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Vector \(\overrightarrow{\mathbf{A}}\) has a magnitude of \(50 \mathrm{~km}\) and points in the positive \(x\) direction. A second vector, \(\overrightarrow{\mathbf{B}}\), has a magnitude of \(120 \mathrm{~km}\) and points at an angle of \(70^{\circ}\) below the \(x\) axis. Which vector has (a) the greater \(x\) component and (b) the greater \(y\) component?

Short Answer

Expert verified
(a) Vector A has the greater x-component; (b) Vector A has the greater y-component.

Step by step solution

01

Understand the Vectors

Vector \( \overrightarrow{\mathbf{A}} \) has a magnitude of 50 km in the positive \( x \) direction, meaning it lies along the \( x \)-axis, so its \( x \)-component is 50 km, and the \( y \)-component is 0 km. Vector \( \overrightarrow{\mathbf{B}} \) has a magnitude of 120 km at an angle of 70° below the \( x \)-axis, which means it has both \( x \) and \( y \) components.
02

Calculate x-components of Vectors

The \( x \)-component of \( \overrightarrow{\mathbf{A}} \) is simply 50 km because it lies along the \( x \)-axis. For \( \overrightarrow{\mathbf{B}} \), use the formula: \( B_x = B \cos(\theta) \). Thus, \( B_x = 120 \cos(70^{\circ}) \approx 120 \cdot 0.3420 \approx 41.0 \) km.
03

Determine Greater x-component

Compare the \( x \)-components. \( A_x = 50 \) km for \( \overrightarrow{\mathbf{A}} \) and \( B_x \approx 41.0 \) km for \( \overrightarrow{\mathbf{B}} \). Therefore, \( \overrightarrow{\mathbf{A}} \) has a greater \( x \)-component.
04

Calculate y-components of Vectors

For \( \overrightarrow{\mathbf{A}} \), the \( y \)-component is 0 km, as earlier mentioned. For \( \overrightarrow{\mathbf{B}} \), use the formula: \( B_y = B \sin(\theta) \). Thus, \( B_y = -120 \sin(70^{\circ}) \approx -120 \cdot 0.9397 \approx -112.8 \) km. The negative sign indicates the direction is below the \( x \)-axis.
05

Determine Greater y-component

Compare the \( y \)-components. \( A_y = 0 \) km for \( \overrightarrow{\mathbf{A}} \) and \( B_y \approx -112.8 \) km for \( \overrightarrow{\mathbf{B}} \). Since 0 km is greater than \(-112.8 \) km, \( \overrightarrow{\mathbf{A}} \) again has the greater \( y \)-component.

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Key Concepts

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

Magnitude and Direction
Vectors are mathematical entities that have both a magnitude and a direction. The magnitude of a vector defines its size or length, while its direction indicates where it points. This is similar to how you might describe a journey: you need to know both how far you are going (magnitude) and in which direction (e.g., north, east). This exercise involves two vectors: \( \overrightarrow{\mathbf{A}} \) and \( \overrightarrow{\mathbf{B}} \).
\( \overrightarrow{\mathbf{A}} \) has a magnitude of \( 50 \text{ km} \) and points in the positive \( x \)-direction, meaning it lies directly along the \( x \)-axis. Imagine standing on a coordinate plane at the origin, and moving 50 km directly to the right; that's the direction specified for \( \overrightarrow{\mathbf{A}} \).
Meanwhile, \( \overrightarrow{\mathbf{B}} \) has a larger magnitude of \( 120 \text{ km} \) but points at an angle of \( 70^{\circ} \) below \( x \)-axis. This means if you start at the origin and move 120 km, but at an angle slanting downward (below \( x \)-axis), you end up in the position denoted by \( \overrightarrow{\mathbf{B}} \). These angles are crucial for breaking the vector into its components later, often using trigonometric functions, which gives us the ability to analyze each dimension separately.
Component Analysis
Component analysis breaks down a vector into its respective parts along the \( x \)- and \( y \)-axes. For any given vector \( \overrightarrow{\mathbf{C}} \), this involves calculating and understanding \( C_x \) and \( C_y \), which represent how much of the vector's magnitude is in the horizontal and vertical directions, respectively.
\( \overrightarrow{\mathbf{A}} \) is straightforward because it is entirely in the positive \( x \)-axis: its \( x \)-component is 50 km, and its \( y \)-component is 0 km as it does not rise or drop on the \( y \)-axis.
For \( \overrightarrow{\mathbf{B}} \), component analysis requires trigonometry. Using formulas derive \( B_x = B \cos(\theta) = 120 \cos(70^\circ) \approx 41.0 \text{ km} \).
Similarly, \( B_y = B \sin(\theta) = -120 \sin(70^\circ) \approx -112.8 \text{ km} \). The negative sign in \( B_y \) highlights that \( \overrightarrow{\mathbf{B}} \) points downward (or below the \( x \)-axis). Understanding these components helps us evaluate their relative sizes and directions.
Vector Addition and Subtraction
Vector addition and subtraction are fundamental operations. They are important for solving problems where you need to find resultant vectors—vectors that result from multiple combined movements or forces.
**Addition of Vectors**
To add vectors, you align them in a sequence, so the tip of one vector meets the tail of the next. The resultant vector is then drawn from the start of the first vector to the end of the last vector. Mathematically, adding vectors involves summing their components: if you have \( \overrightarrow{\mathbf{C}} = \overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}} \), then \( C_x = A_x + B_x \) and \( C_y = A_y + B_y \).
**Subtraction of Vectors**
Subtraction works similarly but involves negating the direction of the second vector before adding. This means flipping it over its tail-tip end. It's like reversing part of your journey then seeing where you end up overall.
In our exercise, understanding how to decompose vectors into components is key. Once broken down, applying addition or subtraction becomes a straightforward arithmetic process.

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Most popular questions from this chapter

Graph Vector \(\overrightarrow{\mathbf{A}}\) has a magnitude of \(27 \mathrm{~m}\) and points in a direction \(32^{\circ}\) above the positive \(x\) axis. Vector \(\overrightarrow{\mathbf{B}}\) has a magnitude of \(35 \mathrm{~m}\) and points in a direction \(55^{\circ}\) below the positive \(x\) axis. Sketch the vectors \(\overrightarrow{\mathbf{A}}, \overrightarrow{\mathbf{B}}\), and \(\overrightarrow{\mathbf{C}}=\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\). Estimate the magnitude and the direction angle of \(\overrightarrow{\mathbf{C}}\) from your sketch.

Suppose that \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\) have nonzero magnitudes. Is it possible for \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) to be zero?

Playing shortstop, you pick up a ground ball and throw it to second base. The ball is thrown horizontally, with a speed of \(22 \mathrm{~m} / \mathrm{s}\), directly toward point \(A\) as shown in Figure 4.37. When the ball reaches the second baseman \(0.45 \mathrm{~s}\) later, it is caught at point B. (a) How far were you from the second baseman? (b) What is the distance of vertical drop, from \(A\) to \(B\) ?

The initial velocity of a projectile has a horizontal component equal to \(5 \mathrm{~m} / \mathrm{s}\) and a vertical component equal to \(6 \mathrm{~m} / \mathrm{s}\). At the highest point of the projectile's flight, what is (a) the horizontal component of its velocity and (b) the vertical component of its velocity? Explain.

(a) If the angle of the chair lift is decreased, will the horizontal distance \(d_{x}\) increase, decrease, or stay the same? Assume that the length of the lift remains the same, \(d=190 \mathrm{~m}\). (b) Find \(d_{x}\) for the angle \(\theta=15^{\circ}\).

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