Question

2/49 on its takeoff roll, the airplane starts from rest and accelerates according to a=a0−kv2, where a0 is the constant acceleration resulting from the engine thrust and −kv2 is the acceleration due to aerodynamic drag. if a0=2 m/s2, k=0.00004 m−1, and v is in meters per second, determine the design length of runway required for the airplane to reach the takeoff speed of 250 km/h if the drag term is (a) excluded and (b) included.

Answer 1

The length of the runway required when (a) drag force is not taken into account is 1.21 km, (b) the drag force is taken into account is 1.33 km.

(a) Express the take off speed in m/s.

$v= (250km/h)(\frac{1000 m/km}{3600 s/h}) =69.44 m/s$

Use the equation of motion,

$v^2=u^2+2as$

Substitute 0 m/s for u, 69.44 m/s for v and 2m/s²for a.

$v^2=u^2+2as\\ (69.44m/s)^2=(0m/s)^2+2(2m/s^2)s\\ s=\frac{(69.44m/s)^2}{4 m/s^2} =1205.5 m\\ =(1205.5 m)(\frac{1 km}{1000 m} )=1.205 km$

The length of the runway required when no drag force acts on the airplane is 1.21 km.

(b) When the drag force acts, the acceleration is given by,

$a=a_0-kv^2$

calculate the acceleration by substituting 2 m/s² for a₀, 0.00004/m for k and 69.44 m/s for v.

$a=a_0-kv^2\\ =(2m/s^2)-(0.00004m^-^1)(69.44 m/s)^2\\ =1.807 m/s^2$

Use the equation of motion,

$v^2=u^2+2as$

Substitute 0 m/s for u, 69.44 m/s for v and 1.807m/s²for a.

$v^2=u^2+2as\\ (69.44m/s)^2=(0m/s)^2+2(1.807 m/s^2)s\\ s=\frac{(69.44 m/s)^2}{2(1.807m/s^2)} =1334.2 m\\ =(1334.2 m)(\frac{1km}{1000m} )\\ =1.334km$

When the acceleration of the plane is reduced by air drag, the length of the runway needed is 1.33 km.