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Chapter 1: Problem 39

The \(x\) vector component of a displacement vector \(\overrightarrow{\mathbf{r}}\) has a magnitude of \(125 \mathrm{m}\) and points along the negative \(x\) axis. The \(y\) vector component has a magnitude of \(184 \mathrm{m}\) and points along the negative \(y\) axis. Find the magnitude and direction of \(\overrightarrow{\mathbf{r}}\). Specify the direction with respect to the negative ic

Short Answer

Expert verified
Magnitude: 222.4 m; Direction: 235.0掳 from negative x-axis.

Step by step solution

01

Understanding Vector Components

The given displacement vector \( \overrightarrow{\mathbf{r}} \) has an \( x \)-component of \( 125 \, \mathrm{m} \) in the negative direction, represented as \( -125 \, \mathrm{m} \). Likewise, it has a \( y \)-component of \( 184 \, \mathrm{m} \) also in the negative direction, represented as \( -184 \, \mathrm{m} \). Thus, the components of the vector are \( (-125, -184) \).
02

Calculate the Magnitude

The magnitude of a vector with components \( (x, y) \) is calculated using the Pythagorean theorem: \[ r = \sqrt{x^2 + y^2} \]. Substitute the values to find the magnitude of \( \overrightarrow{\mathbf{r}} \): \[ r = \sqrt{(-125)^2 + (-184)^2} = \sqrt{15625 + 33856} = \sqrt{49481} \approx 222.4 \, \mathrm{m} \]. This approximation gives us the magnitude of the displacement vector.
03

Calculate the Direction

The direction \( \theta \) with respect to the negative \( x \)-axis is found using the tangent function: \[ \theta = \arctan\left(\frac{|y|}{|x|}\right) \]. Here, \( |y| = 184 \) and \( |x| = 125 \): \[ \theta = \arctan\left(\frac{184}{125}\right) \approx 55.0^\circ \]. Since both components are negative, the vector \( \overrightarrow{\mathbf{r}} \) is in the third quadrant, so the angle is measured counterclockwise from the negative \( x \)-axis, giving us \( 180^\circ + 55.0^\circ = 235.0^\circ \).

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Key Concepts

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

Displacement Vector
A displacement vector represents a change in position from one point to another in space. It has both a magnitude, which tells us how far we've moved, and a direction, which indicates where we've moved. In this problem, the displacement vector \( \overrightarrow{\mathbf{r}} \) has components in the negative \( x \)-axis and \( y \)-axis, meaning it moves both left and downward. This is key for understanding the final direction in relation to a standard coordinate system.
Vector Components
Vector components are essential for breaking down a vector into its perpendicular directions, typically along the \( x \) and \( y \) axes. Here, the displacement vector has components \((-125, -184)\). This notation means the vector moves 125 meters left along the \( x \)-axis and 184 meters down along the \( y \)-axis. Using negative signs shows direction, distinguishing the vector's movement direction clearly.
Pythagorean Theorem
The magnitude of the vector is found using the Pythagorean theorem, which relates the sides of a right triangle. For a vector with components \((x, y)\), the magnitude \( r \) is \( r = \sqrt{x^2 + y^2} \). Substituting \(-125\) for \( x \) and \(-184\) for \( y \), we calculate: \[ r = \sqrt{(-125)^2 + (-184)^2} = \sqrt{15625 + 33856} = \sqrt{49481} \approx 222.4 \, \mathrm{m} \].
This result gives us the overall length of the displacement vector, regardless of direction.
Angle Calculation
To find the direction of the vector, we use angle calculation with trigonometry. The tangent function, specifically, helps us with \( \theta = \arctan\left(\frac{|y|}{|x|}\right) \).
The components are \(|y| = 184\) and \(|x| = 125\): \[ \theta = \arctan\left(\frac{184}{125}\right) \approx 55.0^\circ \].
Since the vector is in the third quadrant (both values negative), the angle is measured counterclockwise from the negative \( x \)-axis, which requires adding \( 180^\circ \). Thus, the complete angle is \( 235.0^\circ \), giving full directional sense in the coordinate plane.

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Most popular questions from this chapter

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