# math

D = 0.08t^2 - 1.6t + 9
Why does this equation above fits this model: A student stands facing a motion detector. He quickly walks toward the detector, slows down, stops and then slowly walks away from the detector. He speeds up as he gets farther away from the detector.

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1. D = 0.08t^2 - 1.6t + 9

is constant acceleration of +0.08 with an initial location of D = 9 and an initial velocity of -1.6
note
when t = 0, D = 9

if you were into calculus I would say more:)

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posted by Damon
2. just look at the graph of the parabola.
It starts off at a certain value, which decreases, stop, turns around, and then moves back up with greater speed.

There are other formulas which would fit the model, but the parabola is the simplest.

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posted by oobleck

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