Teacher Note: Students should identify that position values shifts by a constant displacement, but differences in position (\(\Delta x\)), instantaneous velocity (\(v\)), and acceleration (\(g = -9.8\,\text{m/s}^2\)) remain identical. Physical reality is invariant under a coordinate shift.
AP Physics 1 • Kinematics Mapping Project
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AP Physics 1 • Kinematics
Page 2 of 3
EXEMPLAR ANSWER KEY
2
A position vs. time graph represents an object's spatial location on the vertical axis (\(x\)) as time (\(t\)) marches forward on the horizontal axis. Interpreting these graphs mathematically is a foundational skill for AP Physics 1:
Slope as Velocity: Because velocity is defined as the rate of change of position, the slope of a position-time graph is equivalent to velocity (\(v = \frac{\Delta x}{\Delta t}\) or \(v = \frac{dx}{dt}\)).
Concavity as Acceleration: The curvature represents acceleration. If a curve is concave up (smiley face), acceleration is positive (\(a > 0\)). If it is concave down (frown), acceleration is negative (\(a < 0\)).
Graphic Profile: Non-Linear Position vs. Time Analysis (Slope/Concavity Annotated)
Position, x (m) Time, t (s) A v > 0 (const) B v = 0 (stopped) C a < 0 (concave down)
Stop & Jot 2
Refer to Segment C in the position vs. time graph above. Describe the velocity and acceleration behavior of the object in this interval. Is it speeding up or slowing down? Explain using slope and concavity concepts.
EXEMPLAR RESPONSE:
In Segment C, the slope starts flat (\(\text{slope} \approx 0\)) and curves downward, becoming increasingly steep and negative. This means the velocity is negative (\(v < 0\)) and its magnitude (speed) is increasing. Because the curve is concave down, the acceleration is constant and negative (\(a < 0\)). Since both velocity and acceleration are negative (same sign), they work together, meaning the object is speeding up in the negative direction.
Think-Pair-Share 2
Think: If an object has a positive position (\(x > 0\)), a negative velocity (\(v < 0\)), and a positive acceleration (\(a > 0\)), what does its motion look like in real life, and what would its curve look like on an \(x-t\) graph?
Teacher Note: This object is positioned to the right of the origin, moving to the left, and slowing down. On an \(x-t\) graph, the curve starts high with a steep negative slope and curves upwards (concave up, flattening out) to a flatter negative slope.
AP Physics 1 • Kinematics Mapping Project
© 2026 Space & Vector Learning Labs
AP Physics 1 • Kinematics
Page 3 of 3
EXEMPLAR ANSWER KEY
3
A velocity vs. time graph represents velocity on the vertical axis (\(v\)) and time (\(t\)) on the horizontal axis. It yields two critical physical parameters through geometric and derivative analysis:
Slope is Acceleration: Since acceleration is the rate of change of velocity, the instantaneous acceleration is the slope of the \(v-t\) curve (\(a = \frac{\Delta v}{\Delta t}\) or \(a = \frac{dv}{dt}\)).
Area Under Curve is Displacement: From calculus, \(dx = v \, dt\), which means integration (or geometrically calculating area) of the \(v-t\) plot gives the change in position (displacement, \(\Delta x\)):
\[ \Delta x = \int_{t_1}^{t_2} v(t) \, dt = \text{Bounded Area} \]
Quantitative Profile: Velocity vs. Time Acceleration & Area (Calculus Annotated)
Area (+) = Displacement v = 0 (t = 6s) Velocity, v (m/s) t +4 0 -4 0s 2s 4s 6s
Stop & Jot 3
Calculate the magnitude of the deceleration from \(t = 4\,\text{s}\) onward, and state the exact moment in time when the object reverses direction. Justify your calculations.
EXEMPLAR RESPONSE:
The slope from \(t = 4\,\text{s}\) (where \(v = +4\,\text{m/s}\)) to \(t = 8\,\text{s}\) (where \(v = -4\,\text{m/s}\)) represents acceleration: \[ a = \frac{v_f - v_i}{\Delta t} = \frac{-4\,\text{m/s} - 4\,\text{m/s}}{8\,\text{s} - 4\,\text{s}} = \frac{-8\,\text{m/s}}{4\,\text{s}} = -2\,\text{m/s}^2 \] Thus, the magnitude of deceleration is \(2\,\text{m/s}^2\). The object reverses direction when velocity changes sign (\(v = 0\)), which is exactly at \(t = 6\,\text{s}\) (the horizontal intercept).
Think-Pair-Share 3
Think: How do you calculate total path distance traveled from a \(v-t\) graph versus the net displacement? Focus on how you treat negative area sections below the axis.
Teacher Note: Net displacement is the integral of velocity (positive area minus negative area). Total distance is the integral of speed (the absolute value of velocity), meaning all areas—both positive and negative—must be summed as positive magnitudes: \(\int |v| \, dt\).
AP Physics 1 • Kinematics Mapping Project
© 2026 Space & Vector Learning Labs
Velocity vs. Time graphs yield two central physical variables through linear calculus properties:
1. Slope = Acceleration The rate of change of velocity (\(a = \frac{\Delta v}{\Delta t}\)).
2. Area = Displacement The integral bounding change in position (\(\Delta x = \int v \, dt\)).
Velocity, v Time, t Area = Δx
Refer to Student Packet Page 3
03 • Think-Pair-Share Slide 7 of 8
Dealing with turning points and changing directions.
Think (1 min): How do you calculate total path distance traveled from a \(v-t\) graph versus the net displacement? Focus on how you treat negative area sections below the axis.
Pair & Share (2 mins): Frame a general rule or formula you can share with your classmates to avoid a zero-displacement mistake when an object turns back.
Active Collaboration Phase
Timer Suggestion: 3 Minutes
Summary Slide 8 of 8
Motion Diagram
Spacing directly shows displacement intervals. Growth rate of spacing shows acceleration direction.
x-t Graphs
Slope yields velocity. Curvature (second derivative) determines the sign of acceleration.
v-t Graphs
Slope yields acceleration. Integrating the bounded area yields position displacement.
Next: Individual Formative Exit Ticket
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AP Physics 1 • Kinematics
Page 2 of 2
Teacher Resource
During Stop & Jot 2, focus on Segment C's curvature. Students may struggle to explain that the negative velocity is speeding up because the slope is "getting steeper, but downwards."
Guiding curvature question: "Draw tangent lines at t = 6s, 7s, and 8s on Segment C. Are the slopes getting closer to zero or further from zero? What does that say about speed?"
Think-Pair-Share 2 Intervention: For the paradox (\(x > 0\), \(v < 0\), \(a > 0\)), highlight that the object is in the positive domain but moving backward and slowing down (e.g., a car braking as it approaches the origin from the right).
Stop & Jot 3 requires quantitative calculations. Ensure students use correct units (\(\text{m/s}^2\) for acceleration, \(\text{m}\) for displacement) and show algebraic steps.
Guiding Turning Point Prompt: "What is the physical velocity when an object switches from moving forwards to backwards? Where is that on the vertical axis?"
Think-Pair-Share 3 Intervention: Contrast displacement (\(\int v \, dt\)) and distance (\(\int |v| \, dt\)). Ensure students understand that to calculate distance, they must split the integral at \(t = 6\,\text{s}\) and add the magnitudes of the areas, rather than letting the negative area cancel the positive area.
For Struggling Students: Have them physically act out the position and velocity graphs in the classroom. Use tape on the floor as the axis, with one student calling out time intervals while another moves in accordance with the graphs.
For Advanced Students (AP Physics C prep): Challenge them to write the polynomial function representing Segment C in the position graph (\(x(t) = c_0 + c_1 t + c_2 t^2\)). Ask them to find the exact derivatives to confirm position and velocity at specific points.
AP Physics 1 • Kinematics Mapping Project
© 2026 Space & Vector Learning Labs
