Asked by A

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.

When is the student closest to the detector?

What is his distance from the detector after 2 seconds?

When is he more than 3m from the detector?
Answered by anonymous
1. the student is 9 m away
2. you sub in 2 in the formula: 0.08(2)^2-1.6(2)+9=6.12
3) unknown
Answered by Reiny
His distance is represented by a quadratic function in the form of an upwards parabola.
When is the student closest to the detector? --- at the t of the vertex.
The t of the vertex is -b/(2a) = 1.6/.16 = 10 seconds

What is his distance from the detector after 2 seconds? --- plug in t=2

When is he more than 3m from the detector? --- >
.08t^2 - 1.6t + 9 > 3
.08t^2 - 1.6t + 6 > 0
divide by .08
t^2 - 20t + 75 > 0
(t-5)(t-15) > 0
t < 5 OR t > 15

verification:

www.wolframalpha.com/input/?i=plot+0.08t%5E2+-+1.6t+%2B+9+%3E+3
Answered by A
@Reiny I don't understand the steps you did for When is the student closest to the detector.