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+((t/ts)-1)F1, (1)
touchdown at time t=0 to the final stop at time ts=26s (0 ≤t ≤ts). The mass of the plane is M=80 tonnes (one tonne is 1000kg). We have F0=268 kN and F1=42 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/s^2)
|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=
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