This result highlights the power of calculus in providing precise quantitative information about motion. By using the derivative to find the velocity function and then evaluating it at a specific time, we can determine the particle's speed and direction at that exact moment. This ability to calculate instantaneous velocities is essential in many areas of physics and engineering, where understanding the dynamics of moving objects is critical. For instance, in designing vehicles, engineers need to know how the velocity changes over time to ensure safety and efficiency. Similarly, in astrophysics, understanding the velocities of celestial bodies is crucial for predicting their movements and interactions. The calculation we performed here, substituting t = 3 into the velocity function, is a fundamental technique used in a wide range of applications. It demonstrates how mathematical models, like the position and velocity functions, can be used to make accurate predictions about the physical world. The ability to move between abstract mathematical representations and concrete physical quantities is a cornerstone of scientific and engineering problem-solving.

In summary, we have successfully found the velocity of a particle at any time t and specifically at t = 3 seconds, given its position function $s(t) = 7t^3 + 4t + 9$. We determined that the velocity function is $v(t) = 21t^2 + 4$, and the velocity at t = 3 seconds is 193 m/s. This exercise demonstrates the crucial relationship between position and velocity, and how calculus provides the tools to analyze and understand motion. The concepts and techniques discussed here are fundamental in physics and engineering, providing a foundation for more advanced topics in dynamics and kinematics. By understanding how to differentiate position functions to find velocity and how to evaluate these functions at specific times, we gain valuable insights into the motion of objects and their behavior over time. This knowledge is essential for anyone seeking to understand the physical world and solve real-world problems related to motion and dynamics.