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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