United StatesPhysics C: MechanicsSyllabus dot point

How do position, velocity and acceleration graphs relate to one another through slopes and areas, and how do we read calculus information directly off a motion graph?

Topic 1.3 Representing Motion: relate position, velocity and acceleration graphs through slopes (derivatives) and areas (integrals), and translate between graphical, equation and verbal descriptions of motion.

A focused answer to AP Physics C: Mechanics Topic 1.3, covering how position, velocity and acceleration graphs are linked by slopes (derivatives) and areas (integrals), how to translate between graphs, equations and words, and how to read turning points and concavity, with calculus-based worked examples.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
Slopes are derivatives
Areas are integrals
Turning points and the sign of motion
Translating between representations
Try this

What this topic is asking

The College Board (Topic 1.3) wants you to represent motion with position, velocity and acceleration graphs and to move fluently between graphical, algebraic and verbal descriptions. In AP Physics C the graph relationships are calculus relationships: a slope is a derivative and an area is an integral. The exam routinely hands you one graph and asks you to sketch or read off another, or to extract a numerical answer using slopes and areas.

Slopes are derivatives

Reading a slope is reading a derivative. Where a position-time curve is steep, the object moves fast; where it is flat, the object is momentarily at rest. The concavity of the position graph carries the next derivative: concave up means the velocity is increasing (positive acceleration), and concave down means it is decreasing (negative acceleration). This lets you infer the acceleration from the shape of the position curve alone, without any numbers.

Areas are integrals

Going the other way, the area between a graph and the time axis is an integral:

\Delta v = \int a\,dt \;=\; \text{area under the } a\text{-}t \text{ graph}, \qquad \Delta x = \int v\,dt \;=\; \text{area under the } v\text{-}t \text{ graph}.

Area below the time axis is negative, which is how a velocity-time graph can show an object that moves forward, stops, and reverses: the positive area before the turning point and the negative area after it combine into the net displacement. For motion that returns to the start, the positive and negative areas cancel exactly. Always check which quantity is on the vertical axis before deciding whether you should read a slope or an area.

Turning points and the sign of motion

A turning point is an instant when the velocity passes through zero and changes sign, so the object reverses direction. On the position graph it is a local maximum or minimum (the slope is zero there); on the velocity graph it is where the line crosses the time axis. Distinguish a turning point (velocity zero, object reverses) from a moment of zero acceleration (velocity unchanging for an instant, no reversal). The course wants you to read these features straight off the graph.

Translating between representations

The science practice tested here is translation: given a verbal description ("a car speeds up, cruises, then brakes to a stop"), sketch all three graphs; given a velocity graph, produce the matching position and acceleration graphs; given an equation $x(t)$ , sketch its graph and label the velocity as the slope. The consistent move is to apply the slope-and-area rules and to track signs. Sketches are graded on the correct shape (linear versus curved, concave up versus down) and the correct features (intercepts, turning points), not on artistic precision.

Reading displacement from a velocity graph

A velocity-time graph for a cart consists of three straight segments: $v = 0$ rising to $4.0$ m/s over $0$ to $2.0$ s; constant at $4.0$ m/s from $2.0$ to $5.0$ s; then falling to $0$ from $5.0$ to $6.0$ s. Find the total displacement.

step 1 Identify each region as an area

The displacement is the total signed area under the velocity-time graph. All three regions lie above the time axis, so all areas are positive.

step 2 Compute the first (triangular) area

$0$ to $2.0$ s: a triangle of base $2.0$ s and height $4.0$ m/s, area $= \tfrac{1}{2}(2.0)(4.0) = 4.0$ m.

step 3 Compute the middle (rectangular) area

$2.0$ to $5.0$ s: a rectangle of width $3.0$ s and height $4.0$ m/s, area $= (3.0)(4.0) = 12$ m.

step 4 Compute the last area and total

$5.0$ to $6.0$ s: a triangle of base $1.0$ s and height $4.0$ m/s, area $= \tfrac{1}{2}(1.0)(4.0) = 2.0$ m. Total displacement $= 4.0 + 12 + 2.0 = 18$ m.

Try this

Q1. A position-time graph is a straight line with a positive slope. State what this tells you about the velocity and acceleration. [2 points]

Cue. Constant positive slope means constant positive velocity; a straight line has zero curvature, so the acceleration is zero.

Q2. The area under an acceleration-time graph from $t = 0$ to $t = 3.0$ s is $12$ m/s. If the object started at $v_0 = 2.0$ m/s, calculate its velocity at $t = 3.0$ s. [2 points]

Cue. The area is $\Delta v$ , so $v = v_0 + \Delta v = 2.0 + 12 = 14$ m/s.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (FRQ, translation between representations). A particle moves along the

x

-axis. Its velocity-time graph is a straight line from

v = +6.0

m/s at

t = 0

v = -6.0

m/s at

t = 4.0

s. (a) Determine the acceleration. (b) Determine the time at which the particle momentarily stops. (c) Calculate the net displacement from

t = 0

t = 4.0

s. (d) Describe how the position-time graph would look over this interval.

Show worked answer →

A 4-point FRQ translating between a velocity graph and position and acceleration.

(a) Acceleration (1 point): the slope of the $v$ - $t$ line, $a = \dfrac{-6.0 - 6.0}{4.0 - 0} = -3.0$ m/s squared (constant).
(b) Stops (1 point): $v = 0$ when $6.0 - 3.0t = 0$ , so $t = 2.0$ s.
(c) Net displacement (1 point): the signed area under the line. From $0$ to $2$ s the area is $\tfrac{1}{2}(2.0)(6.0) = +6.0$ m; from $2$ to $4$ s it is $\tfrac{1}{2}(2.0)(-6.0) = -6.0$ m. Net displacement $= +6.0 - 6.0 = 0$ .
(d) Description (1 point): the position rises (positive slope) to a maximum at $t = 2.0$ s, then falls back, returning to the start; the curve is a downward parabola (concave down throughout because the acceleration is negative).

Markers reward using signed area for displacement and identifying the turning point where velocity changes sign.

AP 2021 (style)1 marksSection I (multiple choice). On a position-time graph, the curve is concave up everywhere. Which statement must be true? (A) the velocity is constant (B) the acceleration is positive (C) the object is at rest (D) the speed is decreasing. Justify your reasoning.

Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

The slope of a position-time graph is velocity, and the rate of change of that slope (the concavity) is acceleration. A curve that is concave up everywhere has an increasing slope, so $\dfrac{dv}{dt} > 0$ and the acceleration is positive. It does not require the object to be at rest or the speed to be decreasing; the velocity is changing, not constant. The trap is to read concavity as a statement about position rather than about acceleration.

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Sources & how we know this

AP Physics C: Mechanics Course and Exam Description — College Board (2024)