Problem 3 - Fundamentals of Physics

Question

The \(x\) component of vector \(\vec{A}\) is \(-25.0 \mathrm{~m}\) and the \(y\) component is \(+40.0 \mathrm{~m}\). (a) What is the magnitude of \(\vec{A} ?\) (b) What is the angle between the direction of \(\vec{A}\) and the positive direction of \(x ?\)

Step-by-Step Solution

Answer

(a) The magnitude is approximately 47.2 m. (b) The angle is approximately 122.0°.

Step 1: Understanding the problem

We are given the components of vector \( \vec{A} \): the \( x \)-component is \(-25.0 \mathrm{~m}\) and the \( y \)-component is \(+40.0 \mathrm{~m}\). We need to find the magnitude of the vector \( \vec{A} \) and the angle it makes with the positive \( x \)-axis.

Step 2: Calculating the magnitude of the vector

The magnitude \( |\vec{A}| \) of a vector \( \vec{A} = (A_x, A_y) \) is calculated using the Pythagorean theorem: \[ \left| \vec{A} \right| = \sqrt{A_x^2 + A_y^2} \] Substitute the given values: \[ \left| \vec{A} \right| = \sqrt{(-25.0)^2 + (40.0)^2} \] \[ \left| \vec{A} \right| = \sqrt{625 + 1600} \] \[ \left| \vec{A} \right| = \sqrt{2225} \] \[ \left| \vec{A} \right| \approx 47.2 \mathrm{~m} \]

Step 3: Finding the angle of the vector

The angle \( \theta \) between the vector \( \vec{A} \) and the positive \( x \)-axis can be found using the tangent function: \[ \theta = \tan^{-1} \left( \frac{A_y}{A_x} \right) \] Substitute the given values: \[ \theta = \tan^{-1} \left( \frac{40.0}{-25.0} \right) \] \[ \theta = \tan^{-1} (-1.6) \] \[ \theta \approx -58.0^\circ \] Since the angle is measured counter-clockwise from the positive \( x \)-axis, the vector actually points in the second quadrant. Therefore, the angle with the positive \( x \)-axis is \[ \theta \approx 180^\circ - 58.0^\circ = 122.0^\circ \]

Step 4: Review the results

We calculated that the magnitude of the vector \( \vec{A} \) is approximately \( 47.2 \mathrm{~m} \), and the angle between the vector's direction and the positive \( x \)-axis is approximately \( 122.0^\circ \).

Key Concepts

Vector Components

In the world of vectors, understanding components is crucial. A vector consists of two essential parts: the magnitude (its length) and its direction. Instead of looking at a vector as a whole, we can break it down into its horizontal and vertical components. For any vector \(\vec{A}\), the components are usually represented as \(A_x\) for the horizontal (or x) component, and \(A_y\) for the vertical (or y) component.

• Human navigation: Consider a bird flying across the sky. If you looked at its flight path in terms of horizontal and vertical movements, those would be its components.
• For calculating purposes: It helps to handle vectors one component at a time, especially in physics and engineering.

Understanding these components helps simplify complex vector operations like addition, subtraction, and finding the resultant vector.

Pythagorean Theorem

The Pythagorean Theorem is a cornerstone in mathematics, especially when dealing with vectors. Named after the ancient Greek mathematician Pythagoras, this theorem helps us calculate the magnitude of a vector when we know its components.The formula is derived from the geometry of a right triangle: \[ c = \sqrt{a^2 + b^2} \]where \(c\) is the hypotenuse (the longest side of the triangle), and \(a\) and \(b\) are the other two sides. For vectors, it translates to:\[ \left| \vec{A} \right| = \sqrt{A_x^2 + A_y^2} \]This implies:

• If you know a vector's horizontal and vertical components, you can find its overall magnitude.
• This formula transforms potentially complicated vector problems into simple arithmetic operations.

For example, if a vector has components \(-25.0\) meters and \(40.0\) meters, applying the theorem gives us a magnitude of approximately \(47.2\) meters.

Trigonometry

Trigonometry deals with the relationships between the angles and sides of triangles. When working with vectors, trigonometric functions like sine, cosine, and tangent are invaluable tools.For vectors, we often use:

• Sine and Cosine: These functions help in unraveling vector components from an initial magnitude and angle.
• Tangent: The function is pivotal for calculating the direction angle of a vector in relation to an axis.

When given a vector and needing to find the direction angle, tangent comes to the rescue:\[ \theta = \tan^{-1} \left( \frac{A_y}{A_x} \right) \] This formula tells us how steep or shallow our vector is compared to the x-axis. It's particularly useful in visualizing vector directions in graphical representations.

Angle Calculation

Finding the angle a vector makes with a reference axis is an essential skill in vector analysis. Using the tangent function, we can determine this angle efficiently.Calculating angles involves:

• Understanding Quadrants: The angle value depends on which quadrant the vector lies. Trigonometry divides the plane into four quadrants, influencing the sign and value of the tangent.
• Using the inverse tangent function: \( \theta = \tan^{-1} \left( \frac{A_y}{A_x} \right) \) helps to find the angle initially.

In our given problem, after computing the inverse tangent, the value might indicate whether the vector points upwards, downwards, to the left, or the right. Additionally, knowing the correct quadrant adjusts the angle, ensuring our result is realistic and spatially accurate, as a vector lying in the second quadrant suggests an angle of \(122^\circ\) with the x-axis.