A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is 3.48×10^5 kg, i?

Question

A Boeing 747 jetliner lands and begins to decelerate as it moves along the runway. Given that its mass is 3.48 × 10^5 kg, its initial speed is 25.2 m/s, and the net braking force acting on it is 4.30 × 10^5 N, what will be its speed 7.83 seconds later?

Anonymous · Accepted Answer

Answer = 15.5 m/s
Given the values:
m = 3.48E+5 kg
Vi = 25.2 m/s
F = 4.3E+5 N
t = 7.83 s
Calculate the acceleration of the aircraft as it slows, using Newton’s 2nd Law
a = F/m
a = (4.3E+5 N) / (3.48E+5 kg)
a = -1.24 m/s^2
Use uniform acceleration laws to determine speed
Vf = Vi + at
Vf = (25.2 m/s) + [ (-1.24 m/s^2) * (7.83 s) ]
Vf = (25.2 m/s) + [ -9.675 m/s ]
Vf = 15.5 m/s

Anonymous · Answer

so you have:
m = 3.48×10^5 kg
Vi = 25.2 m/s
Fbrakes = -4.30×10^5 N (its negative because it being applied in the opposite direction of the direction the jetliner is travelling in)
Δt = 7.83s
so you know: F = ma
you have F and m so find the acceleration of the jetliner:
-4.30×10^5 = 3.48×10^5(a)
a = -1.235m/s^2
you can now find the Vf after 7.83s with this: Vf – Vi / Δt = a
Vf = (-1.235m/s^2)*(7.83s) + 25.2m/s
Vf = 15.52m/s
therefore, the speed after 7.83s is 15.52m/s.
now you know the acceleration of the jetliner. Now use the eq’n:
d = 1/2aΔt^2 + ViΔt
sub in everything you have and solve for d:
d = (1/2)(-1.235)(7.83^2) + (25.2)(7.83)
d = 159.46m
therefore, the jetliner travelled 159.46m after 7.83s.

Nash Gleason

A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is 3.48×10^5 kg, i?

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