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Chapter 2: Problem 31

A Cessna aircraft has a liftoff speed of \(120 \mathrm{~km} / \mathrm{h}\). (a) What minimum constant acceleration does the aircraft require if it is to be airborne after a takeoff run of \(240 \mathrm{~m}\) ? (b) How long does it take the aircraft to become airborne?

Short Answer

Expert verified
The minimum constant acceleration the aircraft requires is approximately \(2.32 \mathrm{~m/s^2}\) and it will take approximately \(14.4 \mathrm{~s}\) for the Cessna aircraft to become airborne.

Step by step solution

01

Understand the problem and identify given data

We are given that the lift-off speed (final velocity) \(v = 120 km / h = 33.3 m/s\) (since 1 km = 1000 m and 1 h = 3600 s), the required distance for lift-off (displacement) \(d = 240 m\), the initial velocity \(u = 0 m/s\) (from rest), and we want to find the acceleration \(a\) in part (a), and time \(t\) in part (b).
02

Apply Kinematic equation to calculate acceleration

For constant acceleration, we use the kinematic equation \(v^2 = u^2 + 2ad\). Since the Cessna starts from rest, \(u = 0\), so the equation simplifies to \(v^2 = 2ad\). By plugging values and rearranging for \(a\), we get \(a = \frac{v^2}{2d} = \frac{(33.3)^2}{2 * 240} = 2.32 \mathrm{~m/s^2}\).
03

Use the Kinematic equation to calculate time

To calculate the time to become airborne, we use the equation \(v = u + at\). Again, as the plane starts from rest (\(u = 0\)), we are left with \(v = at\). Rearranging for time, we get \(t = \frac{v}{a} = \frac{33.3}{2.32} = 14.4 \mathrm{~s}\).

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Key Concepts

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

Constant Acceleration
When a Cessna aircraft prepares for takeoff, it experiences a period of constant acceleration. This means the aircraft's speed increases at a steady rate until it reaches the liftoff speed necessary for becoming airborne. The constant acceleration is crucial because it allows for the prediction of how the aircraft will move during the takeoff process using mathematical models called kinematic equations.

In constant acceleration, the rate at which velocity changes over time remains the same. For the Cessna in our example, reaching that steady acceleration is essential to ensure it can attain the required liftoff speed over the given displacement.
Kinematic Equations
Kinematic equations are mathematical formulas used to describe the motion of objects moving with uniform acceleration without considering the forces that cause such motion. In the context of a Cessna's takeoff, these equations relate important variables like initial velocity, final velocity, displacement, acceleration, and time.

For our Cessna's scenario, there are two key kinematic equations used. The first relates velocity, acceleration, and displacement: \(v^2 = u^2 + 2ad\). The second connects velocity, acceleration, and time: \(v = u + at\). These formulas are indispensable for pilots and engineers to determine the necessary conditions for a safe takeoff.
Initial Velocity
The initial velocity, often denoted by \(u\), is the speed of the aircraft when it begins to accelerate along the runway. In most cases, and specifically for the Cessna in our example, this initial velocity is zero because the aircraft starts from rest.

The initial velocity is a key variable in kinematic equations. Knowing the initial velocity allows us to predict how the takeoff will progress, especially in combination with information about acceleration and displacement.
Final Velocity
Final velocity, denoted by \(v\), is the speed the aircraft must reach to become airborne. For the Cessna, this speed is 120 km/h which translates to 33.3 m/s. Final velocity is the clue to determining if the takeoff run is sufficient for the aircraft to lift off, given the parameters of constant acceleration and runway length.

In the kinematic equations, final velocity is what we solve for to find either the acceleration necessary over a given distance, or the time it will take to reach that velocity.
Displacement
Displacement refers to the distance the Cessna covers on the runway before becoming airborne, and it is a vector quantity that measures the overall change in position. A displacement of 240 meters is given for our Cessna, which is the straight-line distance from the start of acceleration to the point of lift-off.

Understanding displacement is essential in aviation as it ensures that the runway is long enough to allow the aircraft to reach the required takeoff speed without overshooting the available space.
Airborne Time
Airborne time, in the context of our problem, refers to the duration the Cessna requires to go from a stationary position to flying speed. It is directly related to acceleration and final velocity and is calculated using kinematic equations. For the Cessna, the airborne time is the time it takes to reach a velocity of 33.3 m/s with constant acceleration.

The time to become airborne is a critical factor for pilots as it affects the takeoff phase which is one of the most critical phases of flight. Knowing this time allows for proper planning and ensures the safety of the takeoff procedure.

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Most popular questions from this chapter

A person takes a trip, driving with a constant speed of \(89.5 \mathrm{~km} / \mathrm{h}\), except for a \(22.0\)-min rest stop. If the person's average speed is \(77.8 \mathrm{~km} / \mathrm{h}\), (a) how much time is spent on the trip and (b) how far does the person travel?

A race car moves such that its position fits the relationship $$ x=(5.0 \mathrm{~m} / \mathrm{s}) t+\left(0.75 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3} $$ where \(x\) is measured in meters and \(t\) in seconds. (a) Plot a graph of the car's position versus time. (b) Determine the instantaneous velocity of the car at \(t=4.0 \mathrm{~s}\), using time intervals of \(0.40 \mathrm{~s}, 0.20 \mathrm{~s}\), and \(0.10 \mathrm{~s}\). (c) Compare the average velocity during the first \(4.0 \mathrm{~s}\) with the results of part (b).

A ball is thrown vertically upward with a speed of 25.0 \(\mathrm{m} / \mathrm{s}\). (a) How high does it rise? (b) How long does it take to reach its highest point? (c) How long does the ball take to hit the ground after it reaches its highest point? (d) What is its velocity when it returns to the level from which it started?

A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for \(30.0 \mathrm{~min}\) at \(80.0 \mathrm{~km} / \mathrm{h}, 12.0 \mathrm{~min}\) at \(100 \mathrm{~km} / \mathrm{h}\), and \(45.0 \mathrm{~min}\) at \(40.0 \mathrm{~km} / \mathrm{h}\) and spends \(15.0 \mathrm{~min}\) eating lunch and buying gas. (a) Determine the average speed for the trip. (b) Determine the distance between the initial and final cities along the route.

An object moves with constant acceleration \(4.00 \mathrm{~m} / \mathrm{s}^{2}\) and over a time interval reaches a final velocity of \(12.0 \mathrm{~m} / \mathrm{s}\). (a) If its original velocity is \(6.00 \mathrm{~m} / \mathrm{s}\), what is its displacement during the time interval? (b) What is the distance it travels during this interval? (c) If its original velocity is \(-6.00 \mathrm{~m} / \mathrm{s}\), what is its displacement during this interval? (d) What is the total distance it travels during the interval in part (c)?
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