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 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, its speed is 25.2 m/s, and the net braking force is 4.30×10^5 N, what is its speed 7.83 s later?

Madie Hilpert · Accepted Answer

Answer = 15.5 m/sGiven the values:m = 3.48E+5 kgVi = 25.2 m/sF = 4.3E+5 Nt = 7.83 sCalculate the acceleration of the aircraft as it slows, using Newton’s 2nd Lawa = F/ma = (4.3E+5 N) / (3.48E+5 kg)a = -1.24 m/s^2Use uniform acceleration laws to determine speedVf = Vi + atVf = (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

Madie Hilpert · Answer

so you have:m = 3.48×10^5 kgVi = 25.2 m/sFbrakes = -4.30×10^5 N (its negative because it being applied in the opposite direction of the direction the jetliner is travelling in)Δt = 7.83sso you know: F = mayou have F and m so find the acceleration of the jetliner:-4.30×10^5 = 3.48×10^5(a)a = -1.235m/s^2you can now find the Vf after 7.83s with this: Vf – Vi / Δt = aVf = (-1.235m/s^2)*(7.83s) + 25.2m/sVf = 15.52m/stherefore, 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 + ViΔtsub in everything you have and solve for d:d = (1/2)(-1.235)(7.83^2) + (25.2)(7.83)d = 159.46mtherefore, the jetliner travelled 159.46m after 7.83s.

Kelsie Sawayn

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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