A light plane attains an air speed of \(480 \mathrm{~km} / \mathrm{h}\). The pilot sets out for a destination \(810 \mathrm{~km}\) to the north but discovers that the plane must be headed \(21^{\circ}\) east of north to fly there directly. The plane arrives in \(1.9 \mathrm{~h}\). What was the vector wind velocity?

Short Answer

Expert verified
The vector wind velocity is found to be a certain speed in a direction that is east of north.

Step by step solution

01

Determine the velocity vector of the plane

The velocity of the plane is \(480 \mathrm{~km} / \mathrm{h}\). Since the plane is headed \(21^{\circ}\) east of north, we break this vector into its northward (\(N\)) and eastward (\(E\)) components by using sine and cosine. \(N = 480 \mathrm{~km} / \mathrm{h} \times \cos(21^{\circ})\), \(E = 480 \mathrm{~km} / \mathrm{h} \times \sin(21^{\circ})\).
02

Determine the resultant velocity vector of the plane

The plane travels \(810 \mathrm{km}\) due north in \(1.9 \mathrm{h}\), regardless of its own heading. This gives us an effective velocity due north of \(810 \mathrm{km}/1.9 \mathrm{h}\). The eastward component of this velocity is 0, since the total distance traveled east is effectively 0. Thus, the velocity vector of the plane relative to the ground is \(N = 810 \mathrm{km} / 1.9 \mathrm{h}, E = 0\).
03

Determine the velocity vector of the wind

The wind velocity is just the difference between the resultant velocity vector of the plane (relative to the ground) and the inherent velocity vector of the plane. Using the component vectors we derived in the previous steps, we can subtract the plan's velocity from its resultant velocity to find the wind velocity: \(N_{wind} = 810 \mathrm{km} / 1.9 \mathrm{h} - 480 \mathrm{km} / \mathrm{h} \times \cos(21^{\circ})\), \(E_{wind} = 0 - 480 \mathrm{km} / \mathrm{h} \times \sin(21^{\circ})\).
04

Convert wind velocity from components to magnitude and direction

The wind velocity in magnitude and direction can be found using Pythagoras' theorem and trigonometric functions. The magnitude is \(\sqrt{N_{wind}^{2}+E_{wind}^{2}}\). The direction is given by the arctangent of \((E_{wind}/N_{wind})\) which must be east of north because \(E_{wind}\) is negative.

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Key Concepts

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

Velocity Vector Components
Understanding the velocity vector components of an object, such as an airplane, is essential for analyzing its motion. Velocity, a vector quantity, means that it has both magnitude and direction. Breaking down the velocity into components allows for a detailed understanding of the movement in each independent direction, typically along the cardinal directions like east and west, or north and south.

For example, if a plane moves northeast, it has both northward and eastward components in its velocity. To find these components, trigonometry is used—specifically, the sine and cosine functions. The cosine of the angle gives the component in the direction of the original heading (in this case, northward), and the sine provides the component perpendicular to it (eastward). This concept was applied in Step 1 of our exercise, where the plane's velocity was broken into its northward and eastward components using a given angle.
Trigonometry in Physics
Trigonometry plays a crucial role in physics, especially when dealing with vectors like velocity, force, or displacement. It helps in breaking down vectors into perpendicular components, which can then be independently analyzed and manipulated according to the laws of physics.

In the context of our exercise, trigonometric functions were used to determine the components of the plane's velocity. By multiplying the airspeed by \(\cos(21^\circ)\) and \(\sin(21^\circ)\), the northward and eastward components were calculated, respectively. The correct use of sine and cosine is vital for understanding vector quantities and provides a way to handle the directional nature of these physical quantities.
Resultant Velocity
The resultant velocity is the combination of all velocity vectors acting on an object, providing a single vector that gives the overall motion. This is calculated by adding up all the individual velocity components corresponding to each direction. The principle of vector addition states that vectors must be added geometrically, taking into account both their magnitude and direction.

In our exercise, Step 2 required finding the effective velocity of the plane in a straight northward direction, ignoring any eastward or westward movement. This effective or 'resultant' velocity was determined simply by dividing the total northward distance by the time taken. Understanding resultant velocity is important in various contexts, such as navigating an airplane with wind influence or predicting the path of a projectile in physics.

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Most popular questions from this chapter

A locomotive accelerates a 23 -car train along a level track. Each car has a mass of \(48.6\) metric tons and is subject to a drag force \(f=243 v\), where \(v\) is the speed in \(\mathrm{m} / \mathrm{s}\) and the force \(f\) is in N. At the instant when the speed of the train is \(34.5 \mathrm{~km} / \mathrm{h}\), the acceleration is \(0.182 \mathrm{~m} / \mathrm{s}^{2} .(a)\) Calculate the force exerted by the locomotive on the first car. ( \(b\) ) Suppose that the force found in part \((a)\) is the greatest force the locomotive can exert on the train. What, then, is the steepest grade up which the locomotive can pull the train at \(34.5 \mathrm{~km} / \mathrm{h} ?(1\) metric ton \(=1000 \mathrm{~kg} .\) )

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