Particle Motion Velocity And Displacement Calculation

In physics, understanding the motion of particles is a fundamental concept. This article delves into the analysis of particle motion along a straight line, focusing on calculating velocity and displacement given the acceleration function and initial conditions. We will explore how to use calculus, specifically integration, to derive the velocity and displacement functions from the acceleration function. This involves understanding the relationship between acceleration, velocity, and displacement, and applying the appropriate mathematical techniques to solve problems related to particle motion. By the end of this article, you will be able to analyze the motion of a particle given its acceleration function and initial conditions, and calculate its velocity and displacement at any given time.

A particle moves along a straight line relative to a fixed origin $O$ with acceleration, measured in $m/s^2$, given by the function $a(t) = 8t^3 - 6t$, where $t$ is the time in seconds. If the particle has velocity $v = -4 m/s$ when $t = 2s$, we need to find:

1. The velocity of the particle at any time $t$.
2. The displacement of the particle from the origin $O$ at any time $t$, given that the particle is initially at the origin.

This problem exemplifies a classic application of calculus in physics. It requires us to use the concepts of integration to move from acceleration to velocity, and then from velocity to displacement. The initial conditions provided are crucial for determining the constants of integration, ensuring a unique solution for both the velocity and displacement functions. The problem also highlights the importance of understanding the physical meaning of these functions and how they relate to the motion of the particle.

1. Finding the Velocity Function

The key to finding the velocity function, $v(t)$, lies in understanding that acceleration is the derivative of velocity with respect to time. Mathematically, this is expressed as:

$a(t) = \frac{dv}{dt}$

To find $v(t)$, we need to integrate the acceleration function $a(t)$ with respect to time:

$v(t) = \int a(t) dt = \int (8t^3 - 6t) dt$

Integrating term by term, we get:

$v(t) = 8 \int t^3 dt - 6 \int t dt = 8 \cdot \frac{t^4}{4} - 6 \cdot \frac{t^2}{2} + C$

Simplifying the expression:

$v(t) = 2t^4 - 3t^2 + C$

Here, $C$ is the constant of integration. To find the value of $C$, we use the given initial condition: $v = -4 m/s$ when $t = 2s$. Substituting these values into the velocity function:

$-4 = 2(2)^4 - 3(2)^2 + C$

$-4 = 32 - 12 + C$

$-4 = 20 + C$

Solving for $C$:

$C = -24$

Therefore, the velocity function is:

$v(t) = 2t^4 - 3t^2 - 24$

This velocity function describes the particle's velocity at any time $t$. It's a quartic function, indicating that the velocity changes non-linearly with time. The constant term, -24, reflects the initial conditions of the particle's motion.

2. Finding the Displacement Function

To determine the displacement function, $s(t)$, we need to recognize that velocity is the derivative of displacement with respect to time:

$v(t) = \frac{ds}{dt}$

Therefore, to find $s(t)$, we integrate the velocity function $v(t)$ with respect to time:

$s(t) = \int v(t) dt = \int (2t^4 - 3t^2 - 24) dt$

Integrating term by term, we get:

$s(t) = 2 \int t^4 dt - 3 \int t^2 dt - 24 \int dt = 2 \cdot \frac{t^5}{5} - 3 \cdot \frac{t^3}{3} - 24t + D$

Simplifying the expression:

$s(t) = \frac{2}{5}t^5 - t^3 - 24t + D$

Here, $D$ is the constant of integration. The problem states that the particle is initially at the origin, which means $s = 0$ when $t = 0$. Substituting these values into the displacement function:

$0 = \frac{2}{5}(0)^5 - (0)^3 - 24(0) + D$

$0 = 0 - 0 - 0 + D$

Solving for $D$:

$D = 0$

Thus, the displacement function is:

$s(t) = \frac{2}{5}t^5 - t^3 - 24t$

This displacement function provides the position of the particle relative to the origin at any given time $t$. It's a quintic function, indicating a more complex relationship between time and position compared to the velocity function. The absence of a constant term confirms that the particle starts at the origin.

In conclusion, this problem demonstrates the power of calculus in analyzing particle motion. By integrating the acceleration function, we successfully derived the velocity function, and by integrating the velocity function, we obtained the displacement function. The initial conditions provided were crucial in determining the constants of integration, leading to unique solutions for both functions. The velocity function, $v(t) = 2t^4 - 3t^2 - 24$, describes the particle's instantaneous velocity at any time $t$, while the displacement function, $s(t) = \frac{2}{5}t^5 - t^3 - 24t$, gives the particle's position relative to the origin at any time $t$. This analysis provides a comprehensive understanding of the particle's motion along the straight line.

Understanding particle motion is not just an academic exercise; it has practical applications in various fields, including engineering, aerospace, and even video game design. The principles discussed here form the foundation for more complex analyses of motion in higher dimensions and under the influence of various forces. Mastering these concepts is essential for anyone pursuing a career in physics or related fields.

By understanding the relationship between acceleration, velocity, and displacement, and by applying the techniques of calculus, we can effectively analyze and predict the motion of particles in a wide range of scenarios. The ability to solve problems like this is a testament to the power of mathematics in describing and understanding the physical world.