math

d = 0.08t^2 - 1.6t + 9
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.

How far is the student from the detector when he starts to walk?

Are my calculations correct?

d= 0.08t^2 -1.6t+9
d= 0.08(0)^2-1.6(0)+9=9
Therefore the student is 9m from the detector when he starts to walk

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asked by A
  1. yes

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    posted by oobleck

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