Posted by **Lindsay** on Tuesday, October 16, 2007 at 4:30pm.

I am given a velocity versus time graph for a runner. For it, I had to figure out the average velocity for the entire 16 s interval (for this, I got 2 m/s). Now they want me to find the displacement of the runner in the first 15.4 s.

I know that it must be somewhere around 30 (since 2*15=30), but how can I factor in the .4 of a second?

- Physics -
**bobpursley**, Tuesday, October 16, 2007 at 5:21pm
Isn't the area underneath the velocity plot distance? Velocity*time= distance?

I think this is what they are wanting you to do, measure the area underneath. An easy way to do that is to count the squares on the graph paper.

- Physics -
**Lindsay**, Tuesday, October 16, 2007 at 5:42pm
Great I got it, thanks. :)

- Physics -
**Emily**, Tuesday, October 16, 2007 at 5:57pm
okay

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