/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 An airplane has a velocity relat... [FREE SOLUTION] | 91Ó°ÊÓ

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An airplane has a velocity relative to the ground of \(210 \mathrm{m} / \mathrm{s}\) toward the east. The pilot measures his airspeed (the speed of the plane relative to the air) to be \(160 \mathrm{m} / \mathrm{s}\) What is the minimum wind velocity possible?

Short Answer

Expert verified
The minimum wind velocity is \(50 \mathrm{m/s}\) east.

Step by step solution

01

Understanding Given Data

The airplane's velocity relative to the ground is given as \(210\, \mathrm{m/s}\) east. The airspeed (velocity relative to the air) is given as \(160\, \mathrm{m/s}\). We need to find out the wind velocity.
02

Setting up the Problem

The ground speed of the airplane is the vector sum of the airspeed and the wind speed. We'll denote the wind speed as \(\mathbf{v_w}\). The equation relating airspeed, ground speed, and wind speed is: \(\mathbf{v_g} = \mathbf{v_a} + \mathbf{v_w}\), where \(\mathbf{v_g} = 210\, \mathrm{m/s}\) and \(\mathbf{v_a} = 160\, \mathrm{m/s}\).
03

Using Vector Magnitudes

Since the ground speed is larger than the airspeed, and knowing the planes fly in relatively straight paths, it's clear the wind is assisting the eastward velocity. Thus, we rearrange and compute: \[ \mathbf{v_w} = \mathbf{v_g} - \mathbf{v_a} = 210 \text{ m/s east} - 160 \text{ m/s east}. \]
04

Calculate Wind Velocity

Solving the above equation gives us the minimum wind velocity: \(\mathbf{v_w} = 210 - 160 = 50\, \mathrm{m/s}\). This is the minimum wind speed possible, going east to achieve the observed ground speed.

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Key Concepts

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

Velocity Calculation
Velocity is a vector quantity. It gives us both the speed and direction of an object. When calculating velocity, it’s vital to consider both magnitude and direction, since velocity can vary with changes in either. In the context of airplanes, velocity calculations often involve vector addition due to multiple forces, such as wind, influencing the motion.
This exercise illustrates how the velocity of an airplane relative to the ground is affected by wind. Here, the plane has both an airspeed and a ground speed. These two velocities are related through vector addition with the wind being the connecting vector. Understanding how to calculate the resulting velocity by combining these vectors helps in determining how various forces influence movement in navigation.
Ground Speed
Ground speed is the actual speed of an aircraft relative to the ground. It accounts for all factors affecting flight, such as wind.
In our exercise, the ground speed is given as 210 m/s. This means that when all influences, primarily the wind, are combined, the plane moves eastward at this speed.
Calculating ground speed requires an understanding of the vector nature of velocity. It is the result of vector addition of airspeed and wind speed. If you know either airspeed or wind speed, you can determine the other if the ground speed is known. This notion is crucial for pilots who need precise information to maintain course and timing.
Wind Speed
The wind speed is the vector that is added to the airspeed to obtain the ground speed. Wind can significantly impact how fast and in what direction an airplane moves relative to the ground.
In this scenario, the wind velocity is calculated as the difference between the ground speed and airspeed. With a given ground speed of 210 m/s and airspeed of 160 m/s, the wind assists the plane by moving it at 50 m/s to the east.
Understanding wind speed is essential for navigating flights effectively. It determines whether a pilot needs to adjust the airplane's course or speed to reach the intended destination accurately. Wind calculations are critical for efficiency and safety in flight.
Airspeed
Airspeed refers to how fast an airplane is moving relative to the air around it. It is distinct from ground speed because it doesn’t consider the impact of wind. Accurately knowing the airspeed is crucial for pilots to ensure the aircraft is flying safely and efficiently.
The exercise specifies an airspeed of 160 m/s. This means the airplane, regardless of the wind, will move through the air at this speed.
For navigation and flight control, knowing the airspeed helps pilots maintain desired speeds, avoid stalling, and manage fuel efficiently. By understanding the relationship with ground speed and wind speed, pilots can anticipate necessary adjustments to reach their destination on time.

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Most popular questions from this chapter

A projectile is launched at \(t=0\) with initial speed \(v_{\mathrm{i}}\) at an angle \(\theta\) above the horizontal. (a) What are \(v_{x}\) and \(v_{y}\) at the projectile's highest point? (b) Find the time \(t\) at which the projectile reaches its maximum height. (c) Show that the maximum height \(H\) of the projectile is $$H=\frac{\left(v_{i} \sin \theta\right)^{2}}{2 g}$$
A ball is thrown from a point \(1.0 \mathrm{m}\) above the ground. The initial velocity is \(19.6 \mathrm{m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) above the horizontal. (a) Find the maximum height of the ball above the ground. (b) Calculate the speed of the ball at the highest point in the trajectory.
A ball is thrown horizontally off the edge of a cliff with an initial speed of \(20.0 \mathrm{m} / \mathrm{s} .\) (a) How long does it take for the ball to fall to the ground 20.0 m below? (b) How long would it take for the ball to reach the ground if it were dropped from rest off the cliff edge? (c) How long would it take the ball to fall to the ground if it were thrown at an initial velocity of \(20.0 \mathrm{m} / \mathrm{s}\) but \(18^{\circ}\) below the horizontal?
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You have been employed by the local circus to plan their human cannonball performance. For this act, a spring-loaded cannon will shoot a human projectile, the Great Flyinski, across the big top to a net below. The net is located \(5.0 \mathrm{m}\) lower than the muzzle of the cannon from which the Great Flyinski is launched. The cannon will shoot the Great Flyinski at an angle of \(35.0^{\circ}\) above the horizontal and at a speed of $18.0 \mathrm{m} / \mathrm{s} .$ The ringmaster has asked that you decide how far from the cannon to place the net so that the Great Flyinski will land in the net and not be splattered on the floor, which would greatly disturb the audience. What do you tell the ringmaster? ( Wheractive: projectile motion)
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