To lift off the surface of the earth, the aircraft accelerates from rest to a speed of 200 km / h.

To lift off the surface of the earth, the aircraft accelerates from rest to a speed of 200 km / h. The thrust force of the engines is 5000 N, the force of air resistance and the force of friction together are 1500 N. The mass of the aircraft is 6000 kg. Determine the length and time of the plane’s movement on the ground.

Given:

v = 200 km / h = 51.3 m / s – the speed to which the aircraft accelerates from rest to lift off from the earth’s surface;

F1 = 5000 Newton – engine thrust force;

F2 = 1500 Newton – the sum of the air resistance force and the friction force;

m = 6000 kilograms – aircraft mass.

It is required to determine l (meter) – the length and t (seconds) – the time the aircraft moves on the ground.

Let us find the resultant of all forces acting on the plane as it moves along the ground:

F = F1 – F2 = 5000 – 1500 = 3500 Newton.

Then the aircraft acceleration will be equal to:

a = F / m = 3500/6000 = 35/60 = 0.6 m / s2 (the result has been rounded to one decimal place).

Since, according to the condition of the problem, the plane was accelerated from a state of rest, the movement time will be equal to:

t = v / a = 51.3 / 0.6 = 85.5 seconds.

The path will be:

S = a * t ^ 2/2 = 0.6 * 85.5 ^ 2/2 = 0.6 * 7310/2 = 2193 meters.

Answer: the plane accelerated in 85.5 seconds and covered a distance of 2193 meters.