Displacement as Function of Time Graph Question

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No, that is not it.
 
Ok, well thanks for the help. The assignment is now due.
 
It's this. That crunch cut it short didn't it.

Hand-drawn acceleration versus time graph with axes labelled t on the horizontal and 0 marked on the vertical; two rectangular blocks dip below the zero line, one short and one wider.
 
Also perhaps this could be the scenario. The crunch represents its change from negative acceleration to positive for a brief moment then back to zero.

Hand-drawn acceleration versus time graph titled 'Acceleration time graph', with a horizontal t axis and a vertical axis marked 0; a short segment dips below zero near the origin, a brief spike rises above zero, then a rectangular dip below zero appears later.
 
If the acceleration only lasted a short time, like you have drawn, then at the end of that short time the object would continue on with a steady speed until some acceleration did again act on it.
 
IS IT THIS?? This I think represents it because it can't instantly go to zero velocity right?

Hand-drawn acceleration versus time graph with the title "Acceleration time graph", a horizontal t axis, and a line labelled 0; the curve sits below zero, with a short sloped segment near the start and a rectangular dip later.
 
I think this is accurately representing it, i remember you saying something about the vertical line to the right. Its this isn't it?

Hand-drawn acceleration versus time graph with a horizontal t axis and a 0 marked on the vertical axis; the curve sits below zero, with a short slanted segment near the origin and a rectangular dip further right.
 
physicsnobrain said:
Also perhaps this could be the scenario. The crunch represents its change from negative acceleration to positive for a brief moment then back to zero.
You are getting the idea. A short sharp force acted to cause the object to lose all of its speed in a very brief moment. But it's not above the axis, that would cause it to rise up faster.
 
NascentOxygen said:
You are getting the idea. A short sharp force acted to cause the object to lose all of its speed in a very brief moment. But it's not above the axis, that would cause it to rise up faster.

So I remove the vertical line like you said before?
Hand-drawn acceleration versus time graph with a horizontal t axis and a vertical axis marked 0, showing a short segment just below zero, a rising step, and a rectangular dip below zero.
 
This is the graph I refer to. You show the spike of acceleration in the right place, but it should not be above the line, it should be below. You need it to act so it slows the object rapidly to stop it. Like bumping your head on an overhead steel beam! <crunch!>

Hand-drawn acceleration versus time graph titled 'Acceleration time graph', with a horizontal t axis and a line at 0; a short step below zero near the origin, a brief upward spike above zero, and a rectangular dip below zero later on.
 
NascentOxygen said:
This is the graph I refer to. You show the spike of acceleration in the right place, but it should not be above the line, it should be below. You need it to act so it slows the object rapidly to stop it. Like bumping your head on an overhead steel beam! <crunch!>

Hand-drawn acceleration versus time graph with axes labelled a (vertical) and t (horizontal), a zero line marked 0; a short step below zero near the origin, a brief spike above zero, and a rectangular block below zero later in time.


Ahhh, so its slow acting. SO like this:

Hand-drawn acceleration versus time graph with the title "Acceleration time graph", a horizontal t axis, and a 0 marked on the vertical axis. The curve sits below zero, with a small step near the start and a rectangular dip between roughly the middle and right of the plot.
 
physicsnobrain said:
Ahhh, so its slow acting. SO like this:
When you bump your head on an overhead beam, it doesn't feel like a gentle slow-acting stop to me!

Your graph is improving. Not far to go now ...
 
physicsnobrain said:
Ahhh, it must be this then.

Hand-drawn acceleration versus time graph with a horizontal axis labelled t and a horizontal line marked 0; the trace sits below zero with a brief dip and a rectangular step segment.
Not quite, but you are close. It's not as complicated as you show. But you have had enough help with this assignment that I think that should do for now. If you think about it more carefully, you might be able to get 100%. Post again if you come up with a better plot.

Remember that I said that vertical line on the extreme right is questionable, for the acceleration vs time plot, as I explained earlier. I would omit it.