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The eye of a hurricane passes over Grand Bahama Island in a direction \(60.0^{\circ}\) north of west with a speed of \(41.0 \mathrm{~km} / \mathrm{h}\). Three hours later the course of the hurricane suddenly shifts due north, and its speed slows to \(25.0 \mathrm{~km} / \mathrm{h}\). How far from Grand Bahama is the hurricane \(4.50 \mathrm{~h}\) after it passes over the island?

Short Answer

Expert verified
The hurricane is around 125.8 km far from Grand Bahama island 4.50 hours after it passes over the island.

Step by step solution

01

Calculate the distance covered in the northwestern path

Calculate the distance covered when the hurricane is moving at a speed of \(41.0 \mathrm{~km}/\mathrm{h}\) for 3 hours, using the relation distance = speed × time: \(d1 = 41.0 \mathrm{km/h} × 3 \mathrm{h} = 123.0 \mathrm{km}\)
02

Break down the northwestern movement into northward and westward components.

This distance needs to be split into northward and westward components. These components can be obtained using trigonometric principles: The westward component \(w1 = d1 \times \cos(60.0^{\circ})\), and the northward component \(n1 = d1 \times \sin(60.0^{\circ})\)
03

Calculate the distance covered in the northward path

Calculate the distance covered during 1.5 hours when the hurricane is moving north at \(25.0 \mathrm{km/h}\): \(d2 = 25.0 \mathrm{km/h} × 1.5 \mathrm{h} = 37.5 \mathrm{km}\)
04

Combine northward components and find the total northward distance.

The total northward movement may be found by adding the northward distance covered during northwestern movement and the distance covered in northward shift: \(N = n1 + d2\)
05

Calculate the resultant distance from Grand Bahama

Finally, find the total distance \(D\) from Grand Bahama using the Pythagorean theorem as the total northward distance \(N\) and the westward distance \(w1\) are perpendicular to each other: \(D = \sqrt{N^{2} + w1^{2}}\)

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Key Concepts

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

Trigonometry in Hurricane Path Analysis
Trigonometry is vital in understanding the path of the hurricane as it moves in different directions. When the hurricane initially travels at an angle of \( 60^{\circ} \) north of west, we can break this diagonal path into two simpler straight-line movements. One of these moves north, and the other moves west. This breakdown is useful as each component aligns with the cardinal directions.

To find these directional movements, we use trigonometric functions. The cosine function helps calculate the westward component, while the sine function is used for the northward component. If the distance covered is \( d_1 \), then the westward component is given by \( w_1 = d_1 \times \cos(60^{\circ}) \), and the northward component is \( n_1 = d_1 \times \sin(60^{\circ}) \).

This application of trigonometry is essential in converting an angled path into understandable segments.
Using the Pythagorean Theorem to Find Distance
The Pythagorean Theorem can be a powerful tool to find distances in problems involving right-angled triangles. In our hurricane scenario, after determining the components of its path, we can assess the total distance by considering a right-angled triangle.

The theorem states that 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. If \( N \) is the total northward distance and \( w_1 \) is the westward distance, the resultant path \( D \) can be calculated using:

\[ D = \sqrt{N^2 + w_1^2} \]

This formula helps us find how far the hurricane is from the starting point, using the combined perpendicular movements.
Breaking Down Vector Components
When analyzing the path of an object like a hurricane, vectors become crucial in understanding both magnitude and direction. By breaking down vectors, we convert complex paths into manageable segments.

A vector pointing diagonally can be split into two "component" vectors along the north and west axis, which match up with our map directions. The overall speed (or magnitude) and direction (or angle) of the hurricane create the basis for these component calculations. This process uses trigonometry: the westward component of the vector uses the cosine function, while the northward uses the sine function, creating a complete depiction of movement.

Breaking down into components gives us clarity and precision in predicting the path and position over time.
Applying Hurricane Path Analysis
Hurricane path analysis is important for predicting movements and preparing for potential impacts. By understanding both speed and direction changes, we can estimate where a hurricane will be at any given time. Initially, our hurricane moves northwest at \( 60^{\circ} \), then shifts direction to move purely north.

We analyze these movements in segments. First, calculate how far the hurricane travels northwest by multiplying speed and time. Then, identify the components using trigonometry. When it shifts northwards, calculate this additional movement and combine it with the previously found north component. Finally, the Pythagorean theorem helps find the straight-line distance from the starting point.

This structured analysis helps in making critical decisions during hurricane events.

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Most popular questions from this chapter

A hiker starts at his camp and moves the following distances while exploring his surroundings: \(75.0 \mathrm{~m}\) north, \(2.50 \times 10^{2} \mathrm{~m}\) east, \(125 \mathrm{~m}\) at an angle \(30.0^{\circ}\) north of east, and \(1.50 \times 10^{2} \mathrm{~m}\) south. (a) Find his resultant displacement from camp. (Take east as the positive \(x\)-direction and north as the positive \(y\)-direction.) (b) Would changes in the order in which the hiker makes the given displacements alter his final position? Explain.

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of \(24.0^{\circ}\) below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline with a constant acceleration of \(4.00 \mathrm{~m} / \mathrm{s}^{2}\) for a distance of \(50.0 \mathrm{~m}\) to the edge of the cliff, which is \(30.0 \mathrm{~m}\) above the ocean. Find (a) the car's position relative to the base of the cliff when the car lands in the ocean and (b) the length of time the car is in the air.

A daredevil is shot out of a cannon at \(45.0^{\circ}\) to the horizontal with an initial speed of \(25.0 \mathrm{~m} / \mathrm{s}\). A net is positioned a horizontal distance of \(50.0 \mathrm{~m}\) from the cannon. At what height above the cannon should the net be placed in order to catch the daredevil?

M A place-kicker must kick a football from a point \(36.0 \mathrm{~m}\) (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is \(3.05 \mathrm{~m}\) high. When kicked, the ball leaves the ground with a speed of \(20.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(53.0^{\circ}\) to the horizontal. (a) By how much does the ball clear or fall short of clearing the crossbar? (b) Does the ball approach the crossbar while still rising or while falling?

\(\mathbf{Q} \mid \mathbf{C}\) Three displacements are \(\overrightarrow{\mathbf{A}}=200 \mathrm{~m}\) due south, \(\overrightarrow{\mathbf{B}}=250 \mathrm{~m}\) due west, and \(\overrightarrow{\mathbf{C}}=150 \mathrm{~m}\) at \(30.0^{\circ}\) east of north. (a) Construct a separate diagram for each of the following possible ways of adding these vectors: \(\overrightarrow{\mathbf{R}}_{1}=\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}} ; \overrightarrow{\mathbf{R}}_{2}=\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}+\overrightarrow{\mathbf{A}} ; \overrightarrow{\mathbf{R}}_{3}=\overrightarrow{\mathbf{C}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{A}}\) (b) Explain what you can conclude from comparing the diagrams.

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