Velocity vs Time Graph: Does Resultant Force Follow y=1/x?

[nokia8650]

On a graph such as below:

http://www.gcsescience.com/Velocity-Time-Graph-Rocket.gif [Broken]

Would the graph of resultant force vs. time be of a shape similar to the curve of "y=1/x" in the positive quadrant? I make this assumption, since the gradient of the v vs. t graph is decreasing, therefore the acceleration is decreasing. I also think that the rate of change of acceleration is decreasing.

Thanks

 

[rosstheboss23]

No, 1/X is curve that starts with a high y value and decreases quickly and then levels off from left to right. Look at the slope...in the v vs. t graph velocity is increasing over time because it is in the positive quadrant. Acceleration is increasing a lot at first and is leveling off at a positive number based on the slope of the graph.

 

[nokia8650]

What would be the shape of the graph then? I think it would be a curve, but don't know the shape.

Thanks

 

[rosstheboss23]

The graph shown is a log x graph. The graph of its slope, aka the acceleration, will start positive maybe (0, 2) then will decrease and level off until it is pratically zero at every value after a certain time. This being the derivative would be the (1/X) curve, but for the derivative not the graph shown...Well guess that does it for this problem. Hope that helps. Ross

 

[Phlogistonian]

nokia8650,

Your original answer is correct. The only criticism I have is that the slope doesn't appear to me to be infinite at the origin.

 

[Phlogistonian]

And it's not a log graph.

 

[nokia8650]

Thanks, yes, I only meant the shape of it, as it is hard to describe! Thanks alot