Position vs. Time graph

HeLLz aNgeL

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 cant be reduced, so i was thinking it might be something like the graph of logx, but starting from the origin... is that correct ?

thanks !

robphy

Do you know of a PHYSICAL EXAMPLE of something experiencing a constant negative force?
Um....

mgb_phys

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.

HeLLz aNgeL

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 ?

Attached Thumbnails

 

HeLLz aNgeL

bump... anyone ??

robphy

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.

HeLLz aNgeL

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 ...

mgb_phys

Closer to 1/x but you're in the right direction.

robphy

Suppose this force had a magnitude of say, um..., i dunno... 9.8 N...

HeLLz aNgeL

o, now im confused .... why would something in free fall have "negetive force" ??? acceleration is positive in that case ....

robphy

Acceleration is a vector.
The "acceleration due to gravity" (or better "the gravitational [vector] field") [tex]\vec g[/tex] 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, [tex]g_y = -9.8{\rm\ m/s^2}[/tex] (or [tex]g_y = -9.8{\rm\ N/kg})[/tex] .

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 [tex]\vec a \cdot \vec v[/tex]).

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.)

HeLLz aNgeL

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

however, i have confusion on the rest... im 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 ???

HeLLz aNgeL

anyone ???? its just a x-t graph people !!!

robphy

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