/*! 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 30 The vector \(-5.2 \overrightarro... [FREE SOLUTION] | 91Ó°ÊÓ

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The vector \(-5.2 \overrightarrow{\mathrm{A}}\) has a magnitude of \(34 \mathrm{m}\) and points in the positive \(x\) direction. Find (a) the \(x\) component and (b) the magnitude of the vector \(\overrightarrow{\mathbf{A}}\).

Short Answer

Expert verified
(a) The \(x\) component of \(\overrightarrow{\mathrm{A}}\) is approximately \\(-6.5385\\), (b) its magnitude is approximately \\ 6.5385 \\.

Step by step solution

01

Understanding the Problem

We have a vector \( -5.2 \overrightarrow{\mathrm{A}} \) with a given magnitude of \( 34 \text{ m} \) pointing in the positive \(x\) direction. We need to find the \(x\) component of the vector (\( \overrightarrow{\mathrm{A}} \)) and its magnitude.
02

Express Vector Relation

The given information is that \( -5.2 \overrightarrow{\mathrm{A}} \) has a magnitude of \( 34 \.\mathrm{m} \). This implies that \[ || -5.2 \overrightarrow{\mathrm{A}} || = 34 \] and the vector points in the positive \(x\) direction.
03

Calculate Magnitude of \( \overrightarrow{\mathbf{A}} \)

To find the magnitude of \( \overrightarrow{\mathrm{A}} \, \, \, || \overrightarrow{\mathrm{A}} || \, \,\) , we use the equation \( || k \overrightarrow{\mathrm{A}} || = |k| || \overrightarrow{\mathrm{A}} || \. \) Using \( k = -5.2 \), we have \( 34 = |-5.2| \, || \overrightarrow{\mathrm{A}} || \, \. \) Therefore, \[ || \overrightarrow{\mathrm{A}} || = \frac{34}{5.2} = 6.5385 \] approximately.
04

Find the \(x\) Component of \(\overrightarrow{\mathrm{A}} \)

Since the vector \( -5.2 \overrightarrow{\mathrm{A}} \) points in the positive \(x\) direction, \( \overrightarrow{\mathrm{A}} \) points in the negative \(x\) direction because it is scaled by \(-5.2\). The \(x\) component of \( \overrightarrow{\mathrm{A}} \) is thus \[ -|| \overrightarrow{\mathrm{A}} || = -6.5385 \] approximately.

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Key Concepts

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

Vector Components
Vectors are mathematical objects that have both magnitude and direction. Understanding vector components is crucial, as they break down a vector into parts that align with specific axes in a coordinate system. For a vector in two dimensions, its components are often labeled as the horizontal (x) and vertical (y) components.

The vector \(\overrightarrow{\text{A}}\) in our exercise has components determined by its direction and magnitude. Given that \(\overrightarrow{A}\) is scaled by -5.2 and points in the negative x-direction, its x-component can be directly affected by this scalar (which we'll delve into later). Throughout your studies, remember these key points about vector components:
  • Vector components are derived from the vector's magnitude and direction.
  • The components align with coordinate axes, usually represented as x, y (and z in 3D).
  • These components are essential for understanding vector operations like addition and subtraction.
Breaking vectors into their components makes complex physics calculations easier, allowing you to handle multi-dimensional data in smaller, manageable pieces.
Vector Direction
Vector direction refers to the orientation of a vector in space and is described using angles or relative positions to coordinate axes.

In the exercise, the direction is given as pointing along the positive x-axis. However, interestingly, \(\overrightarrow{A}\) itself points in the negative x-direction when considering the scalar multiplication by -5.2. Direction can be misunderstood without clear context, so always:
  • Visualize or draw a diagram to help understand the vector's real-world direction.
  • Assess how scalars (like negative numbers) may invert a vector's direction.
  • Use angles or unit vectors for clear representation in more complex cases.
Understanding direction is vital, not just in physics, but also in fields like engineering and computer graphics where accurate representation of vector direction can affect overall accuracy of models.
Scalar Multiplication
Scalar multiplication involves multiplying a vector by a scalar (a constant), which affects the vector's magnitude and potentially its direction.

In the given problem, multiplying \(\overrightarrow{A}\) by -5.2 results in a vector with a magnitude of 34 meters. The scalar -5.2 not only scales the magnitude but also reverses the direction of \(\overrightarrow{A}\). This reversal is due to the negative sign.
  • When you multiply a vector by a positive scalar, only its magnitude changes.
  • A negative scalar changes both the vector's magnitude and flips its direction.
  • Scalar multiplication is crucial in stretching or compressing vectors in simulations.
This operation is fundamental in vector mathematics, helping us scale vectors appropriately in various applications, from physics problems like this one to more advanced engineering tasks.

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Most popular questions from this chapter

Vector A points in the negative \(x\) direction. Vector \(\overrightarrow{\mathrm{B}}\) points at an angle of \(30.0^{\circ}\) above the positive \(x\) axis. Vector \(C\) has a magnitude of \(15 \mathrm{m}\) and points in a direction \(40.0^{\circ}\) below the positive \(x\) axis. Given that \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}=0,\) find the magnitudes of \(\overline{\mathbf{A}}\) and \(\overline{\mathbf{B}}\).

An off-roader explores the open desert in her Hummer. First she drives \(25^{\circ}\) west of north with a speed of \(6.5 \mathrm{km} / \mathrm{h}\) for 15 minutes, then due cast with a speed of \(12 \mathrm{km} / \mathrm{h}\) for \(7.5 \mathrm{minutes}\). She completes the final leg of her trip in 22 minutes. What are the direction and speed of travel on the final leg? (Assume her speed is constant on each leg, and that she returns to her starting point at the end of the final leg.)

You drive a car 680ft to the east, then 340ft to the north. (a) What is the magnitude of your displacement? (b) Using a sketch, estimate the direction of your displacement. (c) Verify your estimate in part (b) with a numerical calculation of the direction.

The Longitude Problem In \(1755,\) John Harrison \((1693-1776)\) completed his fourth precision chronometer, the \(\mathrm{H} 4\) which eventually won the celebrated Longitude Prize. (For the human drama behind the Longitude Prize, see Longitude, by Dava Sobel.) When the minute hand of the H4 indicated 10 minutes past the hour, it extended \(3.0 \mathrm{cm}\) in the horizontal direction. (a) How long was the \(\mathrm{H} 4\) 's minute hand? (b) At 10 minutes past the hour, was the extension of the minute hand in the vertical direction more than, less than, or equal to \(3.0 \mathrm{cm}\) ? Explain. (c) Calculate the vertical extension of the minute hand at 10 minutes past the hour.

A football is thrown horizontally with an initial velocity of \((16.6 \mathrm{m} / \mathrm{s}) \hat{\mathrm{x}} .\) Ignoring air resistance, the average acceleration of the football over any period of time is \(\left(-9.81 \mathrm{m} / \mathrm{s}^{2}\right) \hat{y}\) (a) Find the velocity vector of the ball \(1.75 \mathrm{s}\) after it is thrown. (b) Find the magnitude and direction of the velocity at this time.

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