Answer:
The acceleration of the plane is -4.35 m/s²
Explanation:
Given;
initial speed of the airplane after landing, u = 91 m/s
final speed of the airplane after landing, v = 0
distance traveled, s = 952 m
Apply the following kinematic equation to determine the acceleration of the airplane;
v² = u² + 2as
0 = u² + 2as
0 = 91² + (2 x 952)a
0 = 8281 + 1904a
- 8281 = 1904a
a = - 8281 / 1904
a = -4.35 m/s²
Therefore, the acceleration of the plane is -4.35 m/s².
