An airplane travels a distance 288 m down the runway before taking off. if it starts from rest, moves with constant acceleration, and becomes airborne in time 8.50 s, what is its speed when it takes off?
2006-09-22 16:54:36 · 4 answers · asked by Theresa C 2 in Science & Mathematics ➔ Physics
Use these equations.
average velocity = distance / time = (vfinal + vinitial) /2
vfinal = vinitial + ( acc * time )
Take note tht these only work for constant acceleration problems (which is the case here)
the final answer should be 67.76 m/s
2006-09-22 17:09:05 · answer #1 · answered by dennis_d_wurm 4 · 0⤊ 0⤋
You have the following:
t = 8.50 seconds
x0 = 0 m (initial position)
v0 = 0 m/s (initial velocity)
xt = 288 m (position at time t)
vt = ? m/s (velocity at time t, which is unknown)
a = some constant.
Use the equation
xt = (.5a)t^2 + (v0)t + x0
to find a (you know all the other values).
Once you have a, use the eqaution
vt = v0 + a*t
to find vt.
2006-09-23 00:10:36 · answer #2 · answered by Anonymous · 0⤊ 0⤋
There is not enough information to answer your question because it is not stated how fast the plane is moving when it first becomes air borne.
2006-09-23 00:16:47 · answer #3 · answered by bruinfan 7 · 0⤊ 0⤋
s(t) = t*v(t) + 0.5.a*t^2
t=8.5, s(8.5) = 288, v(8.5) you have to calculate
a is constant
You have two unknowns : v and a so you cant solve you can only express v as function of a
2006-09-23 01:03:01 · answer #4 · answered by gjmb1960 7 · 0⤊ 0⤋