/*! 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 89 A jet is taking off from the dec... [FREE SOLUTION] | 91Ó°ÊÓ

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A jet is taking off from the deck of an aircraft carrier, as shown in the image. Starting from rest, the jet is catapulted with a constant acceleration of \(+31 \mathrm{m} / \mathrm{s}^{2}\) along a straight line and reaches a velocity of \(+62 \mathrm{m} / \mathrm{s}\). Find the displacement of the jet. (a) A jet is being launched from an aircraft carrier. (b) During the launch, a catapult accelerates the jet down the flight deck.

Short Answer

Expert verified
The displacement of the jet is 62 meters.

Step by step solution

01

Understand the Problem

Identify the given variables and what needs to be calculated. The initial velocity of the jet, \( u \), is 0 since it starts from rest. The final velocity, \( v \), is given as \( +62 \mathrm{m/s} \). The constant acceleration, \( a \), is \( +31 \mathrm{m/s}^2 \). We need to find the displacement, \( s \), which is the distance the jet travels on the flight deck.
02

Select the Appropriate Equation

Since we are dealing with constant acceleration and have values for initial velocity, final velocity, and acceleration, we can use the kinematic equation: \[ v^2 = u^2 + 2as \]This equation allows us to solve for the displacement \( s \).
03

Substitute Known Values into the Equation

Substitute the known values into the kinematic equation:\[ (62)^2 = (0)^2 + 2(31)s \]This simplifies to:\[ 3844 = 62s \]
04

Solve for Displacement \( s \)

Rearrange the equation to solve for \( s \):\[ 2(31)s = 3844 \]\[ 62s = 3844 \]Divide both sides of the equation by 62 to solve for \( s \):\[ s = \frac{3844}{62} \]\[ s = 62 \]
05

Interpret the Result

The calculated displacement \( s = 62 \) means that the jet travels a displacement of 62 meters along the flight deck before reaching its final velocity of 62 m/s.

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Key Concepts

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

Constant Acceleration
In physics, constant acceleration refers to a scenario where the velocity of an object changes at a steady rate over time. This means that regardless of the position or time interval observed, the acceleration remains unchanged. Constant acceleration is a fundamental concept that simplifies the analysis of motion by allowing the use of linear equations to predict future positions and velocities.

When a jet starts from rest, as in the given problem, it experiences a constant acceleration due to the force exerted by the catapult on the flight deck. This acceleration causes the jet's speed to increase steadily until it reaches the required final velocity for takeoff. In mathematical terms, if an object's acceleration is given as a constant value, it can be denoted as \(a = 31\,\mathrm{m/s^2}\).

Key points to remember about constant acceleration include:
  • It is a uniform rate of change of velocity.
  • Equations involving constant acceleration can predict displacement and velocity over time.
  • Assumes negligible external forces such as air resistance or friction.
Displacement
Displacement represents how far an object has moved from its initial position, considering only the straight-line distance. Unlike distance, displacement is a vector quantity, meaning it has both magnitude and direction. In the context of the jet taking off, displacement refers to the straight-line path it travels during the acceleration on the flight deck.

To find the displacement, once the acceleration and the velocity changes are known, we can use kinematic equations. In this scenario, displacement was calculated using the equation \( v^2 = u^2 + 2as \), which provided a direct method to determine the distance the jet traveled, based on its initial and final velocities along with constant acceleration. This calculation included substituting known values, leading to a displacement of \( s = 62 \) meters.

Remember that displacement:
  • Measures a change in position in a particular direction.
  • Can be shorter than the actual path traveled if the movement involves detours.
  • Is not affected by any initial or final paths that do not change the direct line of travel.
Kinematic Equations
Kinematic equations are essential tools in physics for describing motion. They relate the variables of motion: displacement, initial velocity, final velocity, acceleration, and time. For situations involving constant acceleration, these equations simplify the process of predicting future motion states.

In our jet's case, the kinematic equation used was: \[v^2 = u^2 + 2as\]Here, \(v\) represents the final velocity, \(u\) the initial velocity, \(a\) the acceleration, and \(s\) the displacement. By substituting the known values into this equation, we can solve for the unknown variable – in this instance, the displacement \(s\).

Important aspects of kinematic equations are:
  • They only apply to motion with constant acceleration.
  • Each equation in the set differs based on the variable to solve for, like time, velocity, or displacement.
  • They offer a systematic way to predict or understand motion without needing real-world experimentation every time.

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Most popular questions from this chapter

From her bedroom window a girl drops a water-filled balloon to the ground, \(6.0 \mathrm{m}\) below. If the balloon is released from rest, how long is it in the air?

A car is traveling at a constant speed of \(33 \mathrm{m} / \mathrm{s}\) on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is \(2.5 \mathrm{km}\) away?

At the beginning of a basketball game, a referee tosses the ball straight up with a speed of \(4.6 \mathrm{m} / \mathrm{s} .\) A player cannot touch the ball until after it reaches its maximum height and begins to fall down. What is the minimum time that a player must wait before touching the ball?

Two soccer players start from rest, \(48 \mathrm{m}\) apart. They run directly toward each other, both players accelerating. The first player's acceleration has a magnitude of \(0.50 \mathrm{m} / \mathrm{s}^{2} .\) The second player's acceleration has a magnitude of \(0.30 \mathrm{m} / \mathrm{s}^{2} .\) (a) How much time passes before the players collide? (b) At the instant they collide, how far has the first player run?

The initial velocity and acceleration of four moving objects at a given instant in time are given in the following table. Determine the final speed of each of the objects, assuming that the time elapsed since \(t=0\) s is 2.0 s. $$ \begin{array}{lcc} & \text { Initial velocity } v_{0} & \text { Acceleration } a \\\\\hline \text { (a) } & +12 \mathrm{m} / \mathrm{s} & +3.0 \mathrm{m} / \mathrm{s}^{2} \\\\\hline \text { (b) } & +12 \mathrm{m} / \mathrm{s} & -3.0 \mathrm{m} / \mathrm{s}^{2} \\\\\hline \text { (c) } & -12 \mathrm{m} / \mathrm{s} & +3.0 \mathrm{m} / \mathrm{s}^{2} \\\\\hline \text { (d) } & -12 \mathrm{m} / \mathrm{s} & -3.0 \mathrm{m} / \mathrm{s}^{2} \\ \hline\end{array}$$

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