An airplane lands with a speed of 91 m/s traveling due west. It comes to rest in 952 m with a constant deceleration. What is the plane’s acceleration?

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.

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