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Magazine Chapter 1: Problem 79
A ship leaves the island of Guam and sails 285 \(\mathrm{km}\) at \(40.0^{\circ}\) north of west. In which direction must it now head and how far must it sail so that its resultant displacement will be 115 \(\mathrm{km}\) directly east of Guam?
Short Answer
Expert verified
Sail approximately 401 km in a direction approximately 37.5掳 south of east.
Step by step solution
01
Set Up the Coordinate System
Let's consider a coordinate system where east is the positive x-axis direction, and north is the positive y-axis direction. We need to find the components of the 285 km journey at 40掳 north of west.
02
Calculate Initial Displacement Components
The initial displacement can be broken into x and y components. The x-component is \( 285 \cos(220^{\circ}) \) and the y-component is \( 285 \sin(220^{\circ}) \), since 40掳 north of west translates to 220掳 from the positive x-axis.
03
Calculate the Necessary Displacement
For the resultant displacement to be 115 km east, the x-component of the displacement must sum to 115 km and the y-component must sum to 0 km. Let the unknown displacement have components \(x_2\) and \(y_2\).
04
Solve for x-component of Required Displacement
The x-component equation is: \( 285 \cos(220^{\circ}) + x_2 = 115 \). Solve for \(x_2\): \( x_2 = 115 - 285 \cos(220^{\circ}) \).
05
Solve for y-component of Required Displacement
The y-component equation is: \( 285 \sin(220^{\circ}) + y_2 = 0 \). Solve for \(y_2\): \( y_2 = -285 \sin(220^{\circ}) \).
06
Calculate Magnitude and Direction of Second Displacement
The magnitude of the displacement \(d\) is \( \sqrt{x_2^2 + y_2^2} \). The direction \(\theta\) is given by \( \theta = \arctan\left(\frac{y_2}{x_2}\right) \).
07
Substitute Values and Compute
Calculate \( x_2 = 115 - 285(-0.766) \) and \( y_2 = -285(0.643) \). Find \(d = \sqrt{x_2^2 + y_2^2}\) and \(\theta = \arctan\left(\frac{y_2}{x_2}\right)\).
08
Interpret the Direction
Convert the angle \(\theta\) to a bearing if necessary. A negative \(y_2\) indicates a south component, and a positive \(x_2\) indicates an east component.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Coordinate System
When dealing with vector displacement in navigation, mapping out your journey on a coordinate system is crucial. In this context, a coordinate system allows us to break down any vector into simpler parts called components.
For the exercise, we used a system where the east direction is aligned with the positive x-axis, while north corresponds to the positive y-axis. It's like having a map where you can easily point out directions. This setup makes it straightforward to assign positive or negative signs to the direction of travel.
For the exercise, we used a system where the east direction is aligned with the positive x-axis, while north corresponds to the positive y-axis. It's like having a map where you can easily point out directions. This setup makes it straightforward to assign positive or negative signs to the direction of travel.
- East as Positive X-axis: It's important because the problem asks for a displacement directly east.
- North as Positive Y-axis: This helps in understanding any movements upwards from the reference line of east.
- Angles Measured Counterclockwise from Positive X-axis: This is a standard practice in trigonometry to maintain consistency. A 40掳 north of west is measured as 220掳 counterclockwise from the east.
Displacement Components Explored
In navigation, breaking down complex directions into simpler parts using the concept of displacement components is necessary. Displacement components refer to splitting a vector into two parts: an x-component (east-west direction) and a y-component (north-south direction).
Consider a ship sailing 285 km at a bearing 40掳 north of west. Here, the displacement vector can be split into:
Consider a ship sailing 285 km at a bearing 40掳 north of west. Here, the displacement vector can be split into:
- X-component: Calculated as \(285 \cos(220^{\circ})\). This describes how far west (negative x-direction) the ship travels.
- Y-component: Calculated as \(285 \sin(220^{\circ})\). This value represents how far north (positive y-direction) it goes.
Utilizing Trigonometry to Solve Problems
Trigonometry is incredibly useful when dealing with vector displacement and angles. Concepts such as sine, cosine, and tangent help break down vectors and solve problems in a systematic way.
In our problem, trigonometry plays a role in both defining the initial path of the ship and calculating the necessary course corrections for the desired final displacement:
In our problem, trigonometry plays a role in both defining the initial path of the ship and calculating the necessary course corrections for the desired final displacement:
- Sine and Cosine: We've already used \(\sin(\theta)\) and \(\cos(\theta)\) to break down the 285 km path into y and x components. These functions are crucial for understanding how much of the ship's movement is toward each axis.
- Inverse Tangent (Arctan): After solving for the new displacement components, the direction \(\theta\) of the required journey is calculated using \(\theta = \arctan\left(\frac{y_2}{x_2}\right)\). This gives the angle of travel relative to our coordinate system.
- Calculating Magnitude: Trigonometry helps find the actual distance the ship needs to sail using the Pythagorean theorem: \(d = \sqrt{x_2^2 + y_2^2}\).
