A plane has a take off speed of 1 point 300 km/h. What is the
acceleration in m/s2 of the
plane if the plane started from
rest and took 45 seconds to
take off? *​

Acceleration = 1.85m/s²

Explanation:

Given the following data;

Final velocity = 300km/h to m/s = 300*1000/3600 = 83.33m/s

Time = 45 seconds

Since the plane started from rest, initial velocity is equal to 0m/s.

To find the acceleration;

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Mathematically, acceleration is given by the equation;

Where,

a is acceleration measured in

v and u is final and initial velocity respectively, measured in

t is time measured in seconds.

Substituting into the equation, we have;

Acceleration = (83.33 - 0)/45

Acceleration = 83.33/45

Acceleration = 1.85m/s²

Therefore, the acceleration of the plane is 1.85m/s².

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.