Question

A 747 jetliner lands and begins to slow to a stop as it moves along the runway. if its mass is 3.5 x 10⁵ kg, its speed is 27 m/s. and the net braking force is 4.3 x 10⁵ n.

(a) what is its speed 7.5s later?

Answer 1

To find the final velocity of the jetliner 7.5 seconds later, use the equation final velocity = initial velocity + (acceleration x time).

Step-by-step explanation:

To find the final velocity of the jetliner 7.5 seconds later, we can use the equation:

final velocity = initial velocity + (acceleration x time)

Given that the mass of the jetliner is 3.5 x 10⁵ kg, its initial velocity is 27 m/s, and the net braking force is 4.3 x 10⁵ N, we can calculate the deceleration using Newton's second law:

deceleration = net force / mass

Once we have the deceleration, we can substitute it into the equation to find the final velocity 7.5 seconds later.