An object moves in one dimensional motion with constant acceleration a = 4.2 m/s^2. at time t = 0 s, the object is at x0 = 3.9 m and has an initial velocity of v0 = 2.2 m/s. how far will the object move before it achieves a velocity of v = 7.2 m/s? your answer should be accurate to the nearest 0.1 m.

The object will move to Xfinal = 7.5m

Explanation:

By relating the final velocity of the object and its acceleration, I can obtain the time required to reach this velocity point:

Vf= a × t ⇒ t= (7.2 m/s) / (4.2( m/s^2)) = 1,7143 s

With the equation of the total space traveled and the previously determined time I can obtain the end point of the object on the x-axis:

Xfinal= X0 + /1/2) × a × (t^2) = 3.9m + (1/2) × 4.2( m/s^2) × ((1,7143 s) ^2) =

= 3.9m + 3.6m = 7.5m

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .