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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 \( 3.15 \, \text{m/s}^2 \) southward.

Step by step solution

01

Identify Given Values

We need to identify the given values in the problem. The initial speed of the jetliner, \( v_i \), is \( 69 \, \text{m/s} \). The final speed, \( v_f \), is \( 6.1 \, \text{m/s} \). The distance on the runway, \( d \), is \( 750 \, \text{m} \).
02

Use the Kinematic Equation

We will use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance:\[ v_f^2 = v_i^2 + 2ad \]We'll solve this equation for \( a \), the average acceleration.
03

Solve for Acceleration

Rearrange the kinematic equation to find acceleration:\[ a = \frac{v_f^2 - v_i^2}{2d} \]Substitute the given values into the equation:\[ a = \frac{(6.1 \, \text{m/s})^2 - (69 \, \text{m/s})^2}{2 \times 750 \, \text{m}} \]
04

Calculate the Acceleration

Calculate the values:\[ a = \frac{37.21 - 4761}{1500} \]\[ a = \frac{-4723.79}{1500} \]Solving this gives:\[ a \approx -3.15 \, \text{m/s}^2 \]
05

Determine Direction and Magnitude

The negative sign indicates that the acceleration is in the opposite direction of the initial motion, which makes sense as the plane is decelerating. Thus, the magnitude of the acceleration is \( 3.15 \, \text{m/s}^2 \), and the direction is southward (opposite the landing direction northward).

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Key Concepts

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

kinematic equation
A kinematic equation is a crucial tool used in physics to describe the motion of objects, especially when acceleration is involved. In this exercise, we use a specific kinematic equation that connects initial velocity, final velocity, acceleration, and distance.

The particular kinematic equation applied here is: \[ v_f^2 = v_i^2 + 2ad \] This formula is instrumental in calculating one of these variables when the others are known. It allows us to solve for the plane's average acceleration as it lands by rearranging the equation to isolate acceleration \(a\): \[ a = \frac{v_f^2 - v_i^2}{2d} \] Understanding how these equations work lays the foundation for solving many problems related to motion, including those involving varying speeds and forces.
initial and final velocities
Initial and final velocities are integral factors in understanding an object's change in motion over time. In this scenario, the jetliner's initial velocity \( v_i \) as it touches down is \( 69 \, \text{m/s} \). Before coming to a complete stop or reaching the intended lower speed at the end of the runway, its final velocity \( v_f \) is \( 6.1 \, \text{m/s} \).

These velocities give us insights into how fast the plane travels initially and how much it must decrease its speed to safely and smoothly land within the given runway distance. The change between these velocities helps us calculate the acceleration needed during landing using the kinematic equations.

In physics, such values are crucial for determining how external forces, such as thrust or drag, affect an object's motion.
distance traveled
Distance traveled is often a key parameter in solving motion-related problems, acting as a constraint or boundary in calculations. For the jetliner's landing process, the available runway length is \( 750 \, \text{m} \), providing the distance within which the plane must slow down.

The distance component in the kinematic equation \( v_f^2 = v_i^2 + 2ad \) highlights how these 750 meters are crucial in solving for acceleration. It’s the runway’s physical length that allows engineers to determine how quickly the speed must change.

Problems involving distance emphasize the importance of understanding environmental constraints and how they impact a moving object's kinematics.
jetliner landing
Jetliner landing is a complex process requiring precise calculations to ensure the aircraft comes to a stop safely. In this exercise, we examine how a jet reduces speed on a specific runway length through the application of physics principles.

When a jetliner lands, reducing speed efficiently within the available space is vital to avoid overshooting the runway. By calculating the average acceleration, in this case \( -3.15 \, \text{m/s}^2 \), we gain insight into the necessary braking force. The negative acceleration indicates deceleration, changing the velocity's direction from northward to effectively slowing the plane southward.

It's these calculated manipulations of accelerative forces that enable safe landings, proving the practical application of motion equations.

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Most popular questions from this chapter

A car is traveling along a straight road at a velocity of \(+36.0 \mathrm{m} / \mathrm{s}\) when its engine cuts out. For the next twelve seconds the car slows down, and its average acceleration is \(\bar{a}_{1} .\) For the next six seconds the car slows down further, and its average acceleration is \(\bar{a}_{2} .\) The velocity of the car at the end of the eighteen-second period is \(+28.0 \mathrm{m} / \mathrm{s}\). The ratio of the average acceleration values is \(\bar{a}_{1} / \bar{a}_{2}=1.50 .\) Find the velocity of the car at the end of the initial twelve-second interval.

Over a time interval of 2.16 years, the velocity of a planet orbiting a distant star reverses direction, changing from \(+20.9 \mathrm{km} / \mathrm{s}\) to \(-18.5 \mathrm{km} / \mathrm{s} .\) Find (a) the total change in the planet's velocity (in \(\mathrm{m} / \mathrm{s}\) ) and (b) its average acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ) during this interval. Include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration.

An Australian emu is running due north in a straight line at a speed of \(13.0 \mathrm{m} / \mathrm{s}\) and slows down to a speed of \(10.6 \mathrm{m} / \mathrm{s}\) in \(4.0 \mathrm{s}\). (a) What is the direction of the bird's acceleration? (b) Assuming that the acceleration remains the same, what is the bird's velocity after an additional \(2.0 \mathrm{s}\) has elapsed?

The Lost Drone. You and your team are exploring the edge of an Antarctic mountain range and you send a drone ahead to help navigate. After takeoff you lose sight of the drone and, a few seconds later, the controls malfunction and the drone stops sending visual images and navigational information except for speed and directional data. Changing speeds erratically, the drone heads west until it makes a drastic turn at the 5 -minute mark to \(35.0^{\circ}\) east of south. After nearly ten minutes, the speed drops to zero and the drone stops sending data. It has crashed. Using the speed directional data, the team draws up the graph shown in the drawing.. How far is the drone from you, and in what direction must you go to retrieve it? Express your result as a geographical direction (i.e., in the form \(30^{\circ}\) north of east, etc.).

(a) What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of \(8.0 \mathrm{m} / \mathrm{s}\) when going down a slope for \(5.0 \mathrm{s} ?\) (b) How far does the skier travel in this time?

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