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Chapter 4: Problem 74

After flying for \(18 \mathrm{~min}\) 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 calculated using its northerly and exactly balancing eastward speeds.

Step by step solution

01

Convert Time to Hours

The time is given in minutes, but we need to convert it into hours to use it in the velocity calculations. Divide the time by 60.\[18 \text{ minutes} = \frac{18}{60} \text{ hours} = 0.3 \text{ hours}\]
02

Decompose the Wind Velocity

The wind is blowing at 42 km/h at an angle of 20掳 south of east. Decompose this into east and south components using trigonometric functions.\[v_{wx} = 42 \cos(20^{\circ}) \text{ km/h (east)} \] \[v_{wy} = 42 \sin(20^{\circ}) \text{ km/h (south)} \]
03

Analyze the Resultant Displacement

The pilot ends up 55 km north of the starting point. Because the displacement due east is zero, calculate the velocity of the airplane needed to cancel out the southward component of the wind and produce a net northward displacement.The southward component needs to be the same magnitude but opposite direction as the displacement caused by the plane's northward component. Since this northward component must also account for the 55 km, we know:\[v_{py} \cdot 0.3 - v_{wy} \cdot 0.3 = 55 \] \[v_{py} = v_{wy} + \frac{55}{0.3} \]
04

Compute the Airplane鈥檚 Northerly Component

Substitute the known values from the previous steps.\[v_{py} = 42 \sin(20^{\circ}) + \frac{55}{0.3} \] Calculate this numerical value.
05

Find the Airplane's Speed Relative to the Air

Since the airplane must have both eastward and northward components, its speed relative to air combines the necessary eastward and northward components to oppose the wind and achieve resultant displacement. Use Pythagorean theorem to find airplane speed. \[v_{pa} = \sqrt{v_{px}^2 + v_{py}^2}\] Where \(v_{px} = v_{wx}\), the airplane must exactly balance the east component of wind.
06

Calculate Final Result

Substitute the values into the Pythagorean theorem to find the airplane's speed relative to the air. The eastward component is simply equal to the wind speed eastward component, as the airplane needs to balance the eastward drift. Calculate the final value using \[v_{px} = 42 \cos(20^{\circ}) \] and using the computed \(v_{py}\) to find \(v_{pa}\)

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Key Concepts

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

Vector Decomposition
When faced with problems involving directions, like this airplane scenario, vector decomposition helps simplify what seems complex. The wind in this problem blows at a speed of 42 km/h at an angle south of east. We break this vector into its horizontal (east-west) and vertical (north-south) components using trigonometric functions.

  • The east component is calculated using cosine: \(v_{wx} = 42 \cos(20^{\circ})\) km/h.
  • The south component uses sine: \(v_{wy} = 42 \sin(20^{\circ})\) km/h.
By breaking the wind vector into these parts, we can understand and counteract its effects on the airplane's path more precisely. These calculations are essential for ensuring the plane reaches its intended destination without being blown off course.
Trigonometric Functions
Trigonometric functions like sine and cosine are powerful tools when dealing with vectors and angles. In this exercise, we use them to split the wind's velocity into eastward and southward components.

  • Cosine is used here to find the component of the vector that aligns with the east direction. It is calculated by \(\cos(\theta)\), where \(\theta\) is the angle from the east.
  • Sine helps us find the perpendicular component, flowing southward in this case, calculated by \(\sin(\theta)\).
These functions essentially allow us to take one slanted force and unravel it into clear, understandable parts. This simplification is crucial in calculations, allowing us to manage each direction individually and accurately.
Pythagorean Theorem
The Pythagorean theorem is a mathematical principle that helps us calculate the magnitude of combined force vectors. Here, it's essential in determining the airplane's overall speed relative to the air. After computing the necessary northward and eastward velocities to balance the wind, we need to find the actual resultant speed of the airplane with respect to still air.

The theorem states:
  • Given two perpendicular components \(v_{px}\) and \(v_{py}\), the resultant speed \(v_{pa}\) can be found using \(v_{pa} = \sqrt{v_{px}^2 + v_{py}^2}\).
This calculation gives a comprehensive speed, not just in one direction but combining both to reflect the actual effort of the airplane against the wind.
Displacement Analysis
Displacement analysis involves understanding how far and in what direction an object moves. In this scenario, the key displacement is 55 km directly north, happened over the time frame of the flight. The airplane's relative velocity needs to be carefully analyzed to ensure this northward destination is met.

  • The south component caused by the wind needs to be canceled out by the airplane's northward travel.
  • Moreover, the computed northward component, \(v_{py}\), lets us compute what additional speed beyond the wind's interference is needed to reach this displacement effectively.
With these calculations, we assess and determine the precise speed and direction adjustments necessary for the aircraft to cover its intended path without deviation. This analysis thus ensures reliable and accurate navigation.

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Most popular questions from this chapter

A stone is catapulted at time \(t=0\), with an initial velocity of magnitude \(18.0 \mathrm{~m} / \mathrm{s}\) and at an angle of \(40.0^{\circ}\) above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at \(t=1.10\) s? Repeat for the (c) horizontal and (d) vertical components at \(t=1.80 \mathrm{~s}\), and for the (e) horizontal and (f) vertical components at \(t=5.00 \mathrm{~s}\).

A watermelon seed has the following coordinates: \(x=-5.0 \mathrm{~m}\), \(y=9.0 \mathrm{~m}\), and \(z=0 \mathrm{~m}\). Find its position vector (a) in unit- vector notation and as (b) a magnitude and (c) an angle relative to the positive direction of the \(x\) axis. (d) Sketch the vector on a right-handed coordinate system. If the seed is moved to the \(x y z\) coordinates \((3.00 \mathrm{~m}\), \(0 \mathrm{~m}, 0 \mathrm{~m}\) ), what is its displacement (e) in unit-vector notation and as (f) a magnitude and \((\mathrm{g})\) an angle relative to the positive \(x\) direction?

From the origin, a particle starts at \(t=0 \mathrm{~s}\) with a velocity \(\vec{v}=7.0 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}\) and moves in the \(x y\) plane with a constant acceleration of \(\vec{a}=(-9.0 \hat{\mathrm{i}}+3.0 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}^{2} .\) At the time the particle reaches the maximum \(x\) coordinate, what is its (a) velocity and (b) position vector?

A defense air force plane, diving with constant speed at an angle of \(52.0^{\circ}\) with the vertical, drops a shell at an altitude of \(720 \mathrm{~m}\). The shell reaches the ground \(6.00 \mathrm{~s}\) after its release. (a) What is the speed of the plane? (b) How far does the shell travel horizontally during its flight? What are the (c) horizontal and (d) vertical components of its velocity just before reaching the ground? Assume an \(x\) axis in the direction of the horizontal motion and an upward \(y\) axis.

At \(t_{1}=2.00 \mathrm{~s}\), the acceleration of a particle in counterclockwise circular motion is \(\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(4.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). It moves at constant speed. At time \(t_{2}=5.00 \mathrm{~s}\), the particle's acceleration is \((4.00\) \(\left.\mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(-6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). What is the radius of the path taken by the particle if \(t_{2}-t_{1}\) is less than one period?
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