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.

Question

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?

Answer 1

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

Answer 2

Thanks you helped me so much!