Analyzing Particle Motion: Paths, Distance, Velocity, And Acceleration

Hey guys! Let's dive into some physics fun! Today, we're going to explore particle motion. We'll be looking at how to figure out the paths, distances, velocities, and accelerations of points when we're given their position vectors. It's like being a detective for moving objects! We'll start with the position vectors and work our way through to understand how things move. So, let's get started and unravel the secrets of motion together. This will be a fun ride, and by the end, you'll be able to analyze particle motion like a pro! I will explain to you step by step with the process of the solution so you can understand it better. I will include some figures to visualize the process.

Understanding the Position Vectors and Paths

Okay, so first things first: we're given two position vectors. Think of a position vector like an arrow that points from a starting point (usually the origin) to where a particle is located at a specific time. These vectors change as time goes on, which means our particle is moving! We've got two different particles here, each with its own position vector. The first one, ${\/vec{r_1}(t) = \alpha \hat{i} + \beta t \hat{j}}$, describes the position of our first particle. Here, ${\alpha}$ and ${\beta}$ are constants. The ${\hat{i}}$ and ${\hat{j}}$ are unit vectors that point along the x and y axes, respectively. Essentially, the first particle has a constant x-coordinate (${\alpha}$) and a y-coordinate that changes linearly with time (${\beta t}$). This is super important because it tells us that this particle moves in a straight line, parallel to the y-axis, but with a constant velocity in the y direction. Our second particle is described by the position vector: ${\/vec{r_2}(t) = R \cos(\omega t) \hat{i} + R \sin(\omega t) \hat{j} + k \hat{y}}$. This one's a bit more interesting! Here, ${R}$, ${\omega}$, and ${k}$ are also constants. The ${\hat{i}}$ and ${\hat{j}}$ still represent the unit vectors along the x and y axes. But hey, it seems we have a typo here because ${k}$ must be a variable, not a constant. But that's okay, we can fix it. The x and y coordinates of this particle vary with sine and cosine functions. This means the particle is moving in a circular path in the x-y plane, with a radius of ${R}$ and an angular velocity of ${\omega}$. The ${k}$ variable represents that this vector is in the y-axis. I will include the figures to understand this better. Now, to find the paths of these particles, we need to think about what the equations of their paths look like. For the first particle, since the x-coordinate is constant, it looks like it will move in a straight line. For the second particle, the sine and cosine functions will trace out a circle.

Particle 1 Path Analysis

Let's break down the first particle's path. Its position vector is ${\/vec{r_1}(t) = \alpha \hat{i} + \beta t \hat{j}}$. As we discussed, the x-coordinate is always ${\alpha}$, and the y-coordinate is ${\beta t}$, which changes linearly with time. This tells us the particle is moving in a straight line. The equation for this line is super simple: ${x = \alpha}$. This means no matter what the time is, the particle's x-coordinate will always be ${\alpha}$. Only the y-coordinate changes with time. Since the y-coordinate is always changing, the particle is moving in a vertical line with a constant speed of ${\beta}$. In our case, the y-coordinate is equal to ${\beta t}$, which means we have a linear relationship. The graph of ${x = \alpha}$ is just a vertical line through the point ${(\alpha, 0)}$. It's a fundamental concept in physics, and it helps you understand how objects move in space. The line moves with time in the ${y}$ direction.

Particle 2 Path Analysis

Now, let's explore the second particle, with the position vector ${\/vec{r_2}(t) = R \cos(\omega t) \hat{i} + R \sin(\omega t) \hat{j} + k \hat{y}}$. The x-coordinate is ${R \cos(\omega t)}$, and the y-coordinate is ${R \sin(\omega t)}$. These are the parametric equations for a circle! To find the equation of the path, we can use a trick: square both the x and y components and add them together. So, ${x^2 + y^2 = (R \cos(\omega t))^2 + (R \sin(\omega t))^2 = R^2 \cos^2(\omega t) + R^2 \sin^2(\omega t) = R^2(\cos^2(\omega t) + \sin^2(\omega t))}$. Since ${\cos^2(\omega t) + \sin^2(\omega t) = 1}$, this simplifies to ${x^2 + y^2 = R^2}$. And that, my friends, is the equation of a circle centered at the origin, with a radius of ${R}$. The path is a circle in the x-y plane, and the particle moves around the circle with a constant speed, because the angular velocity ${\omega}$ is constant. The ${k}$ variable represents that the vector is along the y-axis. The particle is tracing a circular path in a plane, completing one full rotation for every ${2\pi/\omega}$ seconds. This concept is crucial when studying rotational motion, and it is widely used in physics.

Calculating the Distance Between the Particles

Alright, let's calculate the distance between the particles at any given time. The distance between two points in space is calculated using the distance formula, which is derived from the Pythagorean theorem. We can find this by subtracting the position vectors and finding the magnitude of the resulting vector. So, the relative position vector ${\/vec{r}_{12}(t)}$ is ${\/vec{r}_2(t) - \vec{r}_1(t)}$. Let's calculate that now: ${ \vec{r}_{12}(t) = (R \cos(\omega t) \hat{i} + R \sin(\omega t) \hat{j} + k \hat{y}) - (\alpha \hat{i} + \beta t \hat{j}) }$. Grouping the ${\hat{i}}$ and ${\hat{j}}$ terms together, we get: ${ \vec{r}_{12}(t) = (R \cos(\omega t) - \alpha) \hat{i} + (R \sin(\omega t) - \beta t) \hat{j} + k \hat{y} }$. The distance ${d(t)}$ between the particles at any time ${t}$ is the magnitude of this relative position vector. This is a vector, and we have to find the magnitude of the vector. We calculate the magnitude of the vector ${\/vec{r}_{12}(t)}$ as follows: ${ d(t) = \sqrt{((R \cos(\omega t) - \alpha)^2 + (R \sin(\omega t) - \beta t)^2 + k^2)} }$. This is the function that describes the distance between the two particles as a function of time. We can also expand this expression and simplify it further, but the key is that this formula provides us with the distance between the particles at any time.

Finding the Velocity and Acceleration

Time to find the velocity and acceleration. Velocity tells us how fast a particle is moving and in what direction. Acceleration tells us how the velocity is changing over time. To find the velocity, we take the derivative of the position vector with respect to time. For acceleration, we take the derivative of the velocity with respect to time. Let's do this step by step. I will include the process to solve the problem for both particles.

Particle 1: Velocity and Acceleration

Let's start with particle 1, ${\/vec{r_1}(t) = \alpha \hat{i} + \beta t \hat{j}}$. To find the velocity ${\/vec{v}_1(t)}$, we take the derivative of ${\/vec{r_1}(t)}$ with respect to time: ${ \vec{v}_1(t) = \frac{d}{dt} (\alpha \hat{i} + \beta t \hat{j}) = 0 \hat{i} + \beta \hat{j} }$. So, the velocity vector for the first particle is ${\/vec{v}_1(t) = \beta \hat{j}}$. This means the particle is moving with a constant velocity ${\beta}$ along the y-axis. The acceleration ${\/vec{a}_1(t)}$ is the derivative of the velocity with respect to time: ${ \vec{a}_1(t) = \frac{d}{dt} (\beta \hat{j}) = 0 \hat{j} }$. This tells us that the acceleration of the first particle is zero. Since the particle is moving at a constant velocity, there is no change in velocity, so no acceleration.

Particle 2: Velocity and Acceleration

Now, let's look at particle 2, ${\/vec{r_2}(t) = R \cos(\omega t) \hat{i} + R \sin(\omega t) \hat{j} + k \hat{y}}$. To find the velocity ${\/vec{v}_2(t)}$, we take the derivative of ${\/vec{r_2}(t)}$ with respect to time: ${ \vec{v}_2(t) = \frac{d}{dt} (R \cos(\omega t) \hat{i} + R \sin(\omega t) \hat{j} + k \hat{y}) = -R \omega \sin(\omega t) \hat{i} + R \omega \cos(\omega t) \hat{j} + 0 \hat{y} }$. The velocity vector is ${\/vec{v}_2(t) = -R \omega \sin(\omega t) \hat{i} + R \omega \cos(\omega t) \hat{j}}$. Now, let's find the acceleration ${\/vec{a}_2(t)}$ by taking the derivative of the velocity with respect to time: ${ \vec{a}_2(t) = \frac{d}{dt} (-R \omega \sin(\omega t) \hat{i} + R \omega \cos(\omega t) \hat{j}) = -R \omega^2 \cos(\omega t) \hat{i} - R \omega^2 \sin(\omega t) \hat{j} }$. So, the acceleration vector for the second particle is ${\/vec{a}_2(t) = -R \omega^2 \cos(\omega t) \hat{i} - R \omega^2 \sin(\omega t) \hat{j}}$. This shows that the acceleration is always directed towards the center of the circle, which is the origin. Because of the ${\omega^2}$, this shows that the acceleration is proportional to the radius of the circle. This is an example of centripetal acceleration, which is crucial in circular motion. The negative signs indicate that the acceleration vector points towards the center of the circle, opposite the direction of the position vector.

Conclusion: Summarizing the Motion

Alright, let's wrap things up! We've covered a lot of ground today, analyzing the motion of two particles based on their position vectors. We've determined their paths, calculated the distance between them, and found their velocities and accelerations. Remember, the first particle moves in a straight line with a constant velocity, while the second particle moves in a circular path with constant speed. This knowledge is important for understanding the basics of kinematics, and this concept is used in many fields, such as engineering, robotics, and aerospace. Keep practicing, and you'll become a pro at analyzing particle motion. Great job, guys!