AP Physics C: Mechanics · Topic 1.2

Topic 1.2: Displacement, Velocity, and Acceleration

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

Topic 1.2 defines displacement and the two averages exactly as AP Physics 1 does, then adds a learning objective the algebra-based course has no version of. Objective 1.2.C asks for instantaneous position, velocity and acceleration as functions of time, and defines the last two as derivatives.

AP Physics: Unit 1 (topics 1.2 Displacement, Velocity, and Acceleration). AP Physics C: Mechanics Unit 1, Topic 1.2. Three learning objectives, the most of any Unit 1 topic. 1.2.A describe a change in an object's position, with 1.2.A.1 (object model: size, shape and internal configuration ignored, treated as a single point with extensive properties such as mass and charge) and 1.2.A.2 (displacement is the change in an object's position, relevant equation delta-x = x - x_0). 1.2.B describe the AVERAGE velocity and acceleration of an object, with 1.2.B.1 (averages consider initial and final states over an interval), 1.2.B.2 (average velocity is displacement over the interval, v_avg = delta-x/delta-t), 1.2.B.3 (average acceleration is change in velocity over the interval, a_avg = delta-v/delta-t), 1.2.B.4 (an object is accelerating if either the magnitude and/or direction of its velocity are changing) and 1.2.B.5 (an average over a very small interval yields a value very close to the instantaneous value). 1.2.C describe the INSTANTANEOUS position, velocity, and acceleration of an object as a function of time, with 1.2.C.1 (as the time interval used to calculate the average approaches zero, the average approaches the value at that instant, called the instantaneous value), 1.2.C.1.i (instantaneous velocity is the derivative of position; relevant equations v = dr/dt and v_x = dx/dt), 1.2.C.1.ii (instantaneous acceleration is the derivative of velocity; relevant equations a = dv/dt and a_x = dv_x/dt) and 1.2.C.2 (time-dependent functions and instantaneous values of position, velocity and acceleration can be determined using differentiation and integration). NO boundary statement in either course. Suggested skills 1.B, 2.B, 2.C, 3.A, 3.C; AP Physics 1's Topic 1.2 suggests 1.C, 2.B, 2.C, 3.C instead. AP Physics 1's Topic 1.2 has only two learning objectives: its 1.2.A.1 through 1.2.B.5 are word-for-word identical to this course's, and it has NO counterpart to 1.2.C at all. AP Physics 1's objective 1.2.B is worded 'Describe the velocity and acceleration of an object'; the Physics C wording inserts 'average'. Topic 1.2 prints seven equations across five statements and NONE of them appears on the C: Mechanics Table of Information. That sheet's kinematics block is exactly five lines: v_x = v_x0 + a_x t; x = x_0 + v_x0 t + (1/2)a_x t^2; v_x^2 = v_x0^2 + 2a_x(x - x_0); delta-x = integral of v_x(t) dt; delta-v_x = integral of a_x(t) dt, the two integrals printed with no limits. The sheet does print omega = d-theta/dt and alpha = d-omega/dt in its rotational column, so it defines the rotational rates as derivatives and not the translational ones, the exact mirror of the integrals, which are printed translationally and not rotationally.

What Topic 1.2 requires

Three learning objectives, which makes Topic 1.2 the largest of Unit 1's five by that measure. Unit 1 has eight objectives in total and three of them are here.

Objective	Statement	Essential knowledge
1.2.A	Describe a change in an object's position.	1.2.A.1, 1.2.A.2
1.2.B	Describe the average velocity and acceleration of an object.	1.2.B.1 to 1.2.B.5
1.2.C	Describe the instantaneous position, velocity, and acceleration of an object as a function of time.	1.2.C.1 with two sub-statements, and 1.2.C.2

What each statement says:

1.2.A.1 When using the object model, the size, shape, and internal configuration are ignored. The object may be treated as a single point with extensive properties such as mass and charge.
1.2.A.2 Displacement is the change in an object's position, with the relevant equation $\Delta x = x - x_0$ .
1.2.B.1 Averages of velocity and acceleration are calculated considering the initial and final states of an object over an interval of time.
1.2.B.2 Average velocity is the displacement of an object divided by the interval of time in which that displacement occurs, with $\vec{v}_{\text{avg}} = \Delta\vec{x}/\Delta t$ .
1.2.B.3 Average acceleration is the change in velocity divided by the interval of time in which that change in velocity occurs, with $\vec{a}_{\text{avg}} = \Delta\vec{v}/\Delta t$ .
1.2.B.4 An object is accelerating if either the magnitude and/or direction of the object's velocity are changing.
1.2.B.5 Calculating average velocity or average acceleration over a very small time interval yields a value that is very close to the instantaneous velocity or instantaneous acceleration.
1.2.C.1 As the time interval used to calculate the average value of a quantity approaches zero, the average value of that quantity approaches the value of the quantity at that instant, called the instantaneous value.
1.2.C.2 Time-dependent functions and instantaneous values of position, velocity, and acceleration can be determined using differentiation and integration.

Suggested skills: 1.B create quantitative graphs with appropriate scales and units, including plotting data; 2.B calculate or estimate an unknown quantity with units from known quantities; 2.C compare physical quantities between two or more scenarios or at different times and/or locations within a single scenario; 3.A create experimental procedures that are appropriate for a given scientific question; and 3.C justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

Topic 1.2 prints no boundary statement in either course. Nothing fences off how far the differentiation goes.

The learning objective AP Physics 1 does not have

Compare the two frameworks statement by statement and the first eight lines are the same text. AP Physics C's 1.2.A.1, 1.2.A.2 and 1.2.B.1 through 1.2.B.5 are word for word AP Physics 1's, printing the same three equations. Then AP Physics C keeps going and AP Physics 1 stops.

Learning objective 1.2.C has no counterpart in AP Physics 1. Not a reworded counterpart, not a narrower one: the algebra-based framework's Topic 1.2 ends at 1.2.B.5 and its Topic 1.2 has two objectives against this course's three.

There is a second, quieter change that tells you why. The two courses word objective 1.2.B differently:

	AP Physics 1	AP Physics C: Mechanics
1.2.B	Describe the velocity and acceleration of an object.	Describe the average velocity and acceleration of an object.

AP Physics C inserted the word "average" into an objective it otherwise left alone. It could do that because the instantaneous case had somewhere else to go. In AP Physics 1, objective 1.2.B has to cover both senses of velocity, and the instantaneous one arrives only as the limiting statement at 1.2.B.5. Here the two senses are separated into their own objectives, and the second one is defined with calculus.

That is the whole shape of the difference, and it is worth stating plainly rather than dressing up. The physics of displacement and averages is identical between the courses. What AP Physics C adds is a definition of the instantaneous quantities that does not depend on shrinking an interval by hand.

The limit in words at 1.2.C.1, then written down at 1.2.C.1.i and 1.2.C.1.ii

Statement 1.2.C.1 gives the limit in English before any symbol appears: as the time interval used to calculate the average value of a quantity approaches zero, the average value of that quantity approaches the value of the quantity at that instant, called the instantaneous value.

Read against 1.2.B.5, which says a very small interval yields a value very close to the instantaneous one, the pair is a definition split across two objectives. 1.2.B.5 is the approximation. 1.2.C.1 is the limit the approximation is approaching.

Statement 1.2.C.1.i then says instantaneous velocity is the rate of change of the object's position, which is equal to the derivative of position with respect to time, and prints two relevant equations:

\vec{v} = \frac{d\vec{r}}{dt} \qquad v_x = \frac{dx}{dt}

Statement 1.2.C.1.ii says instantaneous acceleration is the rate of change of the object's velocity, which is equal to the derivative of velocity with respect to time, and prints two more:

\vec{a} = \frac{d\vec{v}}{dt} \qquad a_x = \frac{dv_x}{dt}

Both statements print the vector form and the component form side by side, which is a deliberate pairing rather than duplication. The vector form is the definition; the component form is how you evaluate it. Differentiating a vector means differentiating each component separately, because $\hat{i}$ , $\hat{j}$ and $\hat{k}$ are constant directions:

\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} \quad \Rightarrow \quad \vec{v}(t) = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j}

That is why Topic 1.1 spent its whole objective teaching you to write $\vec{r}$ in components. This is what the notation was for.

One detail worth noticing: acceleration is defined as the derivative of velocity, not of speed. Statement 1.2.B.4 already said an object is accelerating if the magnitude and/or direction of its velocity is changing, and the derivative form inherits that. A velocity vector of constant magnitude turning through an angle has a nonzero $d\vec{v}/dt$ , which is the whole content of centripetal acceleration in Topic 2.10.

Average and instantaneous are answers to different questions

An average is a property of an interval. An instantaneous value is a property of an instant. Statement 1.2.B.1 makes the first half explicit: averages are calculated considering the initial and final states of an object over an interval of time. Nothing in between enters the calculation.

The practical consequences:

An average velocity over an interval can be zero while the object never stopped, because $\Delta \vec{x}$ can be zero for a round trip.
Average speed is not the magnitude of average velocity, because speed accumulates distance and velocity accumulates displacement. Those agree only when the motion never reverses.
Average acceleration ignores everything the acceleration did between the endpoints. A worked example below has an object whose average acceleration over four seconds is $-3.0$ m/s $^2$ at a moment when its instantaneous acceleration is $-6.0$ m/s $^2$ .

There is one clean case where the two coincide. For constant acceleration, the average velocity over an interval equals the instantaneous velocity at the midpoint of the interval, and it also equals $\left( v_0 + v \right)/2$ . That is a property of a linear $v_x(t)$ , not a general rule, and a worked example below shows it failing the moment the position function becomes a cubic.

The reason this matters for the exam rather than only for the vocabulary is that AP Physics C questions hand you a function. Once you have $x(t)$ , both quantities are available and the question decides which one it wants. "How fast is it moving at $t = 3.0$ s" is a derivative. "How fast on average over the first three seconds" is a difference quotient. Reading the wrong one out of a correct function is the most common way to lose the mark after doing the calculus right.

The object model, and the word extensive

Statement 1.2.A.1 is easy to skim, and it is doing something specific: when using the object model, the size, shape, and internal configuration are ignored, and the object may be treated as a single point with extensive properties such as mass and charge.

The word extensive is the point. An extensive property scales with how much stuff there is: mass, charge, volume. The object model lets you throw away geometry while keeping those. So a car can be a dot with a mass, and a planet can be a dot with a mass, and neither idealisation costs you anything as long as the question is about where the dot is.

It costs you something the moment the question is about the object's own shape. That is why the CED introduces the rigid system later, in Unit 5, where different points of one body move in different directions and the object model stops being available. Naming the model here is what makes it possible to say later exactly when you may no longer use it.

For Unit 1 the practical reading is narrower: every position, velocity and acceleration here belongs to a point. When Unit 2 turns to a system of several objects, the point you track becomes the centre of mass, and the sheet prints an integral for it.

Statement 1.2.C.2 runs in both directions

Time-dependent functions and instantaneous values of position, velocity, and acceleration can be determined using differentiation and integration. Two words, two directions, and the second one is easy to lose because Topic 1.2 prints no integral of its own. The integral forms are printed one topic later, at 1.3.A.4.iii and 1.3.A.4.iv.

Going down the chain is differentiation and is unconditional:

x(t) \ \xrightarrow{\ d/dt\ } \ v_x(t) \ \xrightarrow{\ d/dt\ } \ a_x(t)

Going up the chain is integration and costs you a constant at each step:

a_x(t) \ \xrightarrow{\ \int dt\ } \ v_x(t) + C_1 \ \xrightarrow{\ \int dt\ } \ x(t) + C_1 t + C_2

Each constant is fixed by an initial condition, and a problem that gives you an acceleration function and asks for position must supply two of them, usually $v_x(0)$ and $x(0)$ . Losing one is the most reliable way to produce a plausible wrong answer, because the shape of the function will be right.

Read the derivative relations backwards and they also locate features:

$v_x = 0$ marks a turning point in position, an instant where the object reverses. It is not the same as the object being at the origin, and it is not the same as the acceleration being zero.
$a_x = 0$ marks an extreme of velocity. If the acceleration changes sign there, the speed was at a maximum or a minimum.

Both of those are ordinary calculus applied to a physical function, and both appear in the worked examples below.

What the equation sheet prints for Topic 1.2, and the asymmetry in it

Nothing. Topic 1.2 prints seven equations across five essential-knowledge statements, and not one of them appears on the [AP Physics C: Mechanics Table of Information](/formulas/ap-physics-c-mechanics).

Equation	Statement	On the sheet
$\Delta x = x - x_0$	1.2.A.2	no
$\vec{v}_{\text{avg}} = \Delta\vec{x}/\Delta t$	1.2.B.2	no
$\vec{a}_{\text{avg}} = \Delta\vec{v}/\Delta t$	1.2.B.3	no
$\vec{v} = d\vec{r}/dt$	1.2.C.1.i	no
$v_x = dx/dt$	1.2.C.1.i	no
$\vec{a} = d\vec{v}/dt$	1.2.C.1.ii	no
$a_x = dv_x/dt$	1.2.C.1.ii	no

The kinematics block at the head of the Mechanics table is five lines, and here they are in the order the booklet prints them: $v_x = v_{x0} + a_xt$ ; $x = x_0 + v_{x0}t + \frac{1}{2}a_xt^2$ ; $v_x^2 = v_{x0}^2 + 2a_x(x - x_0)$ ; $\Delta x = \int v_x(t)\,dt$ ; $\Delta v_x = \int a_x(t)\,dt$ . Three constant-acceleration equations and two integrals. No derivative of position, and no derivative of velocity.

Now read the rotational column of the same table, and the first two lines are

\omega = \frac{d\theta}{dt} \qquad \alpha = \frac{d\omega}{dt}

The sheet defines the rotational rates as derivatives and leaves the translational ones to you. The mirror image holds for the integrals: the translational column prints both, and the rotational column prints neither. Neither omission is an accident you can exploit, and neither is a hint that the definition is optional. It is a reminder that the sheet is a reference table rather than a syllabus, and that checking whether something is printed is a separate question from whether you are expected to know it.

One caution if you go looking for the Vectors and Calculus tables: `src/data/equations.ts` on this site transcribes the physics equations and the constants boxes only, so a search of that file will not find them. They are printed in the real booklet. The derivative rules you need for this topic, including $\frac{d}{dx}(x^n) = nx^{n-1}$ and the chain rule, are in the Calculus table.

Traps on this topic

Substituting into a constant-acceleration equation when the acceleration is not constant. The three printed equations open with the words "For constant acceleration" at 1.3.A.2. If you have been handed $a_x(t)$ , they are not approximations, they are the wrong equations.

Confusing average speed with the magnitude of average velocity. They differ whenever the motion reverses, and a question that includes a reversal is usually testing exactly that.

Reading $v_x = 0$ as "stopped for good". It marks an instant, and the object is generally accelerating through it. At the top of a vertical throw the velocity is zero and the acceleration is not.

Reading $a_x = 0$ as "not moving" or "constant position". It means constant velocity, and a nonzero constant velocity is fast motion. Where $a_x$ crosses zero, the speed is at an extreme.

Dropping a constant of integration. Two integrations need two initial conditions.

Losing the units when you differentiate a graph. Skill 1.B here is quantitative graphing with scales and units, and units survive differentiation: metres per second divided by seconds is metres per second squared. A slope that does not carry those units is wrong before the arithmetic is checked.

Treating $\left| \vec{v} \right|$ as the thing being differentiated. Acceleration is $d\vec{v}/dt$ , not $d\left| \vec{v} \right|/dt$ . The two agree only for straight-line motion.

If you want the algebra-based version of this topic

Both courses have a Topic 1.2 with an identical title, and this is one place where saying so plainly is more useful than manufacturing a difference. The definitions of displacement, average velocity and average acceleration are the same definitions, in the same words, with the same three equations.

	AP Physics 1 Topic 1.2	AP Physics C Topic 1.2
Title	Displacement, Velocity, and Acceleration	Displacement, Velocity, and Acceleration
Learning objectives	1.2.A, 1.2.B	1.2.A, 1.2.B, 1.2.C
Objective 1.2.B wording	the velocity and acceleration	the average velocity and acceleration
Statements 1.2.A.1 to 1.2.B.5	identical text	identical text
Instantaneous values	1.2.B.5, as a close approximation	1.2.C, as a derivative
Equations printed	three	seven
Suggested skills	1.C, 2.B, 2.C, 3.C	1.B, 2.B, 2.C, 3.A, 3.C
Boundary statement	none	none

The skill lists differ in a way that matches the content. AP Physics C swaps 1.C, qualitative sketches of graphs, for 1.B, quantitative graphs with scales and units including plotting data, and adds 3.A, create experimental procedures. A course that expects you to fit a function to data needs the axes to have numbers on them.

If you are studying for AP Physics 1, the page you want is Displacement, Velocity, and Acceleration, and the average versus instantaneous velocity comparison serves both courses. If you are studying for AP Physics C: Mechanics, the next page is Topic 1.3, which prints the integrals this topic promised.

Watching an average converge on the instantaneous value

An object moves along the $x$ -axis with $x(t) = 2.0t^2$ , where $x$ is in metres and $t$ in seconds. Calculate the average velocity over intervals starting at $t = 3.0$ s of length 1.0 s, 0.1 s, 0.01 s and 0.001 s. Then find the instantaneous velocity at $t = 3.0$ s from the derivative, and show algebraically why the averages behave as they do.

Positions. $x(3.0) = 2.0(9.00) = 18.0$ m. $x(4.0) = 2.0(16.0) = 32.0$ m. $x(3.1) = 2.0(9.61) = 19.22$ m. $x(3.01) = 2.0(9.0601) = 18.1202$ m. $x(3.001) = 2.0(9.006001) = 18.012002$ m.
Average velocities, from $\vec{v}_{\text{avg}} = \Delta\vec{x}/\Delta t$ at 1.2.B.2. Over 1.0 s: $14.0/1.0 = 14.0$ m/s. Over 0.1 s: $1.22/0.1 = 12.2$ m/s. Over 0.01 s: $0.1202/0.01 = 12.02$ m/s. Over 0.001 s: $0.012002/0.001 = 12.002$ m/s.
Instantaneous velocity from 1.2.C.1.i. $v_x = dx/dt = 4.0t$ , so $v_x(3.0) = 12.0$ m/s.
The algebra behind the pattern. Over $[3.0,\ 3.0 + \Delta t]$ the average is $\dfrac{2.0(3.0 + \Delta t)^2 - 2.0(3.0)^2}{\Delta t} = 12.0 + 2.0\,\Delta t$ , so the error is exactly $2.0\,\Delta t$ and shrinks in proportion to the interval.
Read it against the two statements. 1.2.B.5 is the observation that 12.002 is very close to 12.0. 1.2.C.1 is the claim that the sequence has 12.0 as its limit, which is what makes the derivative a definition rather than an estimate.

The averages are 14.0, 12.2, 12.02 and 12.002 m/s against an instantaneous 12.0 m/s. The error is exactly $2.0\,\Delta t$ , so each tenfold shrink of the interval removes one factor of ten from it.

When the average equals the midpoint value, and when it does not

Object A has $x_A(t) = 3.0t^2 + 4.0t$ and object B has $x_B(t) = 0.50t^3$ , both in metres with $t$ in seconds. For each object find the average velocity over the interval from $t = 2.0$ s to $t = 6.0$ s, and compare it with the instantaneous velocity at the midpoint $t = 4.0$ s.

Object A positions. $x_A(2.0) = 3.0(4.0) + 8.0 = 20.0$ m and $x_A(6.0) = 3.0(36.0) + 24.0 = 132$ m.
Object A average. $v_{\text{avg}} = (132 - 20.0)/(6.0 - 2.0) = 112/4.0 = 28.0$ m/s.
Object A instantaneous at the midpoint. $v_A = dx_A/dt = 6.0t + 4.0$ , so $v_A(4.0) = 24.0 + 4.0 = 28.0$ m/s. The two agree, because $a_A = dv_A/dt = 6.0$ m/s $^2$ is constant and a linear $v_x(t)$ averages to its midpoint value.
Object B positions. $x_B(2.0) = 0.50(8.0) = 4.0$ m and $x_B(6.0) = 0.50(216) = 108$ m.
Object B average. $v_{\text{avg}} = (108 - 4.0)/4.0 = 26.0$ m/s.
Object B instantaneous at the midpoint. $v_B = 1.5t^2$ , so $v_B(4.0) = 1.5(16.0) = 24.0$ m/s, which is not 26.0 m/s. Its acceleration $a_B = 3.0t$ is not constant, so the midpoint rule does not apply.
Where object B does reach 26.0 m/s. Solve $1.5t^2 = 26.0$ to get $t = 4.16$ s, which is inside the interval but past its midpoint.

Object A: 28.0 m/s average and 28.0 m/s at the midpoint, equal because its acceleration is constant. Object B: 26.0 m/s average against 24.0 m/s at the midpoint. Object B does reach the average value, at $t = 4.16$ s rather than at $t = 4.0$ s.

Differentiating a position vector, component by component

A particle has position $\vec{r}(t) = \left( 4.0t \right)\hat{i} + \left( 5.0t - 1.0t^2 \right)\hat{j}$ , with $\vec{r}$ in metres and $t$ in seconds. Find $\vec{v}(t)$ and $\vec{a}(t)$ , the speed at $t = 0$ , and the instant and value of the particle's minimum speed.

Differentiate each component, per 1.2.C.1.i. $\vec{v} = d\vec{r}/dt = 4.0\hat{i} + \left( 5.0 - 2.0t \right)\hat{j}$ m/s.
Differentiate again, per 1.2.C.1.ii. $\vec{a} = d\vec{v}/dt = 0\hat{i} - 2.0\hat{j}$ , a constant $2.0$ m/s $^2$ in the $-y$ direction.
Speed at $t = 0$ . $\left| \vec{v}(0) \right| = \sqrt{(4.0)^2 + (5.0)^2} = \sqrt{41} = 6.4$ m/s.
Find the minimum speed. $\left| \vec{v} \right|^2 = (4.0)^2 + (5.0 - 2.0t)^2$ . The first term is fixed, so the speed is smallest when the second term is zero, at $5.0 - 2.0t = 0$ , that is $t = 2.5$ s.
Minimum speed. At $t = 2.5$ s the velocity is $4.0\hat{i}$ m/s alone, so the speed is 4.0 m/s.
Check it against 1.2.B.4. The acceleration is never zero, so the particle is accelerating at every instant, including at $t = 2.5$ s when its speed is momentarily stationary. What is changing there is the direction of $\vec{v}$ , not its magnitude.

$\vec{v} = 4.0\hat{i} + (5.0 - 2.0t)\hat{j}$ m/s and $\vec{a} = -2.0\hat{j}$ m/s $^2$ . The speed is 6.4 m/s at $t = 0$ and reaches a minimum of 4.0 m/s at $t = 2.5$ s, where the particle is still accelerating.

Frequently asked questions

What is the difference between AP Physics C Topic 1.2 and AP Physics 1 Topic 1.2?

The titles are identical and so are the first eight essential-knowledge statements, word for word. AP Physics C then adds a third learning objective, 1.2.C, which AP Physics 1 has no counterpart to: describe the instantaneous position, velocity, and acceleration of an object as a function of time. Under it, instantaneous velocity is defined as the derivative of position and instantaneous acceleration as the derivative of velocity. AP Physics C also inserts the word average into objective 1.2.B, because the instantaneous case now has its own objective to live in.

How does AP Physics C define instantaneous velocity?

As a derivative. Essential knowledge 1.2.C.1.i states that instantaneous velocity is the rate of change of the object's position, which is equal to the derivative of position with respect to time, and prints two equations for it: the vector form v = dr/dt and the component form vx = dx/dt. Statement 1.2.C.1 gives the underlying limit in words first: as the time interval used to calculate an average approaches zero, the average approaches the value of the quantity at that instant. AP Physics 1 stops at the approximation, that a very small interval gives a value very close to the instantaneous one.

Are dx/dt and dv/dt on the AP Physics C equation sheet?

No. The AP Physics C: Mechanics Table of Information prints five kinematics lines: the three constant-acceleration equations, and the two integrals for displacement and change in velocity. Neither derivative definition is printed, and neither is the definition of displacement or of average velocity or average acceleration. The same table does print the rotational derivatives, omega = d-theta/dt and alpha = d-omega/dt, in its rotational column, so the omission on the translational side is specific rather than a general policy about derivatives.

What is the difference between average and instantaneous velocity?

An average is a property of an interval and an instantaneous value is a property of an instant. Average velocity is the displacement divided by the time interval, so it depends only on the initial and final positions and ignores everything in between. Instantaneous velocity is the derivative of position at one moment. They coincide at the midpoint of an interval only when the acceleration is constant. For a position function that is cubic in time, the average over an interval is reached at some instant inside it, but not at the midpoint.

How do you find position from an acceleration function in AP Physics C?

Integrate twice, and use one initial condition at each step. Integrating the acceleration gives the velocity up to a constant, which is fixed by the initial velocity; integrating the velocity gives the position up to a second constant, fixed by the initial position. Essential knowledge 1.2.C.2 states that time-dependent functions and instantaneous values of position, velocity, and acceleration can be determined using differentiation and integration, and Topic 1.3 prints the two integral forms. Losing a constant of integration produces an answer with the right shape and the wrong values.

What does the object model mean in AP Physics C?

Essential knowledge 1.2.A.1 states that when using the object model, the size, shape, and internal configuration are ignored, and the object may be treated as a single point with extensive properties such as mass and charge. Extensive means the property scales with how much matter there is, so mass and charge survive the idealisation while geometry does not. The model stops being available when a question depends on the body's own shape, which is why the CED introduces the rigid system later, for rotation, where different points of one body move in different directions.

Can an object be accelerating while its speed is not changing?

Yes. Essential knowledge 1.2.B.4 states that an object is accelerating if either the magnitude and/or direction of its velocity are changing. Acceleration is defined as the derivative of the velocity vector, not of the speed, so a velocity of constant magnitude that is turning has a nonzero acceleration. Uniform circular motion is the standard case: the speed never changes and the acceleration points to the centre at every instant.