velocity graph changed to position vs time graph?

Question

seconds - velocity
0-0
1-2.5
2-2.5
3-1.25
4-0
5- -1.25
6- -2.5
7- -3.75
8- -3.75
8.5- 0
9- 0
10- 0

Fist number is time 2nd is velocity at time, what would the positions be at the times?

Thanks

Answer 1

Numerical integration is required:
http://en.wikipedia.org/wiki/Numerical_integration

Answer 2

First off, it can't be done precisely because the velocity/time graph gives no indication of the actual location of the object. Mathematically (if you will), this means that we have no reference position to start the graph. What we can do is figure out roughly how the position is changing, so we'll assume that the object starts at position 0. From 0 to 1 second, the velocity went from 0 to 2.5 m/s (I assume). We have no information about the shape of the velocity curve in that interval, so we'll assume it's linear (meaning constant acceleration). In that case, the average velocity will be the simple average of the initial and final velocities: 1.25 m/s. With an average velocity of 1.25 m/s in 1 second, the object will move 1.25 m. From 1 second to 2 seconds, the velocity doesn't change. With the same assumptions as above, the average velocity will be 2.5 m/s. In one second, the object will move 2.5 m. From second 2 to second 3, the assumed average velocity is 1.875 m/s, so it moves 1.875 m. You continue in this way. Now, we can get our position/time data: first column seconds, second column meters. 0 0 - initial position 1 1.25 - initial position 0 + 1.25 meters moved 2 3.75 - initial position 1.25 + 2.5 meters moved 3 5.625 - initial position 3.75 + 1.875 meters moved And it goes on from there. Note that I have paid no attention to significant digits - you'll probably want to rectify that.