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.
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:)
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.
To understand why the given equation fits the described motion of the student, we need to analyze the equation's variables and coefficients.
The equation D = 0.08t^2 - 1.6t + 9 represents the distance (D) traveled by the student over time (t).
Let's break down the equation:
- The term 0.08t^2 represents the acceleration of the student. Since the coefficient is positive (0.08), it indicates that the student's acceleration is increasing with time squared. This means the student is accelerating faster and faster as time progresses. For example, after 1 second, the acceleration will be 0.08 times 1 squared which is 0.08. After 2 seconds, the acceleration will be 0.08 times 2 squared which is 0.32, and so on.
- The term -1.6t represents the rate of decrease in speed or deceleration. The negative coefficient (-1.6) indicates that the student is slowing down as time progresses. It is linear, meaning the student reduces their speed by 1.6 units every second.
- The constant term 9 represents the initial distance from the motion detector when the student starts walking. It indicates the starting point or the distance before the student begins moving.
Now, let's link this equation to the described motion of the student:
1. "A student stands facing a motion detector." - Before the student starts moving, the initial distance (D) is 9 units from the motion detector. This aligns with the constant term in the equation.
2. "He quickly walks toward the detector, slows down, stops, and then slowly walks away from the detector." - The term 0.08t^2 represents the increasing acceleration as time progresses. This aligns with the initial quick walking towards the detector. The term -1.6t (deceleration) represents the point where the student slows down, stops, and then starts walking away from the detector.
3. "He speeds up as he gets farther away from the detector." - Since the acceleration term (0.08t^2) is always positive, it indicates that the student's speed increases as time goes on, particularly when the distance from the detector increases.
Therefore, the equation D = 0.08t^2 - 1.6t + 9 accurately models the described motion of the student, starting from a stationary position, walking towards the detector, slowing down, stopping, and then walking away while gradually increasing speed as the distance from the detector increases.