Chapter 4: Problem 5
Find the \(x\) and \(y\) components of a position vector that has a magnitude of \(r=75.0 \mathrm{~m}\) and an angle relative to the \(x\) axis of (a) \(35.0^{\circ}\) and (b) \(65.0^{\circ}\).
Short Answer
Expert verified
Part (a): x = 61.43 m, y = 43.05 m; Part (b): x = 31.65 m, y = 67.95 m.
Step by step solution
01
Understand the Problem
We need to find the components of a position vector. The position vector has a magnitude of \(r = 75.0\, \text{m}\) and makes an angle \(\theta\) with the \(x\)-axis. We have two scenarios to evaluate: one where \(\theta = 35.0^{\circ}\) and another where \(\theta = 65.0^{\circ}\).
02
Break Down the Solution for Part (a)
For the first scenario where \(\theta = 35.0^{\circ}\):1. The \(x\)-component of the vector can be determined using \(x = r \cdot \cos(\theta)\).2. The \(y\)-component can be found using \(y = r \cdot \sin(\theta)\).Using \(r = 75.0\, \text{m}\) and \(\theta = 35.0^{\circ}\), the equations become:- \(x = 75.0 \cdot \cos(35.0^{\circ})\)- \(y = 75.0 \cdot \sin(35.0^{\circ})\)
03
Calculate Components for Part (a)
Using a calculator:- \(\cos(35.0^{\circ}) \approx 0.819\)- \(\sin(35.0^{\circ}) \approx 0.574\)Plug these values into the equations:- \(x = 75.0 \cdot 0.819 \approx 61.43\, \text{m}\)- \(y = 75.0 \cdot 0.574 \approx 43.05\, \text{m}\)
04
Break Down the Solution for Part (b)
For the second scenario where \(\theta = 65.0^{\circ}\):1. The \(x\)-component is given by \(x = r \cdot \cos(\theta)\).2. The \(y\)-component is \(y = r \cdot \sin(\theta)\).Using \(r = 75.0\, \text{m}\) and \(\theta = 65.0^{\circ}\), the calculations are:- \(x = 75.0 \cdot \cos(65.0^{\circ})\)- \(y = 75.0 \cdot \sin(65.0^{\circ})\)
05
Calculate Components for Part (b)
Using a calculator:- \(\cos(65.0^{\circ}) \approx 0.422\)- \(\sin(65.0^{\circ}) \approx 0.906\)Substitute these into the equations:- \(x = 75.0 \cdot 0.422 \approx 31.65\, \text{m}\)- \(y = 75.0 \cdot 0.906 \approx 67.95\, \text{m}\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Magnitude
The magnitude of a vector refers to its "size" or "length" in the context of geometry and physics. It is a non-negative value typically calculated using the Pythagorean theorem.
For example, if a vector has components along the x and y axes, its magnitude can be found by: \( r = \sqrt{x^2 + y^2} \). In our particular problem, we are given the magnitude directly as
For example, if a vector has components along the x and y axes, its magnitude can be found by: \( r = \sqrt{x^2 + y^2} \). In our particular problem, we are given the magnitude directly as
- \( r = 75.0 \, \text{m} \)
Angle of Vector
The angle of a vector, often denoted as \(\theta\), informs you about the direction of the vector relative to a specific reference axis. In two-dimensional problems, this is usually measured from the positive x-axis.
For our exercise, two angles are explored: 35° and 65°.
For our exercise, two angles are explored: 35° and 65°.
- These angles determine how we project the vector onto the x and y axes, which is necessary for calculating the vector's components.
- Expressing the vector angle in degrees is common, and converting it to radians can sometimes be helpful for calculations involving trigonometric functions.
- Knowing the angle helps in using trigonometric functions effectively to decompose the vector into useful parts.
Trigonometric Functions
Trigonometric functions like sine (\(\sin\)) and cosine (\(\cos\)) are essential tools for splitting a vector into its components.
The functions relate the angles of a triangle to its side lengths, positioning sinusoidal relationships in a circle. In the context of vectors, they are used as follows:
The functions relate the angles of a triangle to its side lengths, positioning sinusoidal relationships in a circle. In the context of vectors, they are used as follows:
- To find the x-component of a vector, you use cosine: \(x = r \cdot \cos(\theta)\).
- To determine the y-component of a vector, you use sine: \(y = r \cdot \sin(\theta)\).
Position Vector
A position vector indicates an object's position in space concerning an origin or reference point. It is a specific kind of vector often described by its components aligned with coordinate axes.
For the exercise, the position vector has the starting point at the origin (0,0) and is described by the given magnitude and angle. These two attributes help in deriving the x and y components:
For the exercise, the position vector has the starting point at the origin (0,0) and is described by the given magnitude and angle. These two attributes help in deriving the x and y components:
- Position vectors can show how far and in what direction an object is from the origin.
- In two-dimensional problems, they are calculated with the standard formula for components, assisting in visualizing the vector's path or direction on a plane.