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=
(a) To find the speed v0 of the plane at touchdown, we can equate the work done by the horizontal forces to the change in kinetic energy.
The work done by the engine force F(t) is given by the integral:
W = ∫ F(t) dx
Since F(t) = −F0 + (t/ts − 1)F1, we can rewrite the integral as:
W = ∫ [−F0 + (t/ts − 1)F1] dx
W = −F0x + F1∫ (t/ts − 1) dx
W = −F0x + F1[t*ln(t/ts) − t] + C
Where C is the constant of integration.
Since the work done on the plane is equal to the change in kinetic energy, we have:
W = ΔKE
ΔKE = (1/2)MV^2 − (1/2)Mv0^2
Where V is the final velocity of the plane after it stops.
At touchdown, the plane is assumed to be at rest, so v0 = 0.
Therefore, ΔKE = (1/2)MV^2
Substituting the values and rearranging the equation, we can solve for V:
−F0x + F1[t*ln(t/ts) − t] + C = (1/2)MV^2
Since the plane stops at time t = ts, we can further simplify the equation:
−F0s + F1[ts*ln(1) − ts] + C = (1/2)M(0^2)
−F0s − F1ts + C = 0
−F0s − F1ts = −C
Setting C = 0 (since it is an arbitrary constant), we can solve for s:
s = (-F1ts) / F0
Substituting the given values, we can find the value of s:
s = (-47 kN * 28 s) / 296 kN
Now we can calculate the value of s.
(b) The horizontal acceleration of the plane at the time ts can be determined from the equation F(t) = ma.
Since a = dV/dt, we can rewrite the equation as:
F(t) = m(dV/dt)
−F0 + (t/ts − 1)F1 = (80,000 kg)(dV/dt)
Simplifying the equation:
(t/ts − 1)F1 = (80,000 kg)(dV/dt) + F0
(t/ts − 1)F1 = (80,000 kg)(dV/dt) + (296 kN)
Since F1 = 47 kN, F0 = 296 kN, and ts = 28 s, we can plug in these values to calculate the acceleration:
(t/28 − 1)(47 kN) = (80,000 kg)(dV/dt) + 296 kN
Now, we can solve for the acceleration at the time ts.
To find the acceleration at the time of touchdown (t = 0), we can substitute t = 0 into the equation:
(0/28 − 1)(47 kN) = (80,000 kg)(dV/dt) + 296 kN
Simplifying this equation:
(-1)(47 kN) = (80,000 kg)(dV/dt) + 296 kN
Now, we can solve for the acceleration at the time of touchdown.
(c) To find the distance s that the plane goes between touchdown and its final stop at time ts, we can use the equation:
s = ∫ V dt
Since V = dV/dt, we can rewrite the equation as:
s = ∫ (dV/dt) dt
s = ∫ dV
s = V
Therefore, the distance s is the same as the final velocity V.
(d) The work done by the engines in reverse thrust mode during the emergency landing can be calculated using the equation:
W = ∫ F(t) dx
Since F(t) = −F0 + (t/ts − 1)F1, we can rewrite the integral as:
W = ∫ [−F0 + (t/ts − 1)F1] dx
W = −F0x + F1∫ (t/ts − 1) dx
W = −F0x + F1[t*ln(t/ts) − t] + C
Integrating this equation, we can calculate the work done by the engines.
(e) The heat energy absorbed by the wheels during the emergency landing can be calculated using the equation:
Eheat = ∫ (F(t) + F0) dx
Since F(t) = −F0 + (t/ts − 1)F1, we can rewrite the integral as:
Eheat = ∫ [−F0 + (t/ts − 1)F1 + F0] dx
Eheat = F1∫ (t/ts − 1) dx
Eheat = F1[t*ln(t/ts) − t] + C
Integrating this equation, we can calculate the heat energy absorbed by the wheels.
To solve this problem, we need to analyze the forces acting on the airliner during the emergency landing and apply Newton's laws of motion.
(a) To find the speed v0 of the plane at touchdown, we need to integrate the acceleration function over time:
v(t) = ∫ a(t) dt
Since the only force acting on the airliner is the engine force, we have:
a(t) = F(t) / M
Substituting the given values, we have:
a(t) = (-F0 + (t*ts - 1)*F1) / M
Integrating the above expression from time t = 0 to t = ts, we find:
v0 = ∫[0 to ts] a(t) dt
Solving this integral, we get:
v0 = (-F0 * ts / M) + (F1 * ts^2 / (2M)) - (F1 * ts / M)
v0 = (-296 kN * 28 s / 80,000 kg) + (47 kN * (28 s)^2 / (2 * 80,000 kg)) - (47 kN * 28 s / 80,000 kg)
Calculating the above expression, we find:
v0 = -11.24 m/s
Therefore, the speed of the plane at touchdown is 11.24 m/s.
(b) To find the horizontal acceleration of the plane at the time ts, we can directly substitute t = ts into the acceleration function:
a(ts) = (-F0 + (ts*ts - 1)*F1) / M
Calculating the above expression, we find:
a(ts) = (-296 kN + (28 s * 28 s - 1) * 47 kN) / 80,000 kg
Taking the absolute value, we get:
|a(ts)| = 10.91 m/s^2
Similarly, to find the acceleration at the time of touchdown (t = 0), we substitute t = 0 into the acceleration function:
a(0) = (-F0 + (0*ts - 1)*F1) / M
Calculating the above expression, we find:
a(0) = (-296 kN + (-1) * 47 kN) / 80,000 kg
Taking the absolute value, we get:
|a(0)| = 4.40 m/s^2
Therefore, the horizontal acceleration of the plane at the time ts is 10.91 m/s^2 and the acceleration at the time of touchdown is 4.40 m/s^2.
(c) To find the distance the plane goes between touchdown and its final stop at time ts, we need to calculate the displacement. Displacement is equal to the area under the velocity-time graph. Since the velocity is changing linearly with time, we can use the equation:
s = (v0 + v(ts)) / 2 * ts
Substituting the given values, we have:
s = (-11.24 m/s + 0 m/s) / 2 * 28 s
Calculating the above expression, we find:
s = -157.92 m
However, displacement cannot be negative in this context, so we take the absolute value:
s = 157.92 m
Therefore, the plane goes a distance of 157.92 meters between touchdown and its final stop at time ts.
(d) To find the work done by the engines in reverse thrust mode during the emergency landing, we need to calculate the work-energy principle. The work done by a constant force is given by the equation:
W = F * d
Since the force due to the engines is (-F0), and the displacement is equal to the distance s calculated in part (c), we have:
W = (-F0) * s
Substituting the given values, we have:
W = (-296 kN) * 157.92 m
Calculating the above expression, we find:
W = -46,694.92 kJ
Therefore, the work done by the engines in reverse thrust mode during the emergency landing is -46,694.92 kJ (negative sign indicates work done against the motion).
(e) To find the heat energy absorbed by the wheels during the emergency landing, we need to calculate the work done by the force due to the wheels. The force due to the wheels is (F(t) + F0), and the displacement is equal to the distance s calculated in part (c).
Eheat = (F(t) + F0) * s
Substituting the given values, we have:
Eheat = (F(t) + F0) * s
Calculating the above expression, we find:
Eheat = (F(t) + F0) * s
Therefore, the heat energy absorbed by the wheels during the emergency landing is (F(t) + F0) * s.