Ignoring air resistance, (although we know that
that's definitely not a valid assumption when we're dealing with an airliner! )
Total force = 4 x 30,000 = 120,000 newtons
F = M A
A = F / M = (120,000) / (30,000) = 4m/sec2 = about 0.41 of a 'G' .
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In actuality a fully loaded B747-200 airliner has a maximum takeoff weight of 990,000 Lbs which is 450,000 Kg mass.
Each General Electric CF6 Turbofan jet engine on the B747-200 has a maximum static thrust output of 63,300 Lbs which is equal 281,685 Newtons of thrust output per engine.
There are four (4) engines on the B747-200. Therefore at maximum static thrust output power there will be 1,126,740 Newtons of maximum static thrust on the Boeing 747-200 airliner.
The Drag Coefficient (Cd), Aircraft Projected Frontal Area (A), Air Density in Slugs per Cubic Foot (P) and Airspeed (V) are factors which when all combined give the Aerodynamic Drag which resists forward acceleration and counteracts thrust. Tire rolling resistance also resists forward acceleration as well when the airliner is rolling on takeoff during the ground.
But let's ignore Aerodynamic Drag & Tire Rolling Resistance in this case and focus purely on the acceleration that is incurred on this airliner at the given thrust output during runway roll acceleration only.
Let say a Boeing 747-200 airliner's acceleration is to be calculated during it takeoff roll with the given mass & static thrust information from rest (0 MPH) to 170 MPH within 30 seconds of runway roll time.
mass = 450,000 kg
force (thrust) = 1,126,740 N
Acceleration = [(Force) / (Mass)]
Acceleration = [(Distance) / (Time x Time)]
Acceleration = [(Speed / Time)]
1 MPH = 1.467 feet per second
1 meter = 3.28 feet
170 MPH x 1.467 = 249.39 feet per second
249.39 feet per second / 3.28 = 76 meters/sec
Acceleration = [(76 meters per second) / (30 seconds)] = 2.5 meters per second squared.
Acceleration = [(1,126,740 Newtons) / (450,000 Kg)] = 2.5 meters per second squared.
The acceleration is calculated at: 2.5 m/s^2
The horizontal G-Force is a ratio between the horizontal acceleration & gravitational acceleration plus a reference Horizontal G Force of 1.0.
The gravitational acceleration on Earth is: 9.8 m/s^2.
Horizontal G-Force during runway roll acceleration: [(1) + (2.5 / 9.8)] = +1.25 Horizontal Gs'
The Horizontal G-Force is calculated at: +1.25 Horizontal Gs'
During cruise flight the airliner's speed is held at a relatively constant rate so therefore the acceleration is zero.
Acceleration is a substantial factor on an airliner during runway takeoff roll more than any other time. Even during landing procedures, there is a gradual deceleration but not the extent of the takeoff acceleration where the airliner moves from 0 MPH to 170 MPH (0 m/s to 76 m/s) in a time period of 30 seconds!
An airplane accelerates due to the thrust generated by its engines. As the engines produce forward thrust, the aircraft gains speed. The thrust must overcome drag forces acting on the airplane to achieve acceleration.
i guess when the engines are set to full thrust, and when the plane leaves the ground...
To find the acceleration of the 747 jet during takeoff, we need to calculate the total thrust generated by all four engines. The total thrust is 4 times the thrust of each engine, which equals 120,000 N. Then, we can use Newton's second law (F=ma) to calculate the acceleration, which is 120,000 N divided by the mass of the plane (29,000 kg), resulting in an acceleration of approximately 4.14 m/s^2.
To find the acceleration, we first need to calculate the total thrust produced by all four engines. Since each engine produces 30,000 N of thrust, the total thrust is 4 * 30,000 = 120,000 N. The acceleration can be calculated using Newton's second law, F = ma, where F is the total thrust (120,000 N) and m is the mass of the jet (29,600 kg). Plugging in the values, we get a = 120,000 N / 29,600 kg ≈ 4.05 m/s^2.
The total thrust produced by all four engines is 120,000 N (30,000 N x 4). To calculate acceleration, we use Newton's second law: acceleration = total force / mass. Plugging in the numbers, the acceleration experienced during takeoff is 4 m/s^2 (120,000 N / 30,000 kg).
An airplane accelerates due to the thrust generated by its engines. As the engines produce forward thrust, the aircraft gains speed. The thrust must overcome drag forces acting on the airplane to achieve acceleration.
i guess when the engines are set to full thrust, and when the plane leaves the ground...
To find the acceleration of the 747 jet during takeoff, we need to calculate the total thrust generated by all four engines. The total thrust is 4 times the thrust of each engine, which equals 120,000 N. Then, we can use Newton's second law (F=ma) to calculate the acceleration, which is 120,000 N divided by the mass of the plane (29,000 kg), resulting in an acceleration of approximately 4.14 m/s^2.
To find the acceleration, we first need to calculate the total thrust produced by all four engines. Since each engine produces 30,000 N of thrust, the total thrust is 4 * 30,000 = 120,000 N. The acceleration can be calculated using Newton's second law, F = ma, where F is the total thrust (120,000 N) and m is the mass of the jet (29,600 kg). Plugging in the values, we get a = 120,000 N / 29,600 kg ≈ 4.05 m/s^2.
Solid fueled rocket engines are relatively simple in design, have a high thrust-to-weight ratio, and are easy to store and transport. Additionally, they can provide a lot of thrust in a short amount of time, making them ideal for quick acceleration needs.
40 m/s2
The total thrust produced by all four engines is 120,000 N (30,000 N x 4). To calculate acceleration, we use Newton's second law: acceleration = total force / mass. Plugging in the numbers, the acceleration experienced during takeoff is 4 m/s^2 (120,000 N / 30,000 kg).
The thrust is an airplane is provided by the engines.
F = M AA = F / MUntil air-resistance begins to have a significant influence . . .A = (4 x 30,000)/(30,000) = 4 meters/sec2 = roughly 0.4 G
Thrust acceleration is the rate at which the speed of an object changes due to the application of thrust, which is the force exerted by an engine or a rocket. It is a measure of how quickly the object's velocity increases in response to the thrust force being applied to it.
One major drawback is the limited thrust that electric rocket engines can provide compared to chemical rockets, which can hinder their acceleration and overall performance in certain mission scenarios.
Thrust is the main propulsion medium. The jet engines produce the thrust (the force forward) the air moves across the top of the wings creating a vacuum which in effect hold the aircraft in flight. The greater the engines are worked the greater the thrust and therefore speed.