Question

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

Part (a): x = 61.43 m, y = 43.05 m; Part (b): x = 31.65 m, y = 67.95 m.

Step-by-step solution

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 \(\theta = 65.0^{\circ}\).

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:- \(x = 75.0 \cdot \cos(35.0^{\circ})\)- \(y = 75.0 \cdot \sin(35.0^{\circ})\)

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:- \(x = 75.0 \cdot 0.819 \approx 61.43\, \text{m}\)- \(y = 75.0 \cdot 0.574 \approx 43.05\, \text{m}\)

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:- \(x = 75.0 \cdot \cos(65.0^{\circ})\)- \(y = 75.0 \cdot \sin(65.0^{\circ})\)

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:- \(x = 75.0 \cdot 0.422 \approx 31.65\, \text{m}\)- \(y = 75.0 \cdot 0.906 \approx 67.95\, \text{m}\)

Key concepts

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

• \( r = 75.0 \, \text{m} \)

This value represents how long the vector is, irrespective of its direction. Understanding magnitude is crucial because it is foundational to determining other vector qualities like direction and components.

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°.

• 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.

Thus, the angle is not just a descriptive measure; it plays a pivotal role in vector operations by dictating how vectors are calculated and combined.

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:

• 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: \(y = r \cdot \sin(\theta)\).

By plugging in the given angle into these trigonometric functions, one can determine the precise horizontal and vertical measures of a vector. It's worth noting that understanding these functions allows you to engage with all kinds of vector problems involving direction and magnitude.

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:

• 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.

The characteristics of a position vector are fundamental to understanding movement, directionality, and spatial relations in vector-based physics and mathematics problems.