A cheetah is crouched 20 m to the east of an observer (Fig. 2.6a). At time the cheetah begins to run due east toward an antelope that is 50 m to the east of the observer. During the first 2.0 s of the attack, the cheetah’s coordinate x varies with time according to the equation (a) Find the cheetah’s displacement between and (b) Find its average velocity during that interval. (c) Find its instantaneous velocity at by taking then then (d) ¢t = 0.1 s, 0.01 s, 0.001 s. Derive an t1 = 1.0 s t t 2 = 2.0 s. 1 = 1.0 s 15.0 m>s2 2t 2 x = 20 m + . t = 0 expression for the cheetah’s instantaneous velocity as a function of time, and use it to find at t=1.0s and t=2.0s

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Explanation:

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.