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Chapter 3: Problem 14

The 750,000 -lb jetliner \(A\) has four engines, each of which produces a nearly constant thrust of \(40,0001 b\) during the takeoff roll. A small commuter aircraft \(B\) taxis toward the end of the runway at a constant speed \(v_{B}=15 \mathrm{mi} / \mathrm{hr} .\) Determine the velocity and acceleration which \(A\) appears to have relative to an observer in \(B 10\) seconds after \(A\) begins its takeoff roll. Neglect air and rolling resistance.

Short Answer

Expert verified
Relative velocity is 46.68 ft/s and relative acceleration is 6.868 ft/s虏.

Step by step solution

01

Calculate Acceleration of Jetliner A

The thrust provided by each engine is given as 40,000 lb, and with four engines, the total thrust is \( T = 4 \times 40,000 = 160,000 \) lb. The weight of jetliner \( A \) is 750,000 lb. The acceleration \( a \) can be calculated using Newton's second law, where \( T = m \cdot a \) and \( m = \frac{W}{g} \). We use \( g = 32.2 \) ft/s虏 (acceleration due to gravity in fps units). Hence, \( m = \frac{750,000}{32.2} \approx 23,291 \) lb路s虏/ft. Thus, \( a = \frac{T}{m} = \frac{160,000}{23,291} \approx 6.868 \) ft/s虏.
02

Convert Initial Conditions to Compatible Units

Speed of aircraft \( B \), \( v_B = 15 \) mi/hr, needs to be converted to ft/s. Since \( 1 \text{ mile} = 5280 \text{ ft} \) and \( 1 \text{ hr} = 3600 \text{ s} \), convert: \( v_B = 15 \times \frac{5280}{3600} \approx 22 \text{ ft/s} \).
03

Determine Velocity of Jetliner A After 10 Seconds

Using the equation of motion \( v = u + at \), where initial velocity \( u = 0 \) (since it starts from rest), \( a = 6.868 \) ft/s虏, and \( t = 10 \) seconds, we find \( v_A = 0 + 6.868 \times 10 = 68.68 \) ft/s.
04

Calculate Relative Velocity and Acceleration

Velocity of \( A \) relative to \( B \) is \( v_{AB} = v_A - v_B = 68.68 - 22 = 46.68 \) ft/s. Acceleration of \( A \) relative to \( B \) is \( a_{AB} = a_A - a_B \). Since \( B \) has constant speed, \( a_B = 0 \), leading to \( a_{AB} = 6.868 \) ft/s虏.

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Key Concepts

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

Understanding Newton's Second Law
Newton's Second Law is vital to understanding how motion occurs. It states that the force applied to an object results in acceleration, proportional to the mass of the object. Mathematically, it's expressed as \[ F = m \cdot a \] where \( F \) is the force applied, \( m \) is the mass, and \( a \) is the acceleration produced. In this exercise, the total thrust from the engines (160,000 lb) on the jetliner causes it to accelerate.
  • The first step is to convert the weight of the jetliner into mass by dividing by the acceleration due to gravity. This gives us the mass in units that match our force units, which enables us to compute acceleration.
  • Then, using \( F = m \cdot a \), we solve for \( a \) to determine how quickly the jetliner's velocity increases.
This relationship is fundamental in not only air travel but any scenario where force is applied to move an object.
Calculating Acceleration
Acceleration calculations often involve converting units to ensure consistency and clarity. In this exercise, the force produced by the jetliner's engines is pivotal to finding its acceleration.
  • The first step is understanding that the total thrust from the engines is spread across the entire mass of the jetliner.
  • To find acceleration, use the formula: \[ a = \frac{F}{m} \] where \( F \) is the total thrust and \( m \) is the mass derived from the jetliner's weight.
  • While calculating, it's crucial to remember the acceleration of gravity \( g \) is used here because weight is often given in pounds, a force unit.
Once acceleration is calculated, it can easily be applied to determine how velocity changes over time.
Grasping Basic Kinematics
Kinematics is the branch of physics that deals with motion, and it provides the equations to calculate both velocity and acceleration over time. It's quite applicable here as we deal with the jetliner's takeoff and the observer on the commuter aircraft.
  • We employ the basic equation of motion: \[ v = u + at \] where \( v \) is the final velocity, \( u \) is the initial velocity (0 if starting from rest), \( a \) is the acceleration, and \( t \) is the time elapsed.
  • In this exercise, once the acceleration \( a \) is determined, it is used to find the new speed \( v \) of the jetliner after 10 seconds.
Finally, to understand relative motion, the velocity of jetliner \( A \) is compared to the constant speed of aircraft \( B \). The differences in their velocities give us the relative velocity, that is, how fast \( A \) appears to be moving from the reference frame of \( B \). This gives a comprehensive picture of motion from different perspectives.

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Most popular questions from this chapter

In the design of a spring bumper for a 3500 -lb car, it is desired to bring the car to a stop from a speed of \(5 \mathrm{mi} / \mathrm{hr}\) in a distance equal to 6 in. of spring deformation. Specify the required stiffness \(k\) for each of the two springs behind the bumper. The springs are undeformed at the start of impact.

The position vector of a particle is given by \(\mathbf{r}=\) \(8 t \mathbf{i}+1.2 t^{2} \mathbf{j}-0.5\left(t^{3}-1\right) \mathbf{k},\) where \(t\) is the time in seconds from the start of the motion and where \(\mathbf{r}\) is expressed in meters. For the condition when \(t=\) 4 s, determine the power \(P\) developed by the force \(\mathbf{F}=40 \mathbf{i}-20 \mathbf{j}-36 \mathbf{k} \mathrm{N}\) which acts on the particle.

The nest of two springs is used to bring the 0.5 -kg plunger \(A\) to a stop from a speed of \(5 \mathrm{m} / \mathrm{s}\) and reverse its direction of motion. The inner spring increases the deceleration, and the adjustment of its position is used to control the exact point at which the reversal takes place. If this point is to correspond to a maximum deflection \(\delta=200 \mathrm{mm}\) for the outer spring, specify the adjustment of the inner spring by determining the distance \(s .\) The outer spring has a stiffness of \(300 \mathrm{N} / \mathrm{m}\) and the inner one a stiffness of \(150 \mathrm{N} / \mathrm{m}\).

The 600,000 -lb jet airliner has a touchdown velocity \(v=120 \mathrm{mi} / \mathrm{hr}\) directed \(\theta=0.5^{\circ}\) below the horizontal. The touchdown process of the eight main wheels takes 0.6 sec to complete. Treat the aircraft as a particle and estimate the average normal reaction force at each wheel during this 0.6 -sec process, during which tires deflect, struts compress, etc. Assume that the aircraft lift equals the aircraft weight during the touchdown.

A particle with a mass of \(4 \mathrm{kg}\) has a position vector in meters given by \(\mathbf{r}=3 t^{2} \mathbf{i}-2 t \mathbf{j}-3 \mathbf{k},\) where \(t\) is the time in seconds. For \(t=5\) s determine the angular momentum of the particle and the moment of all forces on the particle, both about the origin \(O\) of coordinates.
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