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After flying for \(15 \mathrm{~min}\) in a wind blowing \(42 \mathrm{~km} / \mathrm{h}\) at an angle of \(20^{\circ}\) south of east, an airplane pilot is over a town that is \(55 \mathrm{~km}\) due north of the starting point. What is the speed of the airplane relative to the air?

Short Answer

Expert verified
The airplane's speed relative to the air is approximately 205.8 km/h.

Step by step solution

01

Determine the Wind's Effect on the Airplane

The given wind speed is 42 km/h blowing south of east at an angle of 20°. Over 15 minutes or 0.25 hours, this means the wind moves:\[42 \text{ km/h} \times 0.25 \text{ h} = 10.5 \text{ km}\] This wind displacement is directed at 20° south of east.
02

Break Down the Wind Displacement into Components

The components of the wind's displacement can be calculated using trigonometry. Using cosine for the eastward component and sine for the southward component:\[d_{east} = 10.5 \cos(20°)\approx 9.86 \text{ km}\]\[d_{south} = 10.5 \sin(20°)\approx 3.59 \text{ km}\]
03

Calculate Airplane's Position Relative to the Ground

The airplane ends up 55 km due north of the start. Therefore, the total northward journey must have countered the southward displacement due to wind. Disregarding other displacements, its net northward position is:\[55 - d_{south} = 55 - 3.59 \approx 51.41 \text{ km}\]
04

Calculate Airplane's Velocity Component

Knowing the airplane covers 51.41 km north over 15 minutes or 0.25 hours, we find:\[51.41 \div 0.25 = 205.64 \text{ km/h}\] This northward speed is at an angle of 90° from that of the wind's east component.
05

Combine Components to Find the Airplane's Speed Relative to Air

The plane's speed relative to the air can be calculated using the Pythagorean theorem with the northward component calculated and the plane's velocity aiming eastward:\[v_{relative} = \sqrt{205.64^2 + 9.86^2}\approx\sqrt{42287.71+97.24}\approx\sqrt{42384.95}\approx 205.8 \text{ km/h}\]

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Key Concepts

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

Vector Components
To solve problems involving motion, we often need to break a vector into its components. A vector has both magnitude (how long it is) and direction (where it points). In the case of the airplane, the wind acting as a vector has a speed (42 km/h) and direction (20° south of east). These vector components tell us exactly how much the vector affects movement in specific directions, like eastward or southward, using trigonometric functions.

We typically resolve the vector into two perpendicular components, one along the x-axis (eastward) and the other along the y-axis (northward), which involve:
  • **Cosine** of the angle for the horizontal (eastward) component, since cosine relates an angle to the adjacent side of a right triangle.
  • **Sine** of the angle for the vertical (southward or northward) component, since sine is linked to the opposite side of the angle.
By breaking vectors into components, we simplify complex movement into more manageable parts, allowing easier calculations.
Trigonometry
Trigonometry is a branch of mathematics that studies the relationships between side lengths and angles of triangles, often employed to break vectors into components. In this exercise, we use trigonometric functions to resolve the wind's effect on the airplane. When we say the wind blows at 20° south of east, we can form a right triangle, where:
  • The hypothenuse represents the total wind displacement.
  • The eastward and southward movements are the triangle's sides.
We use:
  • **Cosine function** to determine the eastward component: \( d_{east} = 10.5 \cdot \cos(20°) \).
  • **Sine function** for the southward component: \( d_{south} = 10.5 \cdot \sin(20°) \).
This approach applies whenever an angle and hypotenuse are known, and we're identifying known sides, demonstrating the beauty of trigonometric relations.
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle in geometry, critical in calculating unknown sides or verifying distance in right triangles. It states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) equals the sum of the squares of the other two sides.

Expressed mathematically, it reads as: \[ c^2 = a^2 + b^2 \]Where \( c \) is the hypotenuse, and \( a \) and \( b \) are the other two sides.
In our airplane problem, after calculating the northward and eastward velocity components (using 205.64 km/h and 9.86 km/h, respectively), we employ the Pythagorean theorem to find the resultant speed. This combination of speeds gives us the airplane's true velocity relative to the air:\[ v_{relative} = \sqrt{205.64^2 + 9.86^2} \approx 205.8 \text{ km/h} \]This method serves to accurately determine resultant velocities when vector components are perpendicular to one another.

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Most popular questions from this chapter

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