2007-09-07 05:05:29 · 4 answers · asked by Riaz J 1 in Science & Mathematics ➔ Engineering
Initial velocity = 0 ft / sec.
Distance = 2900 ft.
Assume constant acceleration until take-off.
d = ½ a t²
a = 2d ÷ t²
a = (2 x 2900) ÷ 35²
a = 5800 ft ÷ 1225 sec² = 4.73 ft / sec²
a = Δv ÷ Δt
a = (Vb - 0) ÷ 35
a = Vb ÷ 35
Vb = 35a (substitute the acceleration computed above)
Vb = 35 * 4.73 = 165 ft / sec = 112 MPH.
This number looks a bit low. If I was on board, I'd be hoping the 747 was nearly empty and that it's really cold outside (i.e. dense air).
2007-09-07 05:29:19 · answer #1 · answered by Thomas C 6 · 2⤊ 0⤋
s=vo*t+(1/2)*a*t^2,vo=0 since it started from rest
s=1/2*a*t^2
2900=(1/2)*a*(35^2)
a=2(2900)/35^2
a= 4.7346938775510204081632653061224 feet per second squared
a=(vf-vo)/t;vo=0 since it started from rest
a=vf/t
solve for vf
vf=a*t=165.71428571428571428571428571429 feet per second
2007-09-09 09:11:16 · answer #2 · answered by ptolemy862000 4 · 0⤊ 0⤋
supply the Vr, and we'll talk. OBTW, Vr is "Velocity of rotation", the speed required for the airctaft to be ABLE to "rotate", that is, point its nose skyward and fly.
2007-09-07 12:20:07 · answer #3 · answered by Stephen H 5 · 0⤊ 1⤋
Please, I don't know what is the speed needed for a 747 for take off, you need this
2007-09-07 12:18:49 · answer #4 · answered by Jorge R 6 · 0⤊ 1⤋