Asked by Mike Rusho
A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is 3.42×105 kg ,its speed is 29.5 m/s,and the net braking force is 4.3*10^5 N.
What is its speed 8.33 s later?
How far has it traveled in this time?
How exactly do I do this? What procedure and equations do I need?
What is its speed 8.33 s later?
How far has it traveled in this time?
How exactly do I do this? What procedure and equations do I need?
Answers
Answered by
Damon
F = m a
-4.3 * 10^5 = 3.42 * 10^5 a
the - sign is because it is braking, not speeding up.
so
a = -4.3/3.42
then
v = Vo + a t where Vo =29.5
v = 29.5 - a * 8.33
and
x = Xo + Vo t + (1/2) a t^2
distance traveled = x - Xo
= 29.5 (8.33) - (1/2)(4.3/3.42)(8.33)^2
-4.3 * 10^5 = 3.42 * 10^5 a
the - sign is because it is braking, not speeding up.
so
a = -4.3/3.42
then
v = Vo + a t where Vo =29.5
v = 29.5 - a * 8.33
and
x = Xo + Vo t + (1/2) a t^2
distance traveled = x - Xo
= 29.5 (8.33) - (1/2)(4.3/3.42)(8.33)^2
Answered by
Mike Rusho
Thanks you helped me so much!
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.