What is the position-vs-time graph for a falling object?

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SUMMARY

The position-vs-time graph for a falling object under the influence of a constant negative force, such as gravity, is a parabolic curve rather than a logarithmic or reciprocal function. When an object experiences a constant negative acceleration, like the gravitational force of -9.8 m/s², the distance covered increases quadratically over time. This is consistent with Newton's laws of motion, which dictate that the position of an object under constant acceleration is described by a quadratic equation. Therefore, the correct graph shape is a parabola opening upwards, starting from the origin.

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Familiarity with the concept of acceleration and its vector nature
  • Basic knowledge of graphing quadratic functions
  • Awareness of gravitational force and its effects on motion
NEXT STEPS
  • Study the mathematical representation of motion under constant acceleration
  • Learn how to derive position-time equations from velocity-time graphs
  • Explore real-world examples of free-fall motion and their graphical representations
  • Investigate the differences between positive and negative acceleration in various contexts
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Physics students, educators, and anyone interested in understanding the dynamics of motion under constant forces, particularly in the context of gravity and free fall.

HeLLz aNgeL
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quick question ... if i have constant negative force being applied to an object, what would the curve be on its position-time graph ...

im pretty sure that its got to go upwards since distance can't be reduced, so i was thinking it might be something like the graph of logx, but starting from the origin... is that correct ?

thanks !
 
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HeLLz aNgeL said:
quick question ... if i have constant negative force being applied to an object, what would the curve be on its position-time graph ...

im pretty sure that its got to go upwards since distance can't be reduced, so i was thinking it might be something like the graph of logx, but starting from the origin... is that correct ?

thanks !

Do you know of a PHYSICAL EXAMPLE of something experiencing a constant negative force?
Um...
 
Depends on what you mean by a negative force.
If the object is stationary and you supply a force in the negative x direction then it will move to the left of the origin.
If you mean the object is moving in a positive direction and you apply a force to slow it down then you will have a curve to the right of the origin.
 
robphy said:
Do you know of a PHYSICAL EXAMPLE of something experiencing a constant negative force?
Um...

lol ... didnt think of that... but i guess it would be in the opposite direction

heres the Force-time graph they gave me... what do you guys think will be the position-time graph for this ?
 

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bump... anyone ??
 
HeLLz aNgeL said:
bump... anyone ??

You still haven't offered a PHYSICAL EXAMPLE... a specific one, from real life.
The "Um..." is a hint that such an example is very easy to find... and that, once you come up with one, the answer to your question might become obvious.
 
i was thinking of something along the lines of a car slowing down ...

that would give me the same graph as logx but started from the origin ...
 
Closer to 1/x but you're in the right direction.
 
Suppose this force had a magnitude of say, um..., i dunno... 9.8 N...
 
  • #10
robphy said:
Suppose this force had a magnitude of say, um..., i dunno... 9.8 N...


o, now I am confused ... why would something in free fall have "negetive force" ? acceleration is positive in that case ...
 
  • #11
Acceleration is a vector.
The "acceleration due to gravity" (or better "the gravitational [vector] field") \vec g is downward pointing vector.
If vertically-upwards is "the direction of increasing y", then acceleration vector in that coordinate system has a negative component.
Thus, g_y = -9.8{\rm\ m/s^2} (or g_y = -9.8{\rm\ N/kg}) .

Physically, this means that the vertically-upward component of velocity is always decreasing.

There may be some misconceptions that you have.
To hopefully clear them up:
- the magnitude of the acceleration is always non-negative (i.e. positive or zero).
- the component of the acceleration depends on the choice of axis.. and thus may be either positive or negative, or else zero, depending on that choice of axis.
- with respect to this last point, sometimes one uses --arguably incorrectly-- the words "accelerating" and "decelerating" to describe whether the speed (i.e. the magnitude of the velocity, not [the component of] the velocity itself) is increasing or decreasing... what this really means is that the acceleration vector is either in the same direction as or opposite the velocity vector (i.e., according to the sign of \vec a \cdot \vec v).

In any case, this is a constant force problem... which via Newton's Law yields a constant acceleration problem. So, what is the general form of a constant-acceleration problem? (It's not log or 1/x or anything that complicated.)
 
  • #12
robphy said:
In any case, this is a constant force problem... which via Newton's Law yields a constant acceleration problem. So, what is the general form of a constant-acceleration problem? (It's not log or 1/x or anything that complicated.)

I'm ok with the fact that it is constant acceleration ...

however, i have confusion on the rest... I am guessing this is constant negetive acceleration, or deceleration ...

and if an object is decelerating, the distance it covers will gradually decrease with time, which leads me to my logx graph...

im looking for the position vs. time graph for this ...

is this explanation right ?
 
  • #13
anyone ? its just a x-t graph people !
 
  • #14
HeLLz aNgeL said:
its just a x-t graph people !

That's how I feel.

What is the position-vs-time graph of a free-falling object (the simplest example for your F-vs-t graph)?

If you are unsure, take a ball an throw it upwards, and study its motion.
If that's too fast, try Galileo's experiment:
http://www.pbs.org/wgbh/nova/galileo/expe_inpl_1b.html
 

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