A jetliner has a gross weight of 250,000 kg. On its takeoff roll it accelerates from rest to 50m/s in a distance of 3000m. Assume constant acceleration and no friction. (Hint: one part of this problem can’t be solved with energy methods!)
What is the kinetic energy of the airliner after the takeoff roll?
Determine the force of thrust of the aircraft’s engines.
Determine the time required to the acceleration to occur.
Determine the power output of the aircraft’s engines.
1. The kinetic energy is by definition
$Ec = m*v^2/2$ (m is the mass, v is the speed)
$Ec = 25*10^4*50*50/2 =3.125*10^8$ Joule
2. The acceleration (a) of the jetliner comes from
$v^2 = 2*a*d$, (d is the distance)
$a =v^2/2/d =50*50/2/3000 =0.4167 m/s^2$
The Force of thrust comes from the second principle
$F = m*a =250000*0.4167 =104166.7 N$
3. Considering the acceleration constant, the time (t) comes from the equation
$v = 0+a*t$
$t = v/a = 50/0.4167 =120 seconds$
4. The power output (P) at the takeoff (speed v) is
$P = F*v = 104166.7*50 =5208335 W = 7081.4 HP$