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?
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
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
@Reiny I don't understand the steps you did for When is the student closest to the detector.
To find the answers to these questions, we need to analyze the given equation that represents the motion of the student in relation to time. The equation is:
d = 0.08t^2 - 1.6t + 9
1. When is the student closest to the detector?
To find the moment when the student is closest to the detector, we need to determine the minimum value of the distance function. This occurs at the vertex of the quadratic equation. The x-coordinate of the vertex can be found using the formula:
t = -b / (2a)
In this case, a = 0.08 and b = -1.6. Substituting these values, we get:
t = -(-1.6) / (2 * 0.08)
t = 20 / 0.16
t = 125
Therefore, the student is closest to the detector at 125 seconds.
2. What is the student's distance from the detector after 2 seconds?
To find the student's distance from the detector after 2 seconds, we substitute t = 2 into the equation:
d = 0.08(2)^2 - 1.6(2) + 9
d = 0.32 - 3.2 + 9
d = 5.12
Therefore, the student is 5.12 meters away from the detector after 2 seconds.
3. When is the student more than 3m from the detector?
To find the moments when the student is more than 3 meters away from the detector, we need to solve the equation d > 3. Rearranging the equation, we get:
0.08t^2 - 1.6t + 9 > 3
To solve this quadratic inequality, we simplify the equation:
0.08t^2 - 1.6t + 6 > 0
We can either factor this expression or use the quadratic formula to find the solutions:
t = (-b ± √(b^2 - 4ac)) / (2a)
Using the quadratic formula, we get:
t = (-(-1.6) ± √((-1.6)^2 - 4(0.08)(6))) / (2 * 0.08)
t = (1.6 ± √(2.56 - 1.92)) / 0.16
t = (1.6 ± √0.64) / 0.16
t = (1.6 ± 0.8) / 0.16
Simplifying further:
t = 2 or t = 6
Therefore, the student is more than 3 meters away from the detector at 2 seconds and at 6 seconds.