/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 (II) A particle's position as a ... [FREE SOLUTION] | 91Ó°ÊÓ

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(II) A particle's position as a function of time \(t\) is given $$ \vec{\mathbf{r}}=\left(5.0 t+6.0 t^{2}\right) \mathrm{m} \mathbf{i}+\left(7.0-3.0 t^{3}\right) \mathrm{m} \mathbf{j} . \text { At } t=5.0 \mathrm{s} $$ $$ \begin{array}{l}{\text { find the magnitude and direction of the particle's displace- }} \\ {\text { ment vector } \Delta \vec{\mathrm{r}} \text { relative to the point } \mathbf{r}_{0}=(0.0 \mathrm{i}+7.0 \mathrm{j}) \mathrm{m}}\end{array} $$

Short Answer

Expert verified
The displacement magnitude is approximately 413.61 m, with a direction of -64.8° from the positive x-axis.

Step by step solution

01

Determine the final position vector

The position vector of the particle at time \( t = 5.0 \) s is given by substituting \( t = 5.0 \) into the expression for \( \vec{\mathbf{r}} \):\[\vec{\mathbf{r}} = \left( 5.0 \times 5 + 6.0 \times (5.0)^2 \right) \mathbf{i} + \left( 7.0 - 3.0 \times (5.0)^3 \right) \mathbf{j} \]Simplifying, we find:\[= (25 + 150) \mathbf{i} + (7 - 375) \mathbf{j} \]\[= 175 \mathbf{i} - 368 \mathbf{j}\]
02

Calculate the displacement vector

The displacement vector \( \Delta \vec{\mathbf{r}} \) is the change in the position vector: \[\Delta \vec{\mathbf{r}} = \vec{\mathbf{r}} - \mathbf{r}_{0} = (175 \mathbf{i} - 368 \mathbf{j}) - (0 \mathbf{i} + 7 \mathbf{j})\]Simplifying, we have:\[\Delta \vec{\mathbf{r}} = (175 - 0) \mathbf{i} + (-368 - 7) \mathbf{j}\]\[= 175 \mathbf{i} - 375 \mathbf{j}\]
03

Find the magnitude of the displacement vector

The magnitude \( |\Delta \vec{\mathbf{r}}| \) of the displacement vector is calculated using:\[|\Delta \vec{\mathbf{r}}| = \sqrt{(175)^2 + (-375)^2}\]Calculating further:\[= \sqrt{30625 + 140625}\]\[= \sqrt{171250}\]\[\approx 413.61 \, \text{m}\]
04

Determine the direction of the displacement vector

The direction of the vector can be determined by finding the angle \( \theta \) with respect to the positive \( \mathbf{i} \)-axis using the tangent function:\[\tan \theta = \frac{-375}{175}\]Finding \( \theta \):\[\theta = \arctan\left(\frac{-375}{175}\right)\]\[\theta \approx \arctan(-2.14)\]\[\theta \approx -64.8^\circ \]This angle indicates the particle is moving downwards and to the right from the positive x-axis.

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Key Concepts

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

Position Vector
A position vector is a fundamental concept in physics and mathematics. It defines the position of a particle or a point in a coordinate system.
In our example, the position vector \(\vec{\mathbf{r}}\) describes where a particle is located in relation to a chosen origin point. It usually has both magnitude and direction.
In terms of coordinates, if you know the x and y positions, you can write the position vector like this:
  • \(\vec{\mathbf{r}} = x \mathbf{i} + y \mathbf{j}\)

Here, \(x\) and \(y\) represent the coordinates in a 2D space, while \(\mathbf{i}\) and \(\mathbf{j}\) are the unit vectors along the x and y axes respectively.
For our exercise, the position vector changes with time \(t\), indicating that the particle moves over time.
Magnitude of a Vector
The magnitude of a vector is a measure of its length or size, regardless of its direction. It's like asking, "How far does the vector reach from its starting point?"
Calculated using the Pythagorean theorem, the magnitude is always positive or zero.
To find the magnitude of a vector like \(\Delta \vec{\mathbf{r}} = 175 \mathbf{i} - 375 \mathbf{j}\), you use:
  • \(|\Delta \vec{\mathbf{r}}| = \sqrt{(175)^2 + (-375)^2}\)

This formula squares each component, adds them together, and takes the square root of the result.
For the vector in our exercise:
  • The magnitude is approximately \(413.61 \text{ m}\).
This shows how the vector stretches spatially between its start and endpoint.
Vector Direction
Understanding the direction of a vector helps indicate where a vector is pointing in space.
It is typically defined as an angle relative to the positive x-axis and can be described using trigonometric functions.
For example, to calculate the direction of a vector \(\Delta \vec{\mathbf{r}} = 175 \mathbf{i} - 375 \mathbf{j}\), you find the angle \(\theta\):
  • Use \(\tan \theta = \frac{-375}{175}\)
  • Find \(\theta\) using the inverse tangent function: \(\theta = \arctan\left(\frac{-375}{175}\right)\)

The angle in our exercise was calculated to be approximately \(-64.8^\circ\).
This negative value means the vector points downward below the x-axis.
The direction complements the magnitude by telling not just how far, but where the displacement happens.

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Most popular questions from this chapter

At \(t=0\), a particle starts from rest at \(x=0, y=0\), and moves in the \(x y\) plane with an acceleration \(\overrightarrow{\mathbf{a}}=(4.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2} .\) Determine \((a)\) the \(x\) and \(y\) components of velocity, \((b)\) the speed of the particle, and \((c)\) the position of the particle, all as a function of time. \((d)\) Evaluate all the above at \(t=2.0 \mathrm{~s}\).

You are driving south on a highway at \(25 \mathrm{~m} / \mathrm{s}\) (approximately \(55 \mathrm{mi} / \mathrm{h}\) ) in a snowstorm. When you last stopped, you noticed that the snow was coming down vertically, but it is passing the windows of the moving car at an angle of \(37^{\circ}\) to the horizontal. Estimate the speed of the snowflakes relative to the car and relative to the ground.

In hot pursuit, Agent Logan of the FBI must get directly across a 1200 -m-wide river in minimum time. The river's current is \(0.80 \mathrm{~m} / \mathrm{s}\), he can row a boat at \(1.60 \mathrm{~m} / \mathrm{s}\), and he can run \(3.00 \mathrm{~m} / \mathrm{s}\). Describe the path he should take (rowing plus running along the shore) for the minimum crossing time, and determine the minimum time.

Students shoot a plastic ball horizontally from a projectile launcher. They measure the distance \(x\) the ball travels horizontally, the distance \(y\) the ball falls vertically, and the total time \(t\) the ball is in the air for six different heights of the projectile launcher. Here is their data. $$ \begin{array}{ccc} \hline \text { Time, } & \text { Horizontal distance, } & \text { Vertical distance, } \\ t(\mathrm{~s}) & x(\mathrm{~m}) & y(\mathrm{~m}) \\ \hline 0.217 & 0.642 & 0.260 \\ 0.376 & 1.115 & 0.685 \\ 0.398 & 1.140 & 0.800 \\ 0.431 & 1.300 & 0.915 \\ 0.478 & 1.420 & 1.150 \\ 0.491 & 1.480 & 1.200 \\ \hline \end{array} $$ (a) Determine the best-fit straight line that represents \(x\) as a function of \(t .\) What is the initial speed of the ball obtained from the best-fit straight line? ( \(b\) ) Determine the best-fit quadratic equation that represents \(y\) as a function of \(t .\) What is the acceleration of the ball in the vertical direction?

A football is kicked at ground level with a speed of \(18.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(38.0^{\circ}\) to the horizontal. How much later does it hit the ground?

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