Unit 1 · Topic 1.2 · about 30 minutes · 7 parts · 14 scored checks, 3 of them multiple choice

AP Physics C: Mechanics 1.2: Displacement, Velocity, and Acceleration

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 1: 10 to 15% of the multiple-choice section

What you'll be able to do

Describe a change in an object's position, from two positions or from a position function $x (t)$ .
Describe an object's average velocity and average acceleration over a time interval.
Describe an object's position, velocity and acceleration at any instant as functions of time, moving between them by differentiating and integrating.

Key vocabulary

Term	What it means	The exam trap
Object model	A model that ignores an object's size, shape and internal configuration and treats it as a single point with properties such as mass.	The model is a choice that fits a question about where something is; it stops fitting when the object's turning or change of shape matters.
Displacement	The change in an object's position, $Δ x = x - x_{0}$ , final position minus initial position.	Displacement is a change, so a final position by itself is not a displacement unless the object started at the origin.
Average velocity	The displacement of an object divided by the time interval in which that displacement occurs, ${\vec{v}}_{avg} = \frac{Δ \vec{x}}{Δ t}$ .	An average velocity uses the displacement, not the distance traveled, and the length of the interval, not the clock reading at its end.
Average acceleration	The change in velocity divided by the time interval in which that change occurs, ${\vec{a}}_{avg} = \frac{Δ \vec{v}}{Δ t}$ .	A reversal changes the sign of the velocity, so subtract velocities with their signs, never the speeds.
Instantaneous velocity	The rate of change of position at one instant, the derivative $v_{x} = \frac{d x}{d t}$ , which an average over a shrinking interval approaches.	The velocity at the middle of an interval is generally not the average over it, unless the acceleration is constant.
Instantaneous acceleration	The rate of change of velocity at one instant, the derivative $a_{x} = \frac{d v_{x}}{d t}$ .	An acceleration is the derivative of the velocity, so it takes two derivatives of a position function, not one, to reach it.

A graph of position x in meters against time t in seconds, from 0 to 4 s, for x(t) = t cubed minus 6 t squared plus 9 t. The curve rises to 4 m at 1 s, falls to 0 at 3 s and rises again to 4 m at 4 s. Two secant lines start at the point (2 s, 2 m): one to (3 s, 0 m) with slope negative 2 meters per second and one to (2.5 s, 0.625 m) with slope negative 2.75 meters per second. A tangent line at 2 s has slope negative 3 meters per second.

Part 1 Think first The top of the throw

Commit to a guess first. Being wrong now makes the right idea stick better.

Predict first

You toss a ball straight up. For one instant at the very top, it is not moving up or down. At that instant, is the ball accelerating?

In short. Displacement is an object's change in position; average velocity and average acceleration divide a change by the time interval it took, and the instantaneous values are the derivatives $v_{x} = \frac{d x}{d t}$ and $a_{x} = \frac{d v_{x}}{d t}$ .

Treat the moving thing as a single point that keeps its mass, and find its displacement as $Δ x = x - x_{0}$ , final position minus initial position.
An average over an interval uses only the initial and final states; shrink the interval toward zero and the average becomes the instantaneous value, the derivative.
Differentiate to go from position to velocity to acceleration, and integrate, with a known starting value, to go back.

Part 2 Learn it Four ideas, with a check after each

2a. A point, and its change in position

Before you describe motion, decide what is moving. A car has spinning wheels and passengers who shift in their seats, but if you only care where the car is, you can use the object model: ignore its size, shape and internal configuration and treat it as a single point that keeps its mass.

In Physics C that point's position is usually given as a function of time. A cart might sit at $x (t) = 4 t^{2} - t$ , with $x$ in meters and $t$ in seconds. To find where it is at an instant, substitute that instant.

Displacement is the change in position, final minus initial:

Δ x = x - x_{0}

For that cart from $t = 1$ s to $t = 2$ s, $x_{0} = 3$ m and $x = 14$ m, so $Δ x = + 11$ m. The sign is a direction: positive means the cart ended up farther along the positive axis. Displacement ignores the path in between. A cart that goes out and comes back to where it started has zero displacement, however far it traveled.

Quick check · Fill the statement

A sled moves along a straight, horizontal track with east as positive. Its position is $x (t) = 2 t^{2} - 8 t + 3$ , with $x$ in meters and $t$ in seconds.

Modeled as an object, the sled's is ignored while its is kept. Its displacement from $t = 0$ to $t = 3$ s is .

2b. Averages use only the two ends

An average over an interval is calculated from the initial and final states alone, whatever happened in between.

Average velocity is displacement divided by the time interval in which it happens, and average acceleration is the change in velocity divided by its time interval:

{\vec{v}}_{avg} = \frac{Δ \vec{x}}{Δ t} {\vec{a}}_{avg} = \frac{Δ \vec{v}}{Δ t}

Both are vectors, so in one dimension both carry a sign. Find each change as final minus initial with the signs kept. A ball moving at $+ 4$ m/s that comes back at $- 4$ m/s has the same speed as before, but its velocity changed by $- 8$ m/s.

That is why an object is accelerating whenever the size or the direction of its velocity changes. A car rounding a curve at a steady 20 m/s is accelerating, because the direction of its velocity keeps turning.

Quick check · Calculate

A tennis ball flies toward a wall at 30 m/s and bounces straight back at 20 m/s. It touches the wall for 0.010 s. Take the direction toward the wall as positive. What is the ball's average acceleration while it touches the wall?

Your answer m/s²

2c. Shrinking the interval gives the derivative

The graph at the top of this page shows $x (t) = t^{3} - 6 t^{2} + 9 t$ , in meters and seconds. Average velocities over intervals that start at $t = 2$ s are slopes of secant lines, and they change as the interval shrinks:

Average velocity over intervals that start at 2 s, for $x (t) = t^{3} - 6 t^{2} + 9 t$
Interval	$Δ x$ (m)	$Δ t$ (s)	$v_{x, avg}$ (m/s)
2 s to 3 s	−2	1	−2
2 s to 2.5 s	−1.375	0.5	−2.75
2 s to 2.1 s	−0.299	0.1	−2.99
2 s to 2.01 s	−0.029999	0.01	−2.9999

An average over a very small interval is very close to the value at that instant. As $Δ t$ approaches zero, the average approaches the instantaneous velocity, which is the derivative of position:

v_{x} = \frac{d x}{d t}

Here $v_{x} = 3 t^{2} - 12 t + 9$ , and at $t = 2$ s that is $- 3$ m/s, the slope of the tangent line. The same limit applied to velocity gives the instantaneous acceleration, $a_{x} = \frac{d v_{x}}{d t}$ , here $6 t - 12$ .

To tell whether an object is speeding up at an instant, compare the signs of $v_{x}$ and $a_{x}$ there. The same sign means speeding up; opposite signs mean slowing down. The sign of $a_{x}$ alone does not tell you.

Quick check · Decide and justify

A cart moves along a track with its position given by $x (t) = 5 t^{2} - t^{3}$ , with $x$ in meters and $t$ in seconds. Take the direction of increasing $x$ as positive. At $t = 2$ s, is the cart speeding up or slowing down?

Which reason supports your answer?

2d. Integrating to work backward

Differentiation takes you from position to velocity to acceleration. Integration takes you back. Because $a_{x} = \frac{d v_{x}}{d t}$ and $v_{x} = \frac{d x}{d t}$ ,

v_{x} (t) = v_{x 0} + \int_{0}^{t} a_{x} (τ) d τ x (t) = x_{0} + \int_{0}^{t} v_{x} (τ) d τ

The integral alone gives a change: $\int_{t_{1}}^{t_{2}} v_{x} d t$ is the displacement over that interval, the signed area under the velocity graph. To get a position you add the starting position, which is the constant of integration fixed by what you know at one instant.

Suppose a cart has $a_{x} = 6 t$ in m/s² and starts from rest at $x_{0} = 1$ m. Then $v_{x} = 3 t^{2}$ and $x = 1 + t^{3}$ , so at $t = 2$ s it is at 9 m, not 8 m.

The constant-acceleration equations of Topic 1.3, such as $v_{x} = v_{x 0} + a_{x} t$ , hold only when the acceleration does not change. When the acceleration is a function of time, integrate.

Quick check · Calculate

A cart on a straight track has velocity $v_{x} (t) = 6 t - 3 t^{2}$ , with $v_{x}$ in meters per second and $t$ in seconds. At $t = 0$ it is at $x = 2$ m. Take the direction of increasing $x$ as positive. Where is the cart at $t = 3$ s?

Your answer m

Common mistakes

The mistake	Why it is wrong	Do this instead
Deciding that an object with zero velocity has zero acceleration.	Acceleration is the rate of change of velocity. A velocity can pass through zero while it is still changing, the way $v_{x} = 6 t - 3 t^{2}$ is zero at 2 s while $a_{x}$ there is $- 6$ m/s².	Find the acceleration from the derivative of the velocity at that instant, and do not judge it from the size of the velocity.
Leaving out the starting value after integrating.	An integral of velocity gives a displacement, a change in position. The position also depends on where the object started.	Write $x (t) = x_{0} + \int_{0}^{t} v_{x} (τ) d τ$ and use the known position at one instant to fix the constant.
Using a constant-acceleration equation when the acceleration changes with time.	An equation such as $v_{x} = v_{x 0} + a_{x} t$ assumes $a_{x}$ never changes. With $a_{x} = 2 t$ , using the acceleration at the end of the interval treats the whole interval as if it had that final value.	When $a$ or $v$ is given as a function of time, integrate it over the interval.

Part 3 Watch it done A drone climbs and levels off

Watch each step, then do the last one yourself.

A delivery drone rises straight up from a rooftop; take up as positive. Its acceleration is $a_{y} (t) = 6 - 2 t$ , with $a_{y}$ in m/s² and $t$ in seconds, for $0 \leq t \leq 6$ s. At $t = 0$ it is at rest at $y = 10$ m above the ground. How far does the drone rise in the first 3 s?

Step 1 · Integrate the acceleration
$v_{y} (t) = v_{y 0} + \int_{0}^{t} (6 - 2 τ) d τ = 0 + 6 t - t^{2}$ . The drone starts from rest, so the constant is zero. Check: differentiating $6 t - t^{2}$ gives back $6 - 2 t$ .
Step 2 · Read the velocity
At $t = 3$ s, $v_{y} = 18 - 9 = 9$ m/s. The acceleration $6 - 2 t$ is positive before 3 s and negative after, so 9 m/s is the drone's greatest upward velocity, and at 6 s it is back to $v_{y} = 36 - 36 = 0$ , hovering.
Why this step?
Why not average 0 and 9 m/s and multiply by 3 s?
Averaging the starting and ending velocities works only when the acceleration is constant, and here it falls from 6 m/s² to 0 over the first 3 s. The drone gains most of its speed early, so it covers more ground than a steady acceleration would. The displacement is the integral of the velocity.
Your step · Calculate

Finish it. How far does the drone rise from $t = 0$ to $t = 3$ s? Up is positive.

Your answer m

Part 4 Try it New situations

Try it · Rank it

Three carts move along parallel tracks. Each position is in meters with $t$ in seconds: cart A at $x = 3 t^{2}$ , cart B at $x = t^{3} + 5$ and cart C at $x = 8 t - t^{2}$ . Rank the carts by average velocity from $t = 0$ to $t = 2$ s.

Greatest first

Give each one a place, 1 for first. Give two the same place if they are equal.

Cart A
Cart B
Cart C

Your ranking: none yet

Try it · Calculate

A model rocket car's position on a straight track is $x (t) = 2 t^{3}$ , with $x$ in meters and $t$ in seconds. Take the direction of increasing $x$ as positive. What is its average velocity from $t = 1$ s to $t = 3$ s?

Your answer m/s

On paper. The rocket car's positions at $t = 1$ , 2 and 3 s are 2 m, 16 m and 54 m. Plot them, $x$ against $t$ , with scales and units on both axes, and draw a smooth curve through them. Then describe how you would use a motion sensor in your lab to estimate the car's velocity at exactly 2 s: which readings you would take, and why closer readings give a better estimate. There is nothing to submit here; compare your description with the sensor readings in Check yourself.

Part 5 Explain it Derive it, then put numbers in

Nothing you type leaves this page. Check your answer against the model and each scoring point.

Free response is 50% of the exam score. This one is shaped like Question 1, Mathematical Routines.

Free-response practice · Question 1 style

A cart moves along a straight track; take right as positive. Its velocity is $v_{x} (t) = 2 t^{2} - 8 t + 6$ , with $v_{x}$ in meters per second and $t$ in seconds, and at $t = 0$ it is at $x = 0$ .

(a) Derive an expression for the cart's position $x (t)$ .

(b) Calculate the cart's acceleration at $t = 3$ s.

(d) On paper, sketch $v_{x}$ against $t$ from 0 to 3 s. Label the velocity at $t = 0$ , 1, 2 and 3 s.

(e) A classmate says that because the average velocity in part (c) is zero, the cart was at rest the whole time. Use your answers to explain why the classmate is wrong.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) $x (t) = x_{0} + \int_{0}^{t} (2 τ^{2} - 8 τ + 6) d τ = \frac{2}{3} t^{3} - 4 t^{2} + 6 t$ , since $x_{0} = 0$ .

(b) $a_{x} = \frac{d v_{x}}{d t} = 4 t - 8$ , so $a_{x} (3) = 4$ m/s², to the right.

(d) A parabola that starts at 6 m/s at $t = 0$ , crosses zero at 1 s, reaches its lowest value, $- 2$ m/s, at 2 s, and comes back to zero at 3 s. Compare your sketch with that shape and those four labeled values.

(e) The velocity is not zero during the interval: it is $+ 6$ m/s at the start and $- 2$ m/s at 2 s. The cart moves right until 1 s, reaching $x = \frac{8}{3} \approx 2.7$ m, then moves left back to $x = 0$ at 3 s. Its average velocity is zero only because it ended where it started, after traveling about 5.3 m.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Integrates the velocity to $\frac{2}{3} t^{3} - 4 t^{2} + 6 t$ and uses $x (0) = 0$ to set the constant.
Differentiates the velocity to $4 t - 8$ and calculates 4 m/s², with its unit.
Calculates a displacement of 0 and an average velocity of 0 m/s from the positions at 0 and 3 s.
Sketches a curve through 6 m/s at 0 s, zero at 1 s, −2 m/s at 2 s and zero at 3 s, opening upward, with the axes labeled.
Explains that the cart moved right and then back left, so a zero average velocity comes from a zero displacement, not from being at rest.

Part 6 Check yourself A cart with a fan

The case questions use the sensor readings and the fit. Answer them from those.

A physics class sends a fan cart along a level track while one student turns the fan up steadily; take the cart's direction of travel as positive. A motion sensor records the cart's position, and the software fits the readings with $x (t) = 0.2 t^{3} + t$ , with $x$ in meters and $t$ in seconds. Some of the readings:

Motion sensor readings
$t$ (s)	$x$ (m)
0	0
1.0	1.200
2.0	3.600
2.9	7.778
3.0	8.400
3.1	9.058
4.0	16.800

Case question 1 · Multiple choice

Using the sensor readings, what is the cart's average velocity from $t = 0$ to $t = 4.0$ s? The direction of travel is positive.

Case question 2 · Calculate

Using the software's fit, what is the cart's instantaneous acceleration at $t = 3.0$ s? The direction of travel is positive.

Your answer m/s²

Case question 3 · Multiple choice

A student who does not trust the fit wants the best estimate of the cart's velocity at $t = 3.0$ s from the sensor readings alone. Which two readings should the student use?

The rest stand on their own and do not use the case.

Check yourself · Sort it

Tap a card, then say which operation finds what it asks for.

Differentiate

Integrate

Check yourself · Spot the error

A cart starts from rest and has acceleration $a_{x} = 2 t$ , with $a_{x}$ in m/s² and $t$ in seconds. A student finds its velocity at $t = 3$ s. Tap the first line with a mistake.

Part 7 Keep it Bring back earlier ideas

This lesson stands on vectors from Topic 1.1. Pulling them back up is what makes them last.

Bring it back · Multiple choice

Two displacements are $\vec{A} = (3 \hat{i} - 2 \hat{j})$ m and $\vec{B} = (- 1 \hat{i} + 5 \hat{j})$ m. What is $\vec{A} + \vec{B}$ ?

Bring it back · Sort it

Tap a card, then say whether the quantity is a scalar or a vector.

Scalar

Vector

In three lines

Displacement is final minus initial position, $Δ x = x - x_{0}$ , found from a position function by substituting both times.
Average velocity and acceleration use only the two ends of an interval; as the interval shrinks to zero they become $v_{x} = \frac{d x}{d t}$ and $a_{x} = \frac{d v_{x}}{d t}$ .
Integrate to go back, adding the starting value, and save the constant-acceleration equations for a constant acceleration.

Exam strategy for this topic

Section I, Multiple Choice: 42 questions in 85 minutes.
Section II, Free Response: 4 questions in 95 minutes.
This lesson's Explain it is shaped like free-response question 1, Mathematical Routines.
A four-function, scientific, or graphing calculator is allowed on both sections of the exam.

On a free-response question, write the calculus before the numbers: the derivative or the integral with its limits, then the substitution, then the result with its unit. When a question gives the acceleration or the velocity as a function of time, that is your signal to integrate rather than reach for a constant-acceleration equation.

Questions students ask

Do I really need calculus for this topic in AP Physics C?

Yes. The course defines instantaneous velocity and acceleration as derivatives, and expects you to find position, velocity and acceleration functions by differentiating and integrating. You will often take a derivative or an integral of a polynomial in time.

When can I still use the constant-acceleration equations?

Only when the acceleration does not change during the interval, such as an object in free fall near Earth's surface. If the acceleration is a function of time, integrate it instead.

What is the difference between displacement and distance traveled?

Displacement is final position minus initial position, with a sign. Distance adds up every meter moved in either direction. A cart that goes 3 m right and 3 m back has a displacement of 0 and a distance of 6 m.

Where does the constant of integration come from?

Integrating a velocity tells you how the position changes, not where the object began. The constant is the starting position, or any position you know at one instant, and it has to come from the question.

Can I estimate an instantaneous velocity from a data table?

Yes. Take the average velocity over the shortest interval that brackets the instant you want, ideally one centered on it. The smaller the interval, the closer the estimate.

Course alignment, for teachers

AP Physics C: Mechanics topic 1.2, Unit 1: Kinematics.

Skill 1.B: Create quantitative graphs with appropriate scales and units, including plotting data.
Skill 2.B: Calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway.
Skill 2.C: Compare physical quantities between two or more scenarios or at different times and/or locations within a single scenario.
Skill 3.A: Create experimental procedures that are appropriate for a given scientific question.
Skill 3.C: Justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

This lesson maps all 11 essential knowledge statements of the topic to the part that teaches each one: 11 are checked by a scored question. The map is in the lesson file.