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Chapter 4: Problem 2

An aircraft flies \(100 \mathrm{~km}\) in a NE direction, then \(120 \mathrm{~km}\) in an ESE direction and finally \(\mathrm{S}\) for a further \(50 \mathrm{~km} .\) Sketch the vectors representing this flight path. What is the distance from start to finish and also the length of the flight path?

Short Answer

Expert verified
Distance from start to finish: 184 km. Length of the flight path: 270 km.

Step by step solution

01

Understand the Directions

Recognize that NE (Northeast) is a 45-degree angle between North and East. ESE (East-Southeast) is a direction that is 112.5 degrees clockwise with respect to North (or 22.5 degrees southward from East). The direction S is directly along the negative y-axis or 180 degrees from North.
02

Decompose the Vectors

Using trigonometry, decompose each vector to their components. For the first vector: NE, 45掳:- North (y component): \ 100 \cos(45\degree) - East (x component): \ 100 \sin(45\degree).For the second vector: ESE, 112.5掳:- East (x component main direction): \ 120 \cos(22.5掳) - South (y component): \ 120 \sin(22.5掳).For the third vector: South:- South (y component): \(-50\) (no x component).
03

Calculate Vector Components

Calculate the components for each direction.1. NE: - x: \ 100 \cos(45掳) = 70.71 - y: \ 100 \sin(45掳) = 70.712. ESE: - x: \ 120 \cos(22.5掳) = 111.54 - y: \( -120 \sin(22.5掳) = -45.88\)3. S: - y: \ -50 (x component is 0) Combine the x components: 70.71 + 111.54 = 182.25Combine the y components: 70.71 - 45.88 - 50 = -25.17.
04

Determine the Distance from Start to Finish

To find the resultant vector's magnitude from start to finish, use the Pythagorean theorem: \(d = \sqrt{(182.25)^2 + (-25.17)^2} \approx 184.00\).
05

Find the Length of the Flight Path

Add the magnitudes of the individual flight path segments. Total length = 100 + 120 + 50 = 270 km.

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Key Concepts

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

Trigonometric Vector Decomposition
Trigonometric vector decomposition is a method used to break down vectors into their horizontal and vertical components. This technique is particularly useful in physics and engineering for resolving vectors into directions that align with coordinate axes.
This can simplify calculations related to vector addition, subtraction, and analysis. In our case of the aircraft flight path, each leg of the journey is represented as a vector, and we use trigonometry to decompose these vectors based on their direction and magnitude.
  • The Northeast (NE) direction forms a 45-degree angle with the horizontal (East) and vertical (North) directions. To find the components, use the sine and cosine of 45掳: both components will equal to 100 km multiplied by \( \cos(45^\circ) \) and \( \sin(45^\circ) \).
  • The East-Southeast (ESE) direction is 22.5 degrees southward from East. The decomposition here requires using \( \cos(22.5^\circ) \) and \( \sin(22.5^\circ) \) for directions East and South, respectively.
  • Finally, moving South is straightforward: the entire vector component in the y-direction is negative, while the x-direction remains zero because it is purely vertical movement.
Pythagorean Theorem
The Pythagorean theorem is an essential mathematical tool that helps in calculating the magnitude of a resultant vector when you know its perpendicular components.
According to this theorem, in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. We use this principle to find the straight-line distance of the flight from start to finish.
  • First, calculate the total x-component by adding all the x-direction contributions: 70.71 km from the NE direction plus 111.54 km from the ESE direction.
  • Similarly, calculate the total y-component: 70.71 km from NE minus 45.88 km from ESE and 50 km from South.
  • Apply the theorem: \[ d = \sqrt{(182.25)^2 + (-25.17)^2} \approx 184.00 \text{ km} \] to find the direct distance between starting and ending positions.
Resultant Vector
The concept of a resultant vector is vital when analyzing systems where multiple vectors are at play. A resultant vector is essentially a single vector that has the same effect as the combination of the original vectors.
When dealing with directions, like the airplane's flight path, it's critical to determine the resultant vector to understand the overall effect.
  • We found the x and y components individually for each segment. By combining these components, we consolidated all the movements into a single, understandable resultant vector.
  • With a known resultant, we understand not only the net direction of the aircraft but also compare the flight path's practical length to the direct distance.
  • This gives insight into efficiency, allowing pilots and engineers to plan the most effective route.

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Most popular questions from this chapter

Given that \(\overrightarrow{\mathrm{OP}}=(3,1,2)\) and \(\mathrm{OQ}=(1,-2,-4)\) are the position vectors of the points \(\mathrm{P}\) and \(\mathrm{Q}\) respectively, find (a) the equation of the plane passing through \(\mathrm{Q}\) and perpendicular to \(\mathrm{PQ}\); (b) the perpendicular distance from the point \((-1,1,1)\) to the plane.

Given the triangle \(\mathrm{OAB}\), where \(\mathrm{O}\) is the origin, and denoting the midpoints of the opposite sides as \(\mathrm{O}^{\prime}, \mathrm{A}^{\prime}\) and \(\mathrm{B}^{\prime}\), show vectorially that the lines \(\mathrm{OO}^{\prime}, \mathrm{AA}^{\prime}\) and \(\mathrm{BB}^{\prime}\) meet at a point. (Note that this is the result that the medians of a triangle meet at the centroid.)

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The vectors \(a=(1,-1,2), b=(0,1,3)\), \(c=(-2,2,-4)\) are given. (a) Evaluate \(a \times b\) and \(b \times c\) (b) Write down the vectors \(b \times a\) and \(c \times b\) (c) Show that \(c \times a=0\) and explain this result.

If the complex numbers \(z_{1}, z_{2}\) and \(z_{3}\) are represented on the Argand diagram by the points \(\mathrm{P}_{1}, \mathrm{P}_{2}\) and \(\mathrm{P}_{3}\), respectively and $$ \overrightarrow{\mathrm{OP}}_{2}=2 \mathrm{j} \overrightarrow{\mathrm{OP}}_{1} \quad \text { and } \quad \overrightarrow{\mathrm{OP}}_{3}=\frac{2}{5} \mathrm{j} \overrightarrow{\mathrm{P}_{2} \mathrm{P}_{1}} $$ prove that \(\mathrm{P}_{3}\) is the foot of the perpendicular from \(\mathrm{O}\) onto the line \(\mathrm{P}_{1} \mathrm{P}_{2}\)
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