What is its speed 7.5s later? how far has it traveled in this time?

Question

a 747 jetliner lands and begins to slow as it moves along the runway. its mass is 3.5x10^5 kg, its speed is 27.0 m/s, and the net braking force is 4.30 x 10^5 N.

how many forces are acting on the the system if the system is the plane?

Answer 1

<< how many forces are acting on the the system if the system is the plane? >>

There are two types of forces acting on the plane.

Force 1 -- normal force being exerted by the runway on the airplane, and

Force 2 -- frictional force between the runway and the airplane and is directed opposite that of the airplane's direction.

<< what is its speed 7.5s later? >>

Before any other calculations can be done, the plane's acceleration needs to be determined. Using Newton's 2nd Law of Motion,

F = ma

where

F = braking force = 4.3 x 10^5 N

m = mass of the airplane = 3.5 x 10^5 kg

a = acceleration

Substituting values,

4.3 x 10^5 = 3.5 x 10^5(a)

and solving for "a",

a = 1.23 m/sec^2

The next working formula to use is

Vf - Vo = aT

where

Vf = airplane's velocity after 7.5 sec.

Vo = initial speed = 27 m/sec  (given)

a = acceleration = 1.23 m/sec^2 (as calculated above)

T = time = 7.5 sec (given)

Substituting appropriate values,

Vf - 27 = (-1,23)(7.5)

NOTE the negative sign attached to the acceleration. This simply implies that the airplane is slowing down since there was a braking force applied.

Solving for Vf,

Vf = 27 - 9.22

Vf = 17.78 m/sec.

<< how far has it traveled in this time? >>

Working formula is

S = VoT + (1/2)(aT^2

where

S = distance travelled by the plane in 7.5 sec

and all the other terms have been previously defined.

Therefore,

S = (27)(7.5) + (1/2)(-1.23)(7.5^2)

S = 167.91 meters