A plane is traveling at a velocity of 90 m/s. It accelerates at a constant rate of 1.5 m/s2 until its velocity reaches 500 m/s. What distance did the plane cover while it was accelerating?

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

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.