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a boeing 747 has a total mass, including passengers and luggage, of 250,000 kg. How much force do the engines supply to achieve a take off velocity of 990 m/s in 35 seconds.

1 Answer

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Acceleration = (change in speed) / (time for the change)

Acceleration = (990 m/s) / (35 sec) = 28.29 m/s²


Force = (mass) x (acceleration)

Force = (250,000 kg) x (28.29 m/s²) = 7.07 million Newtons

about 1.59 million pounds .

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There you have the Physics and the Math.
Does any of this resemble the real world ?
No. The question is completely bogus.

-- Take-off speed can't be 990 m/s.
That would be about 2,200 mph, almost 3 times the speed of sound.

-- The take-off run can't accelerate at 28.3 m/s² .
That would be almost 3 G's. If it didn't rip the wings off of the 747,
it would surely guarantee early use of most of the passengers' barf bags.

-- The Saturn V first-stage booster that sent Apollo to the moon
had a thrust of 7.5 million pounds. I'm pretty sure that was more
than 4.7 Boeing 747's.

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Ah hah ! I've just been handed a bulletin, or as they say nowadays,
there is breaking news !

I see that the take-off speed has been revised downward to 90 m/s.
That's right at 200 mph which seems reasonable to me, so let's go
back and do the whole thing again:


Acceleration = (change in speed) / (time for the change)

Acceleration = (90 m/s) / (35 sec) = 2.57 m/s² (about 0.26 G)


Force = (mass) x (acceleration)

Force = (250,000 kg) x (2.57 m/s²) = 643 thousand Newtons

about 144 thousand pounds.

about 1.9% of a Saturn V booster.

I like that a lot better.


User Diego Antunes
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