Final answer:
The rate at which a 747 jetliner coasts to a stop after landing, which is a deceleration, can be calculated using Newton's second law. The deceleration, given a net braking force of 4.3 × 10^5 N and a mass of 3.5 × 10^5 kg, is 1.23 m/s^2.
Step-by-step explanation:
The question asks about the rate at which a 747 jetliner coasts to a stop after landing. The rate at which the jetliner comes to a stop can be understood as the deceleration of the airplane, which can be calculated using Newton's second law of motion. Given that the net braking force is 4.3 × 105 N and the mass of the jetliner is 3.5 × 105 kg, the deceleration can be calculated by dividing the net force by the mass (a = Fnet / m).
To find the rate of deceleration, apply the formula to the given values:
a = Fnet / m
a = (4.3 × 105 N) / (3.5 × 105 kg)
a = 1.23 m/s2
The rate of deceleration is 1.23 m/s2, which means the jetliner slows down at this rate until it comes to a complete stop.