Middle school? lol but, the third one.
The answer is:
Both the distance traveled in a given time and the magnitude of the acceleration at a given instant
Hope I Helped!
solution:
y = v0t + ½at²
1150 = 79t + ½3.9t²
0 = 3.9t² + 158t - 2300
from quadratic equations and eliminating the negative answer
t = (-158 + v158² -4(3.9)(-2300)) / 2(3.9)
t = 11.37 s to engine cut-off
the velocity at that time is
v = v0 + at
v = 79 + 3.9(11.37)
v = 123.3 m/s
it rises for an additional time
v = gt
t = v/g
t = 123.3 / 9.8
t = 12.59 s
gaining more altitude
y = ½vt
y = 123.3(12.59) /2
y = 776 m
for a peak height of
y = 776 + 1150
Um it's so dark here I can't find the question lol...
Magnitude of acceleration = (change of speed) / (time for the change) =
(12 m/s - 0) / (3 sec) =
12/3 = <em>4 m/s²</em>
What's a challenge question ? Have we all passed the event horizon
and been spaghettified without knowing it ?