Position-Time graph vs. Velocity-Time graph

[BloodyMinded]

The page layout is included in the picture attachments which are in order.
So far, if I'm correct, the constant velocity between position- 0 and time- 8.2 is 1.5m/s
The part I'm having trouble with is the curves of the line (how to figure them out and how to place them on the velocity-time graph).
What I've done for the curves is: 14-13/10-8.2 = .555 but I don't think that is right.
I would be most appreciative if you guys could help me out.

 

[PeterO]

The page layout is included in the picture attachments which are in order.
So far, if I'm correct, the constant velocity between position- 0 and time- 8.2 is 1.5m/s
The part I'm having trouble with is the curves of the line (how to figure them out and how to place them on the velocity-time graph).
What I've done for the curves is: 14-13/10-8.2 = .555 but I don't think that is right.
I would be most appreciative if you guys could help me out.

In the absence of any conflicting data, it is reasonable to a assume that the curved sections are parabolic - meaning constant acceleration - meaning the v-t graph will be sloping up or down at a constant rate.

I would usually identify the long straight sections of the v-t graph - you have the first one, the next one will last from 10 seconds to 18 seconds Then just join the ends of the straight sections [those sections will probably be disjoint] with straight lines. same for the last bit.

 

[hexhall]

Okay, so I'm learning the same exact thing in my science class. In fact, we just had a test over it today! So, the first one is a position vs. time graph, right? Okay, so the line that you're having trouble with is more of a little "hump" like shape. The "hump" basically means that the speed of the elevator is decreasing a little.

So, you're trying to figure out how to place it on a velocity vs. time graph. Well, to show that you're slowing down on a velocity vs. time graph, you would abandon that small little diagonal line that you drew, and instead, make a straight line.

So, to put it simpler, you have your straight line, and then, make a vertical line going all the way down to 1 m/s.
*The reason you go down to 1 is because the slope of the line in the position vs. time graph is 1/1 or 1.
Then you move the line over 1 second.

(The next part kind of exceeds what your asking, but you could use it to check your work)

After that, you would go down to 0 on the horizontal axis from 10 seconds to 18 seconds. Then, since the slope of the last diagonal line is -1, you would go down (using a vertical line) to -1 on the x-axis and draw a horizontal line from 18 to 25 seconds.

Hope I helped you just a little. It took a while for me to type:)