Question

While taking off, the net external force acting on an airplane during the ground roll segment can be assumed to be constant. The airplane starts from rest. \( S_{LO} \) and \( V_{LO} \) are the ground roll distance and the lift-off speed, respectively. \( \alpha V_{LO} \) (\( \alpha>0 \)) denotes the airplane speed at 0.5 \( S_{LO} \). Neglecting changes in the airplane mass during the ground roll segment, the value of \( \alpha \) is \_\_\_\_\_ (rounded off to two decimal places).

Hint:

To solve this problem, apply the kinematic equations for motion with constant acceleration and use the relationship between speed and distance during the ground roll segment.

Answer 1

From the kinematic equation for motion under constant acceleration, the relationship between distance and speed is: \[ V^2 = 2aS \] where:
\( V \) is the velocity,
\( a \) is the acceleration,
\( S \) is the distance traveled.
The airplane starts from rest, so the equation simplifies for \( S_{LO} \) and \( V_{LO} \) (at lift-off) as: \[ V_{LO}^2 = 2a S_{LO} \] The speed at \( 0.5 S_{LO} \) is \( \alpha V_{LO} \), and using the same relationship for distance and speed: \[ (\alpha V_{LO})^2 = 2a \times 0.5 S_{LO} \] Solving these equations: \[ \alpha^2 = 0.5 \quad \Rightarrow \quad \alpha = \sqrt{0.5} \approx 0.71 \] Thus, the value of \( \alpha \) is approximately 0.71.