How much acceleration does a 747 jumbo jet of mass 30000 kg experience in takeoff when the thrust for each of four engines is 30000n?

Answer 1

F = M x A

120,000/30,000 = 4

so it would be 4m/s2 (doesn't seem a lot but the figures aren't close 747 are more than 30 tons)

Answer 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).