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A jetliner, traveling northward, is landing with a speed of \(69 \mathrm{m} / \mathrm{s}\). Once the jet touches down, it has \(750 \mathrm{m}\) of runway in which to reduce its speed to \(6.1 \mathrm{m} / \mathrm{s}\). Compute the average acceleration (magnitude and direction) of the plane during landing.

Short Answer

Expert verified
The average acceleration is approximately \(3.155 \, \mathrm{m/s^2}\) opposite to the travel direction.

Step by step solution

01

Identify Known Values

We start by listing all known quantities:- Initial velocity, \( v_i = 69 \, \mathrm{m/s} \)- Final velocity, \( v_f = 6.1 \, \mathrm{m/s} \)- Distance available for deceleration, \( d = 750 \, \mathrm{m} \)
02

Recall the Equation of Motion

For constant acceleration, the equation relating final velocity, initial velocity, acceleration, and distance is:\[ v_f^2 = v_i^2 + 2a d \]where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, \( a \) is the acceleration, and \( d \) is the distance covered.
03

Substitute Known Values

Substitute the known values into the motion equation:\[ (6.1)^2 = (69)^2 + 2a(750) \]
04

Solve for Average Acceleration

First, compute the squares of the velocities:\[ 6.1^2 = 37.21 \]\[ 69^2 = 4761 \]Substitute these values back into the equation:\[ 37.21 = 4761 + 1500a \]Rearrange to solve for \( a \):\[ 1500a = 37.21 - 4761 \]\[ a = \frac{37.21 - 4761}{1500} \]Simplify to get the result:\[ a \approx -3.155 \mathrm{m/s^2} \]
05

Determine Magnitude and Direction

The magnitude of acceleration is \(3.155 \, \mathrm{m/s^2}\) and the negative sign indicates that the direction of acceleration is opposite to the direction of motion (i.e., the plane is slowing down).

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Key Concepts

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

Jetliner Landing
When a jetliner comes in for a landing, it transitions from a high-speed flight to a complete stop on the runway. The key challenge during this phase is to safely reduce its speed over a limited distance. The jetliner in the exercise starts with an initial speed of 69 meters per second. Once it touches down, it has 750 meters to decelerate to a final velocity of 6.1 meters per second. This process of slowing down is a crucial aspect of the landing, ensuring the safety of passengers and the efficiency of airport operations.
Reducing speed in a controlled manner over a specified distance involves understanding and calculating the average acceleration. The pilot applies braking forces, which cause negative acceleration or deceleration, to bring the plane to a manageable speed, or to a complete stop.
Equation of Motion
The equation of motion provides a mathematical framework to describe the movement of objects under uniform acceleration. In this context, it is used to calculate the average acceleration of the jetliner during landing. The specific equation applied in this scenario is: \[ v_f^2 = v_i^2 + 2ad \] This equation links four key variables: the initial velocity \( v_i \), the final velocity \( v_f \), the acceleration \( a \), and the distance \( d \).
  • The initial velocity \( v_i \) represents how fast the jetliner is moving when it first touches the runway.
  • The final velocity \( v_f \) is the speed we want the jetliner to achieve at the end of the runway.
  • The distance \( d \) is the length of runway available to bring the jetliner to the desired final speed.
  • The unknown \( a \) is what we solve for, representing the average acceleration needed to achieve the desired speed within the given runway length.
Understanding and applying the equation of motion is crucial for determining how effectively a jetliner can land within space constraints.
Initial and Final Velocity
Velocity refers to both the speed and direction of an object's motion. In the context of the jetliner landing, two key velocities are considered: the initial and the final velocity.
  • Initial Velocity (\(v_i\)): This is the speed of the jetliner as it first makes contact with the runway, which is given as 69 meters per second northward in the exercise.
  • Final Velocity (\(v_f\)): This is the speed the jetliner must be reduced to as it reaches the end of the runway, which in this case is 6.1 meters per second, also northward.
These velocities are central to calculating the necessary deceleration (average acceleration) to safely land the jetliner.
Deceleration
Deceleration refers to the process of reducing an object's speed over time. It is a type of acceleration that occurs in the opposite direction of the initial motion. In this exercise, deceleration is required to bring the jetliner from an initial velocity of 69 m/s to a final velocity of 6.1 m/s.
The magnitude of deceleration indicates how quickly the jetliner's speed decreases. It is computed using the motion equation, resulting in an average deceleration value of approximately -3.155 meters per second squared. The negative sign in this value emphasizes that the jetliner is slowing down, moving against the initial direction of motion.
The effective control of deceleration is crucial during landing to ensure the plane stops within the available runway space, maintaining safety for passengers and crew.

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