An airliner makes an emergency landing with its nose wheel locked in a position perpendicular to its normal rolling position. The forces acting to stop the airliner arise from friction due to the wheels and from the breaking effort of the engines in reverse thrust mode. The force of the engine on the plane is constant, Fengine=−F0. The sum of the horizontal forces on the airliner (in its forward direction) can be written as
F(t)=−F0+(tts−1)F1,(1)
from touchdown at time t=0 to the final stop at time ts= 28 s (0≤t≤ts). The mass of the plane is M=80 tonnes (one tonne is 1000 kg). We have F0= 296 kN and F1= 47 kN. Neglect all air drag and friction forces, except the one stated in the problem.
(a) Find the speed v0 of the plane at touchdown (in m/s).
v0=
(b) What is the horizontal acceleration of the plane at the time ts? What is the acceleration at the time of touchdown? (absolute values; in m/s2)
|a(ts)|=
|a(0)|=
(c) What distance s does the plane go between touchdown and its final stop at time ts? (in meters)
s=
(d) What work do the engines in reverse thrust mode do during the emergency landing?
(magnitude in Joules; the force due to engines is (−F0))
W=
(e) How much heat energy is absorbed by the wheels during the emergency landing?
(magnitude in Joules; the force due to wheels is (F(t)+F0))
Eheat=
To find the answers to these questions, we can use basic equations of motion. Let's break down each part step by step:
(a) To find the speed v0 of the plane at touchdown, we can use the equation:
v = u + at
Here, u is the initial velocity (which we assume to be 0 as the plane starts at rest), a is the acceleration, and t is the time (in seconds).
Since the force acting on the plane changes over time according to equation (1), the acceleration will also change. We can calculate the acceleration at any given time t using:
F(t) = ma
In this case, the force acting is given by equation (1), and the mass of the plane is provided as 80 tonnes, which is equal to 80,000 kg.
So, we can rewrite equation (1) as:
ma = −F0 + (tts−1)F1
Simplifying further:
a = −F0/M + (tts−1)F1/M
For the time of touchdown (t = 0), the acceleration is:
a(0) = −F0/M
Substituting the given values for F0 and M, we get:
a(0) = −296,000 N / 80,000 kg
Now we can use the equation v = u + at, where u = 0, t = 0, and a = a(0) to find the speed at touchdown (v0).
v0 = 0 + a(0) * 0
(b) To find the horizontal acceleration of the plane at the time ts, we can substitute t=ts into the equation for acceleration.
Substituting t=ts into equation (1) for acceleration:
a(ts) = −F0/M + (ts^2 - 1)F1/M
Then, take the absolute value of both a(ts) and a(0).
|a(ts)| = |−F0/M + (ts^2 - 1)F1/M|
|a(0)| = |−F0/M|
(c) To find the distance s that the plane travels between touchdown and its final stop at time ts, we can use the equation of motion:
s = ut + 0.5at^2
For this case, u = 0 (as the plane starts from rest), and a = a(t) (not a constant value). Integrating this equation with respect to time from t=0 to t=ts will give us the required distance s.
(d) To find the work done by the engines in reverse thrust mode during the emergency landing, we can use the work-energy principle. The work done is given by the equation:
W = ∫(−F0)dx
where dx is the displacement and F0 is the force.
To find the displacement, we can integrate the velocity equation v = u + at with respect to time from t=0 to t=ts.
(e) Similarly, to find the heat energy absorbed by the wheels during the emergency landing, we can calculate the work done by the friction force acting on the wheels using the same method as in part (d). The friction force is given by (F(t) + F0).
Following these steps, you should be able to find the values for v0, |a(ts)|, |a(0)|, s, W, and Eheat.