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Hello Daniel,

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 |

$$ - not a meaningful unit of movement.

I hope this helps answer your question.

Andy

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