Velocity graph changed to position vs time graph?

Question

Velocity graph changed to position vs time graph?

Velocity graph changed to position vs time graph?

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?

Answer 1

I can't give you an exact answer because you didn't tell me the units of velocity. It will make a big difference if the velocity is in m/s rather than mph.
But here is an explanation of the method. Assume a straight line connects all the points on the velocity graph. Interpolating between the points will give an aswer...
So, let's use velocity units of m/s for convenience:
The average velocity for the first time interval will be:
(2.5 m/s + 0 m/s)/ 2
= 1.25m/s

The distance traveled in the first interval is:
1.25m/s x (1sec - 0sec)
= 1.25 m
Plot (0sec, 1.25m) as the first point on your Position v. Time graph.

For the second time interval:
average velocity:
(2.5m/s + 2.5m/s)/2
= 2.5m/s (constant velocity during this time interval)

distance traveled in this interval:
2.5m/s x (2 sec - 1 sec)
= 2.5m

To get the total distance (or position) at the end of this interval (the 2 second point), add this 2.5m to the position at the end of the last interval:
so: 1.25m + 2.5m = 3.75m
3.75m is the position at 2 seconds.
Plot (2sec, 3.75m) as the second point on your Position v. Time graph.

Now you continue figuring the distance traveled in each time interval and adding it to the previous total distance traveled.

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.