Final answer:
The 747 jumbo jet experiences an acceleration of approximately 3.43 m/s² during takeoff when the thrust for each of its four engines is 5999 N. This is calculated using Newton's second law of motion, with the total thrust being 23996 N.
Step-by-step explanation:
The acceleration a jumbo 747 jet experiences during takeoff can be calculated using Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and its acceleration (F = ma). Here, we need to find the total thrust provided by all four engines and then use this to find the acceleration.
The total thrust (F) is the sum of the thrust from each of the four engines. Therefore, F = 4 × 5999 N = 23996 N. With the mass (m) given as 6999 kg, the acceleration (a) can be calculated using the formula a = F / m.
Acceleration (a) = Total thrust (F) / Mass (m) = 23996 N / 6999 kg = approximately 3.43 m/s².
Therefore the 747 jumbo jet experiences an acceleration of approximately 3.43 m/s² during takeoff, ignoring air resistance and other counteracting forces.