Can High School Physics Explain Motion with Constant Acceleration and Velocity?

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The discussion focuses on the application of high school physics principles to motion with constant acceleration and velocity. The initial question concerns the z-velocity of an object under constant acceleration, which is clarified to depend on the net acceleration after accounting for gravity. It is emphasized that one must consider the net force to determine the unique acceleration acting on the object. Additionally, the z-position of an object moving at constant velocity does not include a gravitational term, as that scenario is distinct from one involving acceleration. The conversation highlights the importance of correctly interpreting forces and accelerations in physics problems.
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Okay, this is high school physics I know but I still need to ask it.

Say I give something a constant acceleration A straight up into the air (A = [0,0,+Az]). If I ignore drag, then the object's z-velocity as a function of time with this constant acceleration will be

(Az - g)t

(where Az is the z-component of the acceleration) if the initial z-velocity at time t = 0 is 0. Am I correct?

Next question: say that an object is traveling straight up through the air with a constant velocity V0. Is the object's z-position as a function of time equal to

V0t - 0.5gt^2 + Z0

(where Z0 is the initial z position) or is it something else?

Sorry if I didn't word these questions right - I'm still very new to this.
 
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If the object were traveling straight up at a constant velocity you wouldn't have the 0.5gt^2 term, this case would be equivilant to having Az=g in your previous example.
 
Signifier said:
Okay, this is high school physics I know but I still need to ask it.

Say I give something a constant acceleration A straight up into the air (A = [0,0,+Az]). If I ignore drag, then the object's z-velocity as a function of time with this constant acceleration will be

(Az - g)t

(where Az is the z-component of the acceleration) if the initial z-velocity at time t = 0 is 0. Am I correct?

No.

Don`t confuse forces with acceleration. It`s confusing when you say that you ''give something a constant acceleration Az''. If this is really what you mean, then you mean that you applied a force stronger than gravity in such a way that the net acceleration is Az (which I assume positive from your statement ''staright up in the air''). Then the velocity is Az t! Period!
An object has only one acceleration, given by the net force divided by the mass. One adds the forces *first* and then one finds the (unique) acceleration. One never adds acceleration. So if you say that you give an object an acceleration Az, it means you applied a force such that the combination of your force and of the force of gravity yields the acceleration Az. Then v = Az t.

Pat
 
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