Answer:
The distance the plane covered while it was accelerating is 80,633.3 m
Explanation:
Given;
initial velocity of the plane, u = 90 m/s
acceleration of the plane, a = 1.5 m/s²
final velocity of the plane, v = 500 m/s
The distance covered by the plane is given as;
v² = u² + 2ad
where;
d is the distance covered by the plane;
500² = 90² + 2(1.5)d
500² - 90² = 3d
241900 = 3d
d = 241900 / 3
d = 80,633.3 m
Therefore, the distance the plane covered while it was accelerating is 80,633.3 m
