Problem 8 - Fundamentals of Physics

Question

A person walks in the following pattern: \(3.1 \mathrm{~km}\) north, then \(2.4 \mathrm{~km}\) west, and finally \(5.2 \mathrm{~km}\) south. (a) Sketch the vector diagram that represents this motion. (b) How far and (c) in what direction would a bird fly in a straight line from the same starting point to the -8 A person walks in the following pattern: \(3.1 \mathrm{~km}\) north, then \(2.4 \mathrm{~km}\) west, and finally \(5.2 \mathrm{~km}\) south. (a) Sketch the vector diagram that represents this motion. (b) How far and (c) in what direction would a bird fly in a straight line from the same starting point to the same final point?

Step-by-Step Solution

Answer

The bird flies 3.19 km at 41.2° south of west from the starting point.

Step 1: Break down the path into vectors

The motion can be represented in three segments as vectors. First, the person walks 3.1 km north, represented by the vector \( (0, 3.1) \). Then, the person walks 2.4 km west, which can be represented by the vector \( (-2.4, 0) \). Finally, the person walks 5.2 km south, represented by the vector \( (0, -5.2) \).

Step 2: Calculate the resultant displacement vector

Add up the vectors from each part of the path to find the resultant vector. It will be calculated as follows:\[ (0, 3.1) + (-2.4, 0) + (0, -5.2) = (-2.4, 3.1 - 5.2) = (-2.4, -2.1) \]. This means the resultant vector from the starting point is \( -2.4 \) km east and \( 2.1 \) km south.

Step 3: Determine the straight-line distance (magnitude of the vector)

Use the Pythagorean theorem to find the magnitude of the resultant vector, which represents the straight-line distance the bird would fly:\[ \text{Distance} = \sqrt{(-2.4)^2 + (-2.1)^2} = \sqrt{5.76 + 4.41} = \sqrt{10.17} \approx 3.19 \text{ km} \].

Step 4: Calculate the direction of the resultant vector

To find the direction, calculate the angle of the vector relative to the west direction using the arctangent function (tangent angle = opposite/adjacent):\[ \theta = \tan^{-1}\left(\frac{-2.1}{-2.4}\right) \approx \tan^{-1}(0.875) \approx 41.2^\circ \]This means the vector is 41.2° south of the western direction.

Step 5: Sketch the vector diagram

Draw a diagram starting at the origin. Sketch a 3.1 km line segment going north, then a 2.4 km segment going left (west), and finally a 5.2 km segment going down (south). The result is a triangle with the direct vector drawn from the start to the end point.

Key Concepts

Magnitude Calculation

When analyzing the concept of vector displacement, one crucial aspect is calculating the magnitude of the vector. Magnitude is essentially the length of the vector, which in practical terms represents the straight-line distance between the starting and ending points of the motion.

The path described in our exercise forms a triangle when viewed in a diagram, where the magnitude is the direct distance a bird would fly across this triangle. To find this magnitude, we use the Pythagorean theorem, which relates the lengths of the sides in a right-angled triangle. For a vector with components

• Horizontal distance: egin{align}(x) = -2.4 ext{ km} ext{ and Vertical distance: (y) = -2.1 km} ight] \(end{align}

, the theorem is applied as:

\[ ext{Distance} = \sqrt{(-2.4)^2 + (-2.1)^2} \approx 3.19 ext{ km} \] Here's a simple breakdown:

• Square each component: Calculate \)(-2.4)^2\( \approx \)5.76\( \ and \)(-2.1)^2\( \approx \)4.41\( \.
• Add them together: The sum is \)10.17$\.
• Take the square root: Obtain the magnitude of \[ \sqrt{10.17} \approx 3.19 \text{ km} \] \.

Understanding this process helps demystify how distances are calculated in vector terms, ensuring clarity in applying it to more complex problems.

Direction Determination

In physics and real-world applications, knowing not just how far, but also in which direction something moves, is vital. When discussing vector displacement, direction provides a full sense of the vector's impact.

In the given exercise, after finding the resultant vector, it remains important to determine its direction relative to a reference point. Utilizing trigonometric functions is key in this, specifically the arctangent, which relates angles to ratios of side lengths in a right triangle. Here’s how it works:

We use the tangent function because it directly relates to the relationship between the opposite and adjacent sides of our right-angled vector triangle. The formula for direction is given by:

\[ \theta = \tan^{-1}\left(\frac{-2.1}{-2.4}\right) \approx 41.2^\circ \]Breaking down the steps to discover the direction:

• Calculate the ratio: First, determine the ratio of the opposite side (-2.1 km, south) to the adjacent side (-2.4 km, east).
• Use the arctangent function: Apply \(\tan^{-1}\) \ to find the angle, which gives us approximately \(41.2^\circ\).
• Interpret the angle: This angle indicates a direction that is 41.2° south of west.

Using these calculations, one can not only find how far a bird would fly but also identify its precise path direction in a navigational context.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry, essential for problems involving vector displacement. It applies to right-angled triangles, providing a formula to calculate the length of the hypotenuse, the side opposite the right angle.

This theorem is expressed as: \[ c^2 = a^2 + b^2 \]
Where:

• a and b are the lengths of the triangle's two shorter sides.
• c is the hypotenuse's length.

In the context of our exercise, the Pythagorean Theorem helps us determine the shortest straight-line distance between two points, given the vector components:

• Determine \(a\) and \(b\): For our vector with components, we have \(-2.4\) km and \(-2.1\) km.
• Calculate \(c\): The straight-line distance, \(c\), becomes \[ \sqrt{(-2.4)^2 + (-2.1)^2} \approx 3.19 \text{ km} \]

By constructing the vectors as a right-angled triangle, this theorem simplifies the process of finding the resultant displacement vector's magnitude, a skill very useful in physics and engineering. Understanding the Pythagorean Theorem in the vector context equips learners with a practical tool in various academic and real-life scenarios.