Calculating Net Displacement From Position Vector A Physics Guide

by THE IDEN 66 views

In physics, understanding the motion of particles is a fundamental concept. One crucial aspect of describing motion is displacement, which refers to the change in position of an object. Unlike distance, which is the total path length traveled, displacement is a vector quantity that considers only the initial and final positions. This article delves into the calculation of net displacement for a particle, given its position vector as a function of time. We will explore how to determine the net displacement after a specific time interval, focusing on a scenario where the position vector is defined by r⃗=(3t2i^+4t2j^+7k^){\vec{r} = (3t^2 \hat{i} + 4t^2 \hat{j} + 7 \hat{k})} meters at time t. This problem is a classic example of applying vector calculus to understand motion in three dimensions, and it's a cornerstone in introductory mechanics courses. The ability to calculate displacement from a position vector is essential for predicting the trajectory and final location of moving objects, making it a vital skill for physicists and engineers alike. By breaking down the position vector into its components and analyzing how each component changes with time, we can accurately determine the net displacement. This understanding is not only crucial for academic problem-solving but also has practical applications in fields such as robotics, aerospace, and even video game development, where precise motion tracking is necessary. This exploration will not only solidify the understanding of displacement but also highlight the broader implications of vector analysis in physics and related disciplines. The principles discussed here lay the foundation for more advanced topics in kinematics and dynamics, making this a fundamental concept to master for anyone pursuing studies or a career in the sciences or engineering.

Problem Statement

The position vector of a particle is given by

r⃗=(3t2i^+4t2j^+7k^) m\vec{r} = (3t^2 \hat{i} + 4t^2 \hat{j} + 7 \hat{k}) \text{ m}

at a given time t{t}. We aim to find the net displacement of the particle after 10 seconds.

Understanding the Position Vector

The position vector, denoted as r⃗{\vec{r}}, provides a comprehensive description of a particle's location in space at any given time. It is a vector quantity, meaning it has both magnitude and direction. In this case, the position vector is expressed in terms of its components along the three coordinate axes: the x-axis (i^{\hat{i}}), the y-axis (j^{\hat{j}}), and the z-axis (k^{\hat{k}}). Each component is a function of time t{t}, indicating how the particle's position changes along that axis as time progresses. Specifically, the x-component is 3t2{3t^2}, the y-component is 4t2{4t^2}, and the z-component is a constant 7. This representation allows us to track the particle's movement in three-dimensional space. The quadratic dependence on time in the x and y components suggests that the particle's motion is not uniform; it is accelerating along these axes. The constant z-component, on the other hand, indicates that the particle's position along the z-axis remains unchanged throughout its motion. Understanding the position vector is crucial for calculating other kinematic quantities, such as velocity and acceleration, which are derived from the rate of change of position with respect to time. Furthermore, the position vector serves as the foundation for determining the particle's displacement over a given time interval, which is the primary focus of this problem. By analyzing the position vector, we can gain insights into the particle's trajectory, its speed, and the forces acting upon it. This makes the position vector a central concept in the study of mechanics and a vital tool for analyzing the motion of objects in various physical systems. The ability to interpret and manipulate position vectors is essential for solving a wide range of problems in physics and engineering.

Solution

Step 1: Define Displacement

Displacement, in physics, refers to the change in position of an object. It's a vector quantity, meaning it has both magnitude and direction. Mathematically, displacement (Δr⃗{\Delta \vec{r}}) is defined as the difference between the final position vector (r⃗f{\vec{r}_f}) and the initial position vector (r⃗i{\vec{r}_i}): $\Delta \vec{r} = \vec{r}_f - \vec{r}_i$ To calculate the net displacement of the particle after 10 seconds, we need to determine its position at the initial time (t=0{t = 0}) and its position at the final time (t=10 s{t = 10 \text{ s}}). The displacement vector will then be the difference between these two positions. This vector provides crucial information about the overall change in the particle's location during the 10-second interval. It indicates not only how far the particle has moved but also the direction of its movement. Unlike distance, which is the total path length traveled, displacement is concerned only with the start and end points of the motion. Understanding the concept of displacement is fundamental in physics, particularly in the study of kinematics, which deals with the motion of objects without considering the forces that cause the motion. Displacement is a key component in defining other kinematic quantities such as velocity and acceleration. The ability to calculate displacement is essential for predicting the motion of objects, whether they are moving in a straight line or along a curved path. Moreover, the concept of displacement extends beyond simple particle motion; it is also applicable in more complex systems, such as the motion of rigid bodies or the flow of fluids. Therefore, a solid grasp of displacement is crucial for anyone studying physics or engineering, as it forms the basis for understanding more advanced concepts and solving a wide range of problems. The clarity and precision with which displacement is defined and calculated are critical for accurate modeling and analysis of physical phenomena.

Step 2: Calculate the Initial Position Vector

To determine the initial position vector (r⃗i{\vec{r}_i}), we need to evaluate the given position vector r⃗{\vec{r}} at the initial time, which is t=0{t = 0}. Substituting t=0{t = 0} into the expression for r⃗{\vec{r}}, we get: $\vecr}_i = (3(0)^2 \hat{i} + 4(0)^2 \hat{j} + 7 \hat{k}) \text{ m}$ Simplifying this expression, we find $\vec{r_i = (0 \hat{i} + 0 \hat{j} + 7 \hat{k}) \text{ m}$ This result indicates that at the initial time, the particle is located at the point (0, 0, 7) in Cartesian coordinates. The x and y components of the initial position vector are zero, meaning the particle starts on the z-axis. The z-component is 7 meters, indicating the particle's initial height above the xy-plane. This initial position serves as the reference point from which we will measure the particle's displacement over the subsequent 10 seconds. Understanding the initial position is crucial for calculating the change in position, which is the displacement. Without knowing the starting point, it is impossible to determine the net movement of the particle. The initial position vector provides a fixed point in space against which the particle's later positions can be compared. This calculation highlights the importance of understanding the role of time in the position vector. The position vector is a function of time, meaning it describes the particle's location at any given moment. By setting t=0{t = 0}, we effectively freeze time and determine the particle's location at the very beginning of its motion. This concept is fundamental in kinematics, where we analyze the motion of objects as a function of time. The initial position vector is a key parameter in describing the initial conditions of the motion, which are essential for predicting the particle's future trajectory. Therefore, accurately calculating the initial position vector is a crucial step in solving problems involving displacement and motion.

Step 3: Calculate the Final Position Vector

To find the final position vector (r⃗f{\vec{r}_f}), we need to evaluate the given position vector r⃗{\vec{r}} at the final time, which is t=10 s{t = 10 \text{ s}}. Substituting t=10{t = 10} into the expression for r⃗{\vec{r}}, we get: $\vecr}_f = (3(10)^2 \hat{i} + 4(10)^2 \hat{j} + 7 \hat{k}) \text{ m}$ Simplifying this expression, we find $\vec{r_f = (300 \hat{i} + 400 \hat{j} + 7 \hat{k}) \text{ m}$ This result indicates that at t=10 s{t = 10 \text{ s}}, the particle is located at the point (300, 400, 7) in Cartesian coordinates. The x-component of the final position vector is 300 meters, and the y-component is 400 meters. The z-component remains at 7 meters, which means the particle's height above the xy-plane has not changed during the motion. The significant changes in the x and y components indicate that the particle has moved considerably in the xy-plane over the 10-second interval. This final position is crucial for determining the particle's displacement, which is the difference between the final and initial positions. Understanding the final position allows us to visualize the particle's location at the end of its motion and to quantify how far it has moved from its starting point. The calculation of the final position vector highlights the importance of the time variable in the position vector equation. The position vector is a dynamic quantity that changes with time, and by substituting a specific time value, we can pinpoint the particle's location at that instant. This concept is fundamental in physics, particularly in kinematics, where we analyze the motion of objects as a function of time. The final position vector, along with the initial position vector, provides a complete picture of the particle's displacement over a given time interval. This information is essential for calculating other kinematic quantities, such as average velocity and average speed, and for understanding the overall motion of the particle. Therefore, accurately calculating the final position vector is a critical step in solving problems involving displacement and motion analysis.

Step 4: Calculate the Net Displacement

Now that we have both the initial position vector (r⃗i=(0i^+0j^+7k^) m{\vec{r}_i = (0 \hat{i} + 0 \hat{j} + 7 \hat{k}) \text{ m}}) and the final position vector (r⃗f=(300i^+400j^+7k^) m{\vec{r}_f = (300 \hat{i} + 400 \hat{j} + 7 \hat{k}) \text{ m}}), we can calculate the net displacement (Δr⃗{\Delta \vec{r}}) using the formula: $\Delta \vecr} = \vec{r}_f - \vec{r}_i$ Substituting the values, we get $\Delta \vec{r = (300 \hati} + 400 \hat{j} + 7 \hat{k}) \text{ m} - (0 \hat{i} + 0 \hat{j} + 7 \hat{k}) \text{ m}$ Performing the vector subtraction, we find $\Delta \vec{r = (300 \hat{i} + 400 \hat{j} + (7-7) \hat{k}) \text{ m}$ $\Delta \vec{r} = (300 \hat{i} + 400 \hat{j} + 0 \hat{k}) \text{ m}$ Thus, the net displacement of the particle after 10 seconds is (300i^+400j^+0k^) m{(300 \hat{i} + 400 \hat{j} + 0 \hat{k}) \text{ m}}. This result indicates that the particle has moved 300 meters in the x-direction and 400 meters in the y-direction. The z-component of the displacement is zero, meaning there was no net change in the particle's position along the z-axis. The displacement vector provides a concise description of the particle's overall change in position during the 10-second interval. It tells us how far the particle has moved and in what direction, without regard to the actual path taken. The displacement vector is a fundamental concept in physics, particularly in kinematics, as it allows us to quantify the change in position of an object over time. This information is essential for understanding the motion of objects and for making predictions about their future positions. The calculation of the net displacement highlights the importance of vector operations in physics. Vector subtraction is a key tool for determining the difference between two vector quantities, such as position vectors. The result of the subtraction is another vector, which represents the displacement. Understanding how to perform vector operations is crucial for solving a wide range of problems in physics and engineering. The net displacement vector provides a comprehensive description of the particle's overall movement, making it a vital concept for analyzing motion.

Step 5: Calculate the Magnitude of the Displacement

To determine the magnitude of the displacement, which represents the total distance the particle has moved from its initial position, we calculate the magnitude of the displacement vector Δr⃗{\Delta \vec{r}}. The magnitude of a vector A⃗=(Axi^+Ayj^+Azk^){\vec{A} = (A_x \hat{i} + A_y \hat{j} + A_z \hat{k})} is given by: $|\vecA}| = \sqrt{A_x^2 + A_y^2 + A_z^2}$ In our case, the displacement vector is Δr⃗=(300i^+400j^+0k^) m{\Delta \vec{r} = (300 \hat{i} + 400 \hat{j} + 0 \hat{k}) \text{ m}}. Therefore, the magnitude of the displacement is $egin{aligned |\Delta \vec{r}| &= \sqrt{(300)^2 + (400)^2 + (0)^2} \ &= \sqrt{90000 + 160000 + 0} \ &= \sqrt{250000} \ &= 500 \text{ m} \end{aligned}$ Thus, the magnitude of the net displacement of the particle after 10 seconds is 500 meters. This value represents the straight-line distance between the particle's initial and final positions. It provides a measure of the overall change in position, regardless of the path the particle may have taken. The magnitude of the displacement is a scalar quantity, meaning it has only magnitude and no direction. It is a useful measure for quantifying the total distance covered by the particle during its motion. The calculation of the magnitude of the displacement highlights the importance of the Pythagorean theorem in vector analysis. The magnitude of a vector is essentially the hypotenuse of a right triangle formed by the vector's components. In this case, the x and y components of the displacement vector form the legs of the right triangle, and the magnitude of the displacement is the hypotenuse. This geometric interpretation of vector magnitude is a fundamental concept in physics and engineering. Understanding how to calculate the magnitude of a vector is crucial for solving a wide range of problems, including those involving displacement, velocity, acceleration, and force. The magnitude of the net displacement provides a concise and meaningful measure of the particle's overall movement, making it a valuable tool for analyzing motion.

Final Answer

The net displacement of the particle after 10 seconds is 500 m.

Significance of the Result

The result that the net displacement of the particle after 10 seconds is 500 meters provides a clear and concise measure of the particle's overall change in position. This value represents the straight-line distance between the particle's initial and final locations, regardless of the actual path it followed. The significance of this result lies in its ability to quantify the overall effect of the particle's motion over the given time interval. Unlike the total distance traveled, which would depend on the specific trajectory of the particle, the displacement focuses solely on the end points of the motion. This makes displacement a particularly useful concept in physics, as it allows us to analyze the net effect of motion without being concerned with the details of the path taken. The 500-meter displacement indicates that the particle has moved a substantial distance from its starting point. This information can be crucial in various applications, such as determining the range of a projectile, the movement of a robotic arm, or the navigation of a vehicle. Furthermore, the displacement vector, with its components in the x and y directions, provides additional information about the direction of the motion. In this case, the particle has moved both horizontally and vertically, and the relative magnitudes of the x and y components indicate the relative contributions of these motions to the overall displacement. The concept of displacement is fundamental in kinematics, the branch of physics that deals with the motion of objects. It is closely related to other kinematic quantities, such as velocity and acceleration, which are derived from the rate of change of displacement with respect to time. The ability to calculate displacement is essential for solving a wide range of problems in physics and engineering, and it forms the basis for understanding more complex concepts in dynamics and mechanics. The result of 500 meters highlights the importance of understanding vector quantities and their properties. Displacement is a vector quantity, meaning it has both magnitude and direction. This is in contrast to scalar quantities, such as distance, which have only magnitude. The correct calculation and interpretation of vector quantities are crucial for accurate modeling and analysis of physical systems. The net displacement of 500 meters provides a valuable insight into the particle's motion and underscores the significance of the displacement concept in physics and its applications.