An airplane taking off from a runway travels 3600 feet before lifting off. If it starts from rest, moves with constant acceleration, and makes the run in 30 seconds, with what speed does it lift off?
2007-12-08 07:23:59 · 4 answers · asked by mr_fic89 1 in Science & Mathematics ➔ Mathematics
a(t) = ct
v(t) = (c/2)t^2; constant of integration = 0 since at rest at time 0
d(t) = (c/6)t^3
3600 = (c/6)(30)^3
c = 6*3600/30^3 = 4/5
v(30) = 4/10(30)^2 = 360 mph
2007-12-08 07:55:30 · answer #1 · answered by J D 5 · 2⤊ 2⤋
s = ut + (1/2) a t ² ------------but u = 0
s = (1/2) a t ²
3600 = (1/2) a (30 ²)
a = 7200 / 900
a = 8 ft /sec ²
v = u + at
v = a t
v = 8 x 30
v = 240 ft/s at take off.
v = 240 x 3600 / 5280 mph
v = 164 mph
2007-12-08 08:11:50 · answer #2 · answered by Como 7 · 3⤊ 2⤋
v = ds/dt
a = dv/dt
v = at = C
v(0) = 0
v = at
s = (1/2)at^2 + C
s(0) = 0
v = at = 2s/t
v(30) = 2*3600/30 = 240 ft/s ≈ 163.6mph
2007-12-08 08:08:03 · answer #3 · answered by Helmut 7 · 1⤊ 1⤋
that is wrong broski, the velocity at 250m should be 94.494 m/s
2016-05-22 05:00:04 · answer #4 · answered by ? 3 · 0⤊ 0⤋