\(v_A = \frac{80 - 0}{4 - 0}\)
\(v_A = \mathbf{+20\text{ m/s}}\)
Constant positive velocity (moving in positive direction).
Interval B (4s to 7s)
\(v_B = \frac{x_f - x_i}{t_f - t_i}\)
\(v_B = \frac{80 - 80}{7 - 4}\)
\(v_B = \mathbf{0\text{ m/s}}\)
Zero slope implies the object is fully at rest.
Interval C (7s to 10s)
\(v_C = \frac{x_f - x_i}{t_f - t_i}\)
\(v_C = \frac{0 - 80}{10 - 7} = \frac{-80}{3}\)
\(v_C \approx \mathbf{-26.7\text{ m/s}}\)
Constant negative velocity (returning back to origin).
Connecting Speed & Direction (Ideal Answer):
"Speed is a scalar quantity and is always positive, but displacement is a vector that depends on direction. The negative sign in Interval C indicates motion in the opposite direction (returning to the start line). If we did not use vector signs, adding these movements would incorrectly suggest the car continued forward, failing to track the physical return to \(0\text{ m}\)."
MODELING MOTION TEACHER GUIDE PAGE 2 OF 4
KEY
Teacher Guide
STATION 3
STATION 3 KEY
This station reinforces how the area under a velocity curve represents physical displacement while the slope yields uniform acceleration in one dimension.
1. Acceleration Calculation
Calculating slope over interval \(t = 0\text{s}\) to \(t = 5\text{s}\):
\(a = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t_f - t_i}\)
\(a = \frac{0\text{ m/s} - 40\text{ m/s}}{5.0\text{ s} - 0.0\text{ s}}\)
\(a = \frac{-40\text{ m/s}}{5.0\text{ s}}\)
\(a = -8.0\text{ m/s}^2\)
Interpretation: The negative sign confirms the vehicle is decelerating (accelerating opposite to the direction of motion).
Acceleration Vector:
-8.0 m/s²
Magnitude:
8.0 m/s²
2. Total Displacement Calculation
Calculating area of the bounded triangle:
\(\text{Displacement } (d) = \text{Area of Triangle}\)
\(d = \frac{1}{2} \cdot \text{base} \cdot \text{height}\)
\(d = \frac{1}{2} \cdot (5.0\text{ s}) \cdot (40\text{ m/s})\)
\(d = \frac{1}{2} \cdot 200\text{ m}\)
\(d = 100\text{ m}\)
Check via kinematics: \(v_f^2 = v_i^2 + 2ad \implies d = 100\text{ m}\).
Calculated Area:
100 meters
Physical Meaning:
Stopping distance
MODELING MOTION TEACHER GUIDE PAGE 3 OF 4
KEY
Teacher Guide
STATION 4
STATION 4 KEY
1. Exemplar Claim
"Doubling the initial horizontal velocity (\(v_{x0}\)) has absolutely no effect on its total flight time. The projectile hits the ground in the exact same time as before."
2. Exemplar Evidence
"The height of the launch point is constant at \(h = 45\text{ m}\). The initial vertical velocity is \(v_{y0} = 0\text{ m/s}\) because the launch is horizontal. The vertical acceleration is the constant force of gravity, \(g \approx 9.8\text{ m/s}^2\) downwards. The vertical kinematic equation is \(y = v_{y0}t + \frac{1}{2}gt^2\)."
3. Exemplar Reasoning
"Since vertical and horizontal components of projectile motion are completely independent, the vertical descent is calculated strictly from vertical parameters: \(t = \sqrt{2h/g}\). Plugging in height (\(45\text{ m}\)) and gravity (\(9.8\text{ m/s}^2\)) yields exactly \(t \approx 3.03\text{ seconds}\). Since horizontal velocity \(v_{x0}\) is mathematically absent from this formula, it cannot alter descent duration. Doubling \(v_{x0}\) only doubles the horizontal range (\(d_x = v_{x0}t\)), landing the ball twice as far away."
| Criteria | Proficient (3 pts) | Developing (2 pts) | Novice (1 pt) |
|---|---|---|---|
| Claim | Clearly states that flight time remains unchanged. | Incorrectly states time changes but attempts a conceptual defense. | States an incorrect claim with no clear connection to the prompt. |
| Evidence | Identifies vertical height, gravity, and horizontal independence. | Lists math formulas but misses constant/independent factors. | Lists numbers with no relevance or formulas. |
| Reasoning | Uses \(t = \sqrt{2y/g}\) to logically show horizontal velocity has no impact. | Mentions independence but cannot link it mathematically. | Lacks physical arguments or repeats the claim directly. |
MODELING MOTION TEACHER GUIDE PAGE 4 OF 4
