|MadSci Network: Astronomy|
You are correct in saying that the area bounded by a velocity versus time graph gives us a distance.
From NASA's online Moon Fact Sheet http:/ /nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html I see that the Moon's mean orbital velocity is given as 1.023 km/s. With a time for one revolution of 27.3217 days, or about 2360500 seconds, multiplying this velocity by this time gives us a figure of about 2414800 km, which is 2*PI*radius of the Moon's orbit, as we would expect.
The slope of a velocity versus time graph equals the acceleration. It's intuitively obvious that a flat velocity graph indicates no acceleration, as we'd expect when using the mean velocity.
In reality, of course, the Moon's orbital velocity isn't constant, as the orbit is very slightly elliptical (e=0.0549). The area under the graph of this varying velocity against time would once again be equal to the distance travelled, but this time around the perimeter of the elliptical orbit of the Moon.
In this case, the changing slope of the velocity versus time graph would be equal to the changing acceleration of the Moon as it speeds up and slows down around the course of one orbit.
Measuring the displacement of the Moon against time, from some arbitrary point in its orbit over the course of one orbit would (using the mean velocity) result in a straight-line graph, whose slope would be equal to the velocity of the Moon.
We can plot three graphs to combine displacement, velocity and acceleration against time, and each one has a variety of uses. In these instances:
|type of graph||slope measures||area measures|
|displacement v time||velocity||distance*time $$|
|velocity v time||acceleration||distance|
|acceleration v time||rate of change of acceleration||velocity|
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