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In a particle accelerator, the position vector of a particle is initially estimated as \(\vec{r}=6.0 \hat{\mathrm{i}}-7.0 \hat{\mathrm{j}}+3.0 \hat{\mathrm{k}}\) and after \(10 \mathrm{~s}\), it is estimated to be \(\vec{r}=-3.0 \hat{\mathrm{i}}+9.0 \hat{\mathrm{j}}-3.0 \hat{\mathrm{k}}\), all in meters. In unit vector notation, what is the average velocity of the particle?

Short Answer

Expert verified
The average velocity is \(-0.9 \hat{\mathrm{i}} + 1.6 \hat{\mathrm{j}} - 0.6 \hat{\mathrm{k}} \) m/s.

Step by step solution

01

Understand the Problem

We are tasked with finding the average velocity of a particle given its initial and final position vectors over a time interval.
02

Recall the Formula for Average Velocity

The average velocity \( \vec{v}_{avg} \) is given by \( \vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t} \), where \( \Delta \vec{r} \) is the change in position vector and \( \Delta t \) is the time interval.
03

Calculate the Change in Position Vector

Compute the change in the position vector \( \Delta \vec{r} = \vec{r}_{final} - \vec{r}_{initial} \). Substitute the given vectors: \[ \Delta \vec{r} = (-3.0 \hat{\mathrm{i}} + 9.0 \hat{\mathrm{j}} - 3.0 \hat{\mathrm{k}}) - (6.0 \hat{\mathrm{i}} - 7.0 \hat{\mathrm{j}} + 3.0 \hat{\mathrm{k}}) \].
04

Perform the Subtraction

Calculate each component in \( \Delta \vec{r} \): - For \( \hat{\mathrm{i}} \): \( -3.0 - 6.0 = -9.0 \) - For \( \hat{\mathrm{j}} \): \( 9.0 - (-7.0) = 16.0 \) - For \( \hat{\mathrm{k}} \): \( -3.0 - 3.0 = -6.0 \). Thus, \( \Delta \vec{r} = -9.0 \hat{\mathrm{i}} + 16.0 \hat{\mathrm{j}} - 6.0 \hat{\mathrm{k}} \).
05

Calculate Average Velocity

Use the formula \( \vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t} \) with \( \Delta t = 10 \, \mathrm{s} \) to find \( \vec{v}_{avg} = \frac{-9.0 \hat{\mathrm{i}} + 16.0 \hat{\mathrm{j}} - 6.0 \hat{\mathrm{k}}}{10} \).
06

Simplify the Expression

Simplify the expression by dividing each component by 10: - \( \hat{\mathrm{i}} \): \( \frac{-9.0}{10} = -0.9 \) - \( \hat{\mathrm{j}} \): \( \frac{16.0}{10} = 1.6 \) - \( \hat{\mathrm{k}} \): \( \frac{-6.0}{10} = -0.6 \).So, \( \vec{v}_{avg} = -0.9 \hat{\mathrm{i}} + 1.6 \hat{\mathrm{j}} - 0.6 \hat{\mathrm{k}} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Particle Motion
Particle motion refers to the movement of particles, and in physics, this movement is often described using vectors. In the context of a particle accelerator, particles move through space at high speeds.
This movement can be measured over time to understand various parameters, such as velocity and acceleration.
To get a comprehensive grasp of particle motion, it is crucial to comprehend how particles change their positions with respect to time. For example, by knowing the position vector of a particle at different moments, you can determine how far and in what direction it has traveled.
The change in position over a time interval leads to understanding average velocity, which is a key concept in analyzing motion.
  • Helps us determine the speed and direction of a particle.
  • Provides insight into how forces may be acting upon the particle.
  • Allows us to draw conclusions about the behavior of particles in a given environment, such as a particle accelerator.
It is essential to break down these components to see not only the effects but also the causes of such dynamic movements.
Position Vector
A position vector is a mathematical construct that provides a compact way to describe the location of a point in space. When dealing with particle motion in physics, position vectors allow us to succinctly represent where a particle is at any given time.
Position vectors use a coordinate system, like the Cartesian coordinate system, to pinpoint a particle's location in one, two, or three dimensions. For instance, the position vector \(\vec{r}=6.0\hat{i}-7.0\hat{j}+3.0\hat{k}\) is a 3D representation indicating the particle's specific alignment along the x, y, and z axes.
Understanding position vectors is essential because:
  • They simplify the description of movement in space.
  • They allow easy calculation of distances and directions between different points.
  • They form the basis for computing changes in position, leading us to important concepts like displacement and velocity.
When working with position vectors, always pay attention to the magnitude and direction of each component. This careful analysis helps unravel complex motion patterns of accelerated particles.
Vector Subtraction
Vector subtraction is a fundamental operation used to find the difference between two vectors. It involves subtracting the components of one vector from the components of another, often to find out how much one vector has displaced another.
In the context of the given problem involving particles in an accelerator, vector subtraction helps us find the change in position: \(\Delta \vec{r} = \vec{r}_{final} - \vec{r}_{initial}\). This operation gives us the displacement vector, which tells us the change in the particle's position over time.
Here's how vector subtraction works with unit vectors:
  • Subtract the respective components of the initial vector from the final vector.
  • The resulting vector components illustrate the displacement in each direction.
  • A simplified view is often more illustrative, as in \( \Delta \vec{r} = -9.0 \hat{\mathrm{i}} + 16.0 \hat{\mathrm{j}} - 6.0 \hat{\mathrm{k}} \).
Vector subtraction is not only about magnitude but also direction, helping us understand precisely how a particle has moved. This understanding lays the groundwork for determining other parameters like velocity, crucial for analyzing motion dynamics.

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Most popular questions from this chapter

Suppose that a shot putter can put a shot at the world-class speed \(v_{0}=15.00 \mathrm{~m} / \mathrm{s}\) and at a height of \(2.160 \mathrm{~m}\). What horizontal distance would the shot travel if the launch angle \(\theta_{0}\) is (a) \(45.00^{\circ}\) and (b) \(42.00^{\circ} ?\) The answers indicate that the angle of \(45^{\circ}\), which maximizes the range of projectile motion, does not maximize the horizontal distance when the launch and landing are at different heights.

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A rugby player runs with the ball directly toward his opponent's goal, along the positive direction of an \(x\) axis. He can legally pass the ball to a teammate as long as the ball's velocity relative to the field does not have a positive \(x\) component. Suppose the player runs at speed \(3.5 \mathrm{~m} / \mathrm{s}\) relative to the field while he passes the ball with velocity \(\vec{v}_{B P}\) relative to himself. If \(\vec{v}_{B P}\) has magnitude \(6.0\) \(\mathrm{m} / \mathrm{s}\), what is the smallest angle it can have for the pass to be legal?

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