/*! 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 186 A helicopter approaches a rescue... [FREE SOLUTION] | 91Ó°ÊÓ

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A helicopter approaches a rescue scene. A victim \(P\) is drifting along with the river current of speed \(v_{C}=2 \mathrm{m} / \mathrm{s} .\) The wind is blowing at a speed \(v_{W}=\) \(3 \mathrm{m} / \mathrm{s}\) as indicated. Determine the velocity relative to the wind which the helicopter must acquire so that it maintains a steady overhead position relative to the victim.

Short Answer

Expert verified
The helicopter must move at \(1\,\text{m/s}\) opposite to the wind.

Step by step solution

01

Understanding the Problem

We are tasked with determining the velocity of a helicopter relative to the wind to maintain a steady position over a victim drifting with the river current at \(v_C = 2\; \text{m/s}\). The wind speed is \(v_W = 3\; \text{m/s}\).
02

Setting Up the Reference Frames

We need to consider the velocities in terms of different reference frames. Let the downriver direction, where the current flows, be positive. The velocity of the victim relative to the ground is the same as the current's velocity, which is \(v_C = 2\; \text{m/s}\).
03

Finding the Total Ground Velocity

To maintain a steady position over the victim, the helicopter should match the victim's velocity relative to the ground (i.e., \(v_{C}\)). This means that the helicopter's velocity relative to the ground \(v_H\) should also be \(2\; \text{m/s}\) in the same direction.
04

Expressing Helicopter's Velocity Relative to the Wind

Let the velocity of the helicopter relative to the wind be \(v_{HW}\). The total ground velocity of the helicopter, \(v_H = v_{C}\), should equal the sum of its velocity relative to the wind \(v_{HW}\) and the wind's velocity \(v_W\): \[ v_H = v_{HW} + v_W \]Where \(v_W = 3\; \text{m/s}\).
05

Calculating Helicopter's Required Velocity Relative to the Wind

Substitute \(v_H = v_C = 2\; \text{m/s}\) and \(v_W = 3\; \text{m/s}\) into the equation to solve for \(v_{HW}\): \[ 2 = v_{HW} + 3 \]Solving for \(v_{HW}\) gives:\[ v_{HW} = 2 - 3 = -1\; \text{m/s} \].
06

Interpreting the Result

The negative sign indicates that the helicopter should move at \(1\; \text{m/s}\) opposite to the wind to remain stationary relative to the victim.

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Key Concepts

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

Helicopter Dynamics
In understanding the motion of helicopters in various environments, we have to consider several factors such as lift, thrust, and relative velocity. When a helicopter operates, it must maintain balance against gravitational forces and any other environmental conditions. This means its rotor speed and direction play a crucial role in counteracting these forces.
For instance, when a helicopter seeks to maintain a specific position relative to an object (like our earlier scenario with a victim in the river), it not only fights against the pull of gravity but also other forces such as wind or current. Therefore, if it aims to stay overhead of a victim in a river, it will need to match its relative speed to the victim, accounting for external factors like wind speed and river currents.
  • Lift overcomes weight (gravity).
  • Thrust counters drag, which includes factors like wind and river currents.
By understanding these dynamics, pilots can ensure helicopters remain steady and ensure effective operations, especially during rescue missions.
River Current
River currents represent the directional flow of water within a river, typically measured in meters per second (\( ext{m/s}\)). In the context of our exercise, the river current is at a speed of \(2\,\text{m/s}\).
For a helicopter hovering over a victim adrift in this river current, it needs to equal the drifting speed with the current to keep a steady position. This implies moving neither faster nor slower than \(2\,\text{m/s}\) in the same downriver direction.
  • Positive velocity: Matches the current's direction.
  • Negative velocity: Opposes the current's direction.
The combined effect of river currents and helicopter dynamics ensures that the helicopter adjusts its orientation and speed adequately to perform its rescue function effectively.
Wind Speed
Wind speed can significantly impact the dynamics of flying vehicles such as helicopters. It measures how fast the wind is traveling between two points and can be both a help and a hindrance for a hovering helicopter. In our scenario, the wind blows at a speed of \(3\,\text{m/s}\).
The helicopter must manage this additional factor effectively. This involves calculating and adjusting its actual velocity to account for the wind, ensuring that its desired relative position is maintained.
  • Wind can propel the helicopter faster in one direction if not countered.
  • It requires counteraction for precise positioning.
Thus, understanding and managing wind speed is crucial for maintaining the necessary balance and performing tasks efficiently.
Reference Frames
In physics, reference frames are essential for understanding how objects move relative to one another. They provide a perspective from which these motions can be observed and calculated. When dealing with helicopters, river currents, and wind, choosing an appropriate reference frame simplifies the problem.
In our exercise, the helicopter's motion is analyzed with respect to two reference frames:
  • Ground reference frame: Where the helicopter's velocity matches the river's current velocity to remain steady over the victim.
  • Wind reference frame: Where the helicopter's velocity is adjusted considering the wind's influence.
By using these frames, we can break complex scenarios into more manageable computations, ensuring accurate and precise results for dynamic maneuvers.

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