AP Physics 1 · Topic 1.3

Topic 1.3: Representing Motion

Unit 1: Kinematics10-15% of the multiple-choice section

Topic 1.3 asks you to describe one motion using every representation of it: motion diagrams, figures, graphs, equations, and words. Two rules carry most of the work. A tangent slope gives an instantaneous rate of change, and area under a velocity or acceleration graph gives an accumulated change.

AP Physics: Unit 1 (topics 1.3 Representing Motion). AP Physics 1 Unit 1, Topic 1.3. Learning objective 1.3.A asks students to describe the position, velocity, and acceleration of an object using representations of that object's motion, including motion diagrams, figures, graphs, equations, and narrative descriptions. Unit 1 carries 10 to 15 percent of the multiple-choice section.

What Topic 1.3 actually asks for

The CED states one learning objective for Topic 1.3: describe the position, velocity, and acceleration of an object using representations of that object's motion. It then names the five representations that count: motion diagrams, figures, graphs, equations, and narrative descriptions.

The tested skill is translation. Given a velocity-time graph, sketch the position-time graph. Given a paragraph about a car, draw the motion diagram. Given a graph, write the equation. The free-response section makes this explicit: question 2 on the AP Physics 1 exam is the Translation Between Representations question, worth 12 points with a suggested time of 25 to 30 minutes. It asks students to create a visual representation of a scenario, derive relevant equations, and draw graphs relating the quantities, then closes with any one of three tasks: justify whether two earlier parts agree, use the work to predict another situation, or predict how the representations would change if the scenario were altered.

Unit 1 carries 10 to 15 percent of the multiple-choice section, and Topic 1.3 is the part of it that reappears inside every later unit.

Slope rules: tangent lines give instantaneous rates

Two of the four graph rules in the CED are about slope.

Instantaneous velocity is the slope of a line tangent to a point on a position-time graph. The word tangent is doing real work. A tangent touches the curve at one point, so its slope is the velocity at that instant and nowhere else.
Instantaneous acceleration is the slope of a line tangent to a point on a velocity-time graph.

A straight line joining two separate points is a different object with a different meaning: its slope is the average rate of change over that interval. On a position-time graph the line joining $(t_1, x_1)$ and $(t_2, x_2)$ has slope $\frac{x_2 - x_1}{t_2 - t_1}$ , the average velocity across the interval, not the velocity at either endpoint. On a straight segment the two agree, because the tangent and the connecting line coincide. On a curve they do not, and multiple-choice questions are built on that gap.

Acceleration-time graphs have slopes too, but the CED assigns no meaning to that slope in AP Physics 1. What you take from an acceleration-time graph is its area.

Area rules: areas give accumulated change

The other two graph rules are about area.

Displacement over a time interval equals the area under the velocity-time curve for that interval. The CED is precise about what area means here: the area bounded by the function and the horizontal axis for the appropriate interval.
Change in velocity over a time interval equals the area under the acceleration-time curve for that interval.

That phrase, bounded by the function and the horizontal axis, is what makes the signs work. Area above the time axis counts positive, area below counts negative, and the two cancel. An object that moves forward for 3.0 s and then back for 3.0 s at the same speed encloses equal areas above and below the axis, so its displacement is zero while the distance it travelled is not.

For straight-line graphs these areas are rectangles and triangles, so geometry finishes the job. AP Physics 1 is algebra-based, so curved regions get estimated or compared rather than integrated.

The area under a position-time graph means nothing physical. It is a reliable trap answer.

The translation table

Read any motion graph by asking two questions: what does the slope mean here, and what does the area mean here.

Graph	Slope at a point	Area under the curve
Position vs. time	Instantaneous velocity	No physical meaning
Velocity vs. time	Instantaneous acceleration	Displacement
Acceleration vs. time	Not used in AP Physics 1	Change in velocity

Shape carries information too, and reading shape is the fastest route through multiple choice.

On a position-time graph, a straight segment means constant velocity and a curve means the velocity is changing. Concave up means positive acceleration, concave down means negative acceleration.
On a velocity-time graph, a horizontal segment means zero acceleration and a straight sloped segment means constant acceleration.
Where a velocity-time graph crosses the horizontal axis, the velocity changes sign, so the object reverses direction there.
A flat segment on a position-time graph means the object is at rest, not that it sits at the origin.

The three kinematic equations Topic 1.3 lists

For constant acceleration, the CED lists exactly three kinematic equations under Topic 1.3:

v_x = v_{x0} + a_x t

x = x_0 + v_{x0} t + \frac{1}{2} a_x t^2

v_x^2 = v_{x0}^2 + 2 a_x (x - x_0)

A note printed under them says the equations are written to indicate motion in the x-direction, but they can be used in any single dimension as appropriate, which is your permission to rewrite them with y subscripts for vertical motion.

Topic 1.3 also states that near the surface of Earth the vertical acceleration caused by the force of gravity is downward, constant, and has a measured value approximately equal to $a_g = g \approx 10 \, \mathrm{m/s^2}$ . Two numbers are in play, from two different places. That one is printed in the essential knowledge; the AP Physics 1 table of information separately prints $g = 9.8 \, \mathrm{m/s^2}$ . The next section quotes the boundary statement that reconciles them. Every worked example here uses 9.8.

These equations are the algebraic representation of the same motion the graphs show, which is why the CED files them here. For choosing among them, see the kinematic equations guide, and check arithmetic with the kinematics calculator.

Motion diagrams and narrative descriptions

Graphs get the attention, but the CED lists motion diagrams, figures, and narrative descriptions as representations too, and free-response questions ask for them by name.

A motion diagram marks the object's position at equally spaced instants. Because the time between marks is fixed, the spacing between marks is proportional to speed: marks bunched together mean slow and marks spread apart mean fast. Spacing that grows by the same increment from one gap to the next is the signature of constant acceleration, and spacing that grows by a shrinking increment means the acceleration itself is dropping. Adding a velocity arrow at each mark makes direction explicit, and comparing consecutive arrows shows which way the acceleration points.

Narrative description is the representation students skip, and it is where sign errors get caught early. Before computing anything, say the motion in one sentence: the cart rolls forward, slows, stops, then rolls backward faster and faster. A student who can produce that sentence will not then sketch a velocity-time graph that stays positive the whole way through.

Nonuniform acceleration: exactly how far the exam goes

Two boundary statements sit under Topic 1.3. The first sets the limit on changing acceleration: AP Physics 1 does not expect students to quantitatively analyze nonuniform acceleration, but students are expected to be able to qualitatively analyze, sketch appropriate graphs of, and discuss situations in which acceleration is nonuniform.

Read that as two instructions. First, the moment acceleration stops being constant, the three kinematic equations stop applying, so a question that hands you changing acceleration is not asking for a plug-in. Second, you still have to be able to draw it. A car whose acceleration decreases as it speeds up has a velocity-time graph that rises with a shallowing slope, flattening toward a limiting value.

The slope and area rules survive here, because they are definitions rather than consequences of constant acceleration. Displacement is still the area under the velocity-time curve when that curve bends; you just cannot get it from a single formula.

The second boundary statement covers g: for all situations in which a numerical quantity is required for g, the value $g \approx 10 \, \mathrm{m/s^2}$ will be used, but students will not be penalized for correctly using the more precise commonly accepted values of 9.81 or $9.8 \, \mathrm{m/s^2}$ . That is explicit permission, and it is why this site computes with 9.8 throughout: 10 is what the exam will hand you, 9.8 is a more precise value the CED allows, and both earn the point.

Graph mistakes that cost points

Reading a position-time graph as if it were a velocity-time graph. A high point on a position-time graph means far away, not fast. Fast is steep.
Assuming negative slope means slowing down. Negative slope on a position-time graph means moving in the negative direction, at whatever speed. Slowing down is the slope shrinking in size, not its sign.
Judging speeding up from acceleration alone. An object speeds up when velocity and acceleration share a sign. On a velocity-time graph, the curve moving away from the horizontal axis means speeding up, and toward it means slowing down.
Dropping the sign on area below the axis. Signed area gives displacement; area counted all positive gives distance.
Setting acceleration to zero wherever velocity is zero. At the top of a throw the velocity-time graph crosses the axis with its slope unchanged, so the acceleration there is still $9.8 \, \mathrm{m/s^2}$ downward.
Forgetting the starting value. Area under a velocity-time graph gives the change in position, so you still need the initial position.

Slopes and areas on a velocity-time graph

A cart moves along a straight track. Its velocity-time graph is three straight segments: velocity rises from $0$ to $8.0$ m/s between $t = 0$ and $t = 4.0$ s, holds at $8.0$ m/s until $t = 10.0$ s, then falls back to $0$ at $t = 12.0$ s. Find the acceleration on each segment, the total displacement, and the average velocity for the full 12.0 s.

Acceleration is the slope of the velocity-time graph. Segment 1: $a = \frac{8.0 - 0}{4.0 - 0} = 2.0 \, \mathrm{m/s^2}$ . Segment 2: the graph is horizontal, so $a = 0$ . Segment 3: $a = \frac{0 - 8.0}{12.0 - 10.0} = \frac{-8.0}{2.0} = -4.0 \, \mathrm{m/s^2}$ .
Displacement is the area under the velocity-time graph. Segment 1 is a triangle: $\frac{1}{2}(4.0 \, \mathrm{s})(8.0 \, \mathrm{m/s}) = 16 \, \mathrm{m}$ . Segment 2 is a rectangle: $(6.0 \, \mathrm{s})(8.0 \, \mathrm{m/s}) = 48 \, \mathrm{m}$ . Segment 3 is a triangle: $\frac{1}{2}(2.0 \, \mathrm{s})(8.0 \, \mathrm{m/s}) = 8.0 \, \mathrm{m}$ .
Add the areas: $16 \, \mathrm{m} + 48 \, \mathrm{m} + 8.0 \, \mathrm{m} = 72 \, \mathrm{m}$ . Every piece sits above the time axis, so nothing subtracts and the distance travelled is also $72 \, \mathrm{m}$ .
Average velocity is total displacement over total time: $\frac{72 \, \mathrm{m}}{12.0 \, \mathrm{s}} = 6.0 \, \mathrm{m/s}$ . It is not the average of the three segment velocities, and it equals the slope of the line joining the endpoints of the matching position-time graph.
Cross-check with the other area rule. On the acceleration-time graph, segment 1 is a rectangle of height $2.0 \, \mathrm{m/s^2}$ and width $4.0$ s, so its area is $8.0$ m/s, matching the rise from $0$ to $8.0$ m/s. Segment 3 has height $-4.0 \, \mathrm{m/s^2}$ and width $2.0$ s, area $-8.0$ m/s, matching the fall back to zero.

Accelerations are $2.0 \, \mathrm{m/s^2}$ , then $0$ , then $-4.0 \, \mathrm{m/s^2}$ . Total displacement is $72 \, \mathrm{m}$ and the average velocity is $6.0 \, \mathrm{m/s}$ .

A position-time graph with a reversal

An object's position-time graph is three straight segments: position goes from $0$ to $12 \, \mathrm{m}$ between $t = 0$ and $t = 3.0$ s, holds at $12 \, \mathrm{m}$ until $t = 5.0$ s, then runs from $12 \, \mathrm{m}$ down to $-4.0 \, \mathrm{m}$ at $t = 9.0$ s. Find the velocity on each segment, then the average velocity and average speed over the full 9.0 s.

Velocity is the slope of the position-time graph. Each segment is straight, so the tangent slope and the segment slope are the same number. Segment 1: $\frac{12 - 0}{3.0 - 0} = 4.0 \, \mathrm{m/s}$ . Segment 2: the graph is flat, so $v = 0$ ; the object is at rest at $x = 12 \, \mathrm{m}$ , not at the origin. Segment 3: $\frac{-4.0 - 12}{9.0 - 5.0} = \frac{-16}{4.0} = -4.0 \, \mathrm{m/s}$ .
Displacement for the whole trip is final position minus initial position: $-4.0 \, \mathrm{m} - 0 = -4.0 \, \mathrm{m}$ . The object finishes 4.0 m on the negative side of where it started.
Distance is path length with both directions counted positive: $12 \, \mathrm{m}$ out plus $16 \, \mathrm{m}$ back, so $28 \, \mathrm{m}$ .
Average velocity: $\frac{-4.0 \, \mathrm{m}}{9.0 \, \mathrm{s}} = -0.44 \, \mathrm{m/s}$ . Average speed: $\frac{28 \, \mathrm{m}}{9.0 \, \mathrm{s}} = 3.1 \, \mathrm{m/s}$ . They differ because the trip doubles back.
Translate to a velocity-time graph as a check: $+4.0$ m/s from 0 to 3.0 s, $0$ from 3.0 to 5.0 s, then $-4.0$ m/s from 5.0 to 9.0 s. Its signed area is $(4.0)(3.0) + 0 + (-4.0)(4.0) = 12 - 16 = -4.0 \, \mathrm{m}$ , matching the displacement read straight off the position-time graph.

Segment velocities are $4.0 \, \mathrm{m/s}$ , then $0$ , then $-4.0 \, \mathrm{m/s}$ . Over the full 9.0 s the average velocity is $-0.44 \, \mathrm{m/s}$ and the average speed is $3.1 \, \mathrm{m/s}$ .

Frequently asked questions

What does the slope of a position-time graph tell you?

The instantaneous velocity. The CED is specific: the slope of a line tangent to a point on the position-time graph equals the object's instantaneous velocity at that instant. The slope of a straight line drawn between two separate points is the average velocity over that interval instead. On a straight segment the two match; on a curve they do not.

What does the area under a velocity-time graph represent?

Displacement over that time interval, defined by the CED as the area bounded by the function and the horizontal axis for the appropriate interval. Area below the axis counts negative, so a trip out and back can enclose plenty of area and still give zero displacement. The area under an acceleration-time graph gives the change in velocity instead.

How do I tell from a graph whether an object is speeding up?

Compare the signs of velocity and acceleration: same sign means speeding up, opposite signs means slowing down. On a velocity-time graph the visual shortcut is whether the curve moves away from the horizontal axis (speeding up) or toward it (slowing down). Negative acceleration on its own never settles the question.

Does AP Physics 1 test acceleration that is not constant?

Yes, but only qualitatively. A CED boundary statement says students are not expected to analyze nonuniform acceleration quantitatively, while they are expected to analyze it qualitatively, sketch appropriate graphs of it, and discuss it. Expect sketch-and-explain questions, and remember the three kinematic equations do not apply once acceleration changes.

Which equations does Topic 1.3 give me?

Three constant-acceleration equations: v = v0 + at, x = x0 + v0t + (1/2)at^2, and v^2 = v0^2 + 2a(x - x0), printed with x subscripts. A CED note says they may be used in any single dimension as appropriate. On g, watch the two sources: essential knowledge 1.3.A.3 prints a_g = g approximately equal to 10 m/s^2, while the AP Physics 1 table of information separately lists g = 9.8 m/s^2, and a boundary statement says numerical work will use 10 while 9.8 and 9.81 also earn credit. See the full AP Physics 1 formula sheet.