/*! 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 49 On its takeoff roll, the airplan... [FREE SOLUTION] | 91Ó°ÊÓ

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On its takeoff roll, the airplane starts from rest and accelerates according to \(a=a_{0}-k v^{2},\) where \(a_{0}\) is the constant acceleration resulting from the engine thrust and \(-k v^{2}\) is the acceleration due to aerodynamic drag. If \(a_{0}=2 \mathrm{m} / \mathrm{s}^{2}, k=0.00004 \mathrm{m}^{-1}\) and \(v\) is in meters per second, determine the design length of runway required for the airplane to reach the takeoff speed of \(250 \mathrm{km} / \mathrm{h}\) if the drag term is \((a)\) excluded and (b) included

Short Answer

Expert verified
Without drag: 1205.63 meters; with drag: 1107.51 meters.

Step by step solution

01

Convert Takeoff Speed

First, convert the takeoff speed from kilometers per hour to meters per second. The takeoff speed given is 250 km/h. To convert this, \[ v_{takeoff} = 250 \times \frac{1000}{3600} \text{ m/s} = 69.44 \text{ m/s} \]
02

Calculate Runway Length without Drag

Since drag is excluded in part (a), the acceleration is constant, i.e., \( a = a_0 \). Use the equation of motion: \[ v^2 = u^2 + 2as \]where \( v = 69.44 \text{ m/s} \), \( u = 0 \text{ m/s} \), and \( a = 2 \text{ m/s}^2 \). Solving for \( s \): \[ (69.44)^2 = 0 + 2 \times 2 \times s \] \[ s = \frac{69.44^2}{4} = 1205.63 \text{ m} \]
03

Set Up Runway Length with Drag Equation

For part (b), include the drag term. The differential equation for acceleration is: \[ a = \frac{dv}{dt} = a_0 - kv^2 \]Transform this into: \[ a_0 - kv^2 = v \frac{dv}{dx} \]
04

Integrate to Find Runway Length with Drag

Separate variables and integrate: \[ \int \frac{v}{a_0 - kv^2} dv = \int dx \]The left integral is solved using the substitution \( u = a_0 - kv^2 \). Solving the integral gives: \[ -\frac{1}{2k} \ln(a_0 - kv^2) \bigg|_0^{69.44} = x \]Insert the values: \[ -\frac{1}{2 \times 0.00004} \left[ \ln(2 - 0.00004 \times (69.44)^2) - \ln(2) \right] = x \]Calculate the numerical value of the runway length \( x \approx 1107.51 \text{ m} \).
05

Conclusion

The runway length needed without aerodynamic drag is 1205.63 meters. With aerodynamic drag included, the runway length reduces to 1107.51 meters.

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Key Concepts

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

Aerodynamic Drag
Aerodynamic drag is a force that opposes the motion of an object through a fluid, such as air. For airplanes, it is a crucial factor during takeoff, flight, and landing. It is created by the air flowing over the aircraft's surfaces.

In the exercise, aerodynamic drag is represented by the term \(-kv^2\), where \(k\) is a constant that allows us to capture the effect of drag force in simple mathematical terms, and \(v\) is the velocity of the airplane. This term \(-kv^2\) shows that drag increases with the square of velocity, meaning that as an airplane goes faster, the drag force increases significantly.

For the runway length calculation, aerodynamic drag reduces the acceleration and effectively shortens the runway distance needed for takeoff by restraining the speed buildup over time. Including this term in motion equations is essential for realistic assessments of runway requirements.
Differential Equations
Differential equations are equations that involve a function and its derivatives. They are used to model various physical systems, including those with changing velocities or accelerations.

In this exercise, the differential equation \(a = \frac{dv}{dt} = a_0 - kv^2\) describes the changing velocity of the airplane under the influence of both engine thrust and aerodynamic drag. The equation stipulates that acceleration \(a\) is not constant but varies depending on the current speed \(v\) due to the drag force.

By solving differential equations, we can find how variables like velocity and position change over time, which is critical for determining the length of the runway required for safe takeoff when drag is considered.
Constant Acceleration
Constant acceleration occurs in situations when the acceleration of an object does not change over time. In many physics problems, it simplifies calculations because the relation between velocity, time, and distance becomes linear.

In part (a) of the exercise, only the engine thrust is considered, meaning the acceleration is constant and equals \(a_0 = 2 \text{ m/s}^2\). This simplification allows us to use basic equations of motion, like \(v^2 = u^2 + 2as\), for determining the distance needed for the airplane to reach its takeoff speed, without needing to account for aerodynamic drag.

Constant acceleration is a theoretical assumption that helps us understand basic motion scenarios. However, real-world conditions often involve varying forces resulting in non-constant acceleration, as shown when aerodynamic drag is present.
Equation of Motion
Equations of motion are fundamental principles that relate to the motion characteristics of an object moving under particular forces. They can express the relationship among velocity, acceleration, and displacement.

In scenarios with constant acceleration, like part (a) of the exercise, \(v^2 = u^2 + 2as\) becomes very useful. It connects initial velocity \(u\), final velocity \(v\), acceleration \(a\), and displacement \(s\). This equation allows us to calculate how far the airplane will travel while accelerating to a certain speed without considering drag forces.

When differential equations involving drag are used (as in part (b)), integration can provide the necessary displacements. This shows the importance of equations of motion as they lay the groundwork for discussing more complex scenarios involving varying forces.

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Most popular questions from this chapter

A block of mass \(m\) rests on a rough horizontal surface and is attached to a spring of stiffness \(k .\) The coefficients of both static and kinetic friction are \(\mu\) The block is displaced a distance \(x_{0}\) to the right of the unstretched position of the spring and released from rest. If the value of \(x_{0}\) is large enough, the spring force will overcome the maximum available static friction force and the block will slide toward the unstretched position of the spring with an acceleration \(a=\mu g-\frac{k}{m} x,\) where \(x\) represents the amount of stretch (or compression) in the spring at any given location in the motion. Use the values \(m=5 \mathrm{kg}, k=150 \mathrm{N} / \mathrm{m}, \mu=0.40,\) and \(x_{0}=200 \mathrm{mm}\) and determine the final spring stretch (or compression \(x_{f}\) when the block comes to a complete stop.

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A particle which moves with curvilinear motion has coordinates in meters which vary with time \(t\) in seconds according to \(x=2 t^{2}+3 t-1\) and \(y=5 t-2 .\) Determine the coordinates of the center of curvature \(C\) at time \(t=1\) s.

The pilot of a jet transport brings the engines to full takeoff power before releasing the brakes as the aircraft is standing on the runway. The jet thrust remains constant, and the aircraft has a nearconstant acceleration of \(0.4 \mathrm{g}\). If the takeoff speed is \(200 \mathrm{km} / \mathrm{h},\) calculate the distance \(s\) and time \(t\) from rest to takeoff.

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