Particle Motion Analysis S(t) = 24t³ - 32t + 5 Velocity And Intervals
The study of motion is fundamental to physics and mathematics, and understanding how objects move through space and time is crucial in many scientific and engineering disciplines. In this article, we will delve into the motion of a particle along a line, described by the function s(t) = 24t³ - 32t + 5. Our primary goal is to dissect this function, determine the particle's velocity, identify the time intervals during which the particle moves in a specific direction, and ultimately, gain a comprehensive understanding of its movement. We will explore the concepts of velocity, derivatives, and intervals of motion, providing a clear and detailed analysis suitable for students, educators, and anyone interested in the mathematical modeling of motion. This exploration will not only enhance your understanding of calculus and physics but also demonstrate how mathematical tools can be applied to solve real-world problems. So, let's embark on this journey to unravel the intricacies of particle motion, step by step, with detailed explanations and practical examples.
(a) Finding the Velocity Function v(t)
To determine the velocity function v(t) of the particle at any time t ≥ 0, we need to understand the fundamental relationship between position and velocity. In calculus, the velocity of an object is defined as the rate of change of its position with respect to time. Mathematically, this means that the velocity function v(t) is the derivative of the position function s(t). The position function given is s(t) = 24t³ - 32t + 5. To find its derivative, we will apply the power rule, which states that the derivative of tⁿ with respect to t is ntⁿ⁻¹. Applying this rule to each term in the position function will give us the velocity function.
Let's start by differentiating the first term, 24t³. Applying the power rule, we multiply the coefficient (24) by the exponent (3) and reduce the exponent by 1: d/dt (24t³) = 24 * 3 * t² = 72t². Next, we differentiate the second term, -32t. Here, the exponent of t is 1, so the derivative is -32 * 1 * t⁰ = -32. Finally, we differentiate the constant term, 5. The derivative of a constant is always 0, so d/dt (5) = 0. Now, combining these results, we get the velocity function: v(t) = 72t² - 32. This quadratic function represents the instantaneous velocity of the particle at any time t. Understanding the velocity function is crucial as it provides insights into how the particle's speed and direction change over time. The positive and negative values of v(t) indicate the direction of motion, while the magnitude of v(t) represents the speed. By analyzing this function, we can identify when the particle is moving forward (positive velocity) and when it is moving backward (negative velocity), as well as the moments when the particle is at rest (zero velocity).
Detailed Steps in Differentiation
- Identify the position function: s(t) = 24t³ - 32t + 5
- Apply the power rule to each term:
- d/dt (24t³) = 72t²
- d/dt (-32t) = -32
- d/dt (5) = 0
- Combine the results to form the velocity function: v(t) = 72t² - 32
(b) Identifying Time Intervals of Forward and Backward Motion
To identify the time intervals during which the particle is moving in a positive direction (forward) or a negative direction (backward), we need to analyze the velocity function v(t) = 72t² - 32. The particle moves forward when v(t) > 0 and backward when v(t) < 0. To find these intervals, we first need to determine when the velocity is zero, which represents the moments when the particle changes direction. We set v(t) = 0 and solve for t.
So, 72t² - 32 = 0. First, we isolate the t² term by adding 32 to both sides: 72t² = 32. Next, we divide both sides by 72: t² = 32/72. Simplifying the fraction, we get t² = 4/9. Taking the square root of both sides, we find t = ±√(4/9), which gives us t = ±2/3. Since we are considering time t ≥ 0, we only take the positive root, t = 2/3. This critical point, t = 2/3, divides the time axis into intervals where the velocity is either positive or negative. To determine the sign of the velocity in these intervals, we can test a value within each interval.
Let's consider the interval 0 ≤ t < 2/3. We can choose a test value, say t = 0. Plugging this into the velocity function, we get v(0) = 72(0)² - 32 = -32, which is negative. This means that the particle is moving backward in this interval. Now, let's consider the interval t > 2/3. We can choose a test value, say t = 1. Plugging this into the velocity function, we get v(1) = 72(1)² - 32 = 40, which is positive. This means that the particle is moving forward in this interval. Therefore, the particle moves backward in the interval [0, 2/3) and moves forward in the interval (2/3, ∞). Understanding these intervals is essential for a complete understanding of the particle's motion, providing insights into when it accelerates, decelerates, and changes direction. The analysis of the velocity function and its sign is a fundamental concept in calculus and physics, offering a powerful tool for modeling and predicting the motion of objects.
Step-by-Step Interval Analysis
- Set v(t) = 0 and solve for t:
- 72t² - 32 = 0
- 72t² = 32
- t² = 32/72 = 4/9
- t = ±2/3
- Since t ≥ 0, we take t = 2/3
- Divide the time axis into intervals based on the critical point:
- [0, 2/3) and (2/3, ∞)
- Test a value within each interval in v(t):
- For [0, 2/3), let t = 0: v(0) = -32 (negative, backward motion)
- For (2/3, ∞), let t = 1: v(1) = 40 (positive, forward motion)
- Determine the intervals of forward and backward motion:
- Backward motion: [0, 2/3)
- Forward motion: (2/3, ∞)
Conclusion
In this article, we have thoroughly analyzed the motion of a particle along a line, described by the function s(t) = 24t³ - 32t + 5. We successfully found the velocity function, v(t) = 72t² - 32, by differentiating the position function. Furthermore, we identified the time intervals during which the particle moves forward and backward by analyzing the sign of the velocity function. The particle moves backward in the interval [0, 2/3) and forward in the interval (2/3, ∞). This comprehensive analysis demonstrates the power of calculus in understanding and predicting motion. By applying fundamental concepts such as derivatives and interval analysis, we can gain valuable insights into the behavior of moving objects. This knowledge is not only crucial in academic contexts but also has practical applications in various fields, including engineering, physics, and robotics. Understanding the motion of particles and objects is essential for designing efficient systems, predicting their behavior, and ensuring their safe operation. The methods and concepts discussed here provide a solid foundation for further exploration of more complex motion scenarios and mathematical modeling techniques. Continuing to explore these topics will undoubtedly enhance your ability to analyze and solve real-world problems involving motion and dynamics.
