Unit 1 · Topic 1.3 · about 30 minutes · 7 parts · 14 scored checks, 1 of them multiple choice

AP Physics 1: Algebra-Based 1.3: Representing Motion

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

A car pulling away from rest at 2 meters per second squared for 5 seconds, shown four ways: a motion diagram with gaps growing each second, a curved position graph with a tangent line marked slope equals v, a straight velocity graph from 0 to 10 meters per second with the triangle under it shaded as area 25 meters, and a level acceleration graph at 2 with the rectangle under it shaded as area 10 meters per second.

Part 1 Think first Where did the cyclist go?

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

Predict first

A cyclist rides along a straight road. Her position is measured from her house, positive away from it. Her position-time graph rises, stays level for a while, then falls back to zero. What did she do?

In short. Representing motion means describing an object's position, velocity and acceleration with motion diagrams, graphs, equations and words, and moving between them.

Velocity is the slope of a position-time graph, and acceleration is the slope of a velocity-time graph.
Displacement is the area under a velocity-time graph, and change in velocity is the area under an acceleration-time graph.
With constant acceleration, $v_{x} = v_{x 0} + a_{x} t$ , $x = x_{0} + v_{x 0} t + \frac{1}{2} a_{x} t^{2}$ and $v_{x}^{2} = v_{x 0}^{2} + 2 a_{x} (x - x_{0})$ hold; near Earth, gravity gives $a_{g} = g \approx 10 {m/s}^{2}$ downward.

Part 2 Learn it Four short ideas, each checked as you go

2a. One motion, many representations

You can show the same motion as a motion diagram, a figure, a graph, an equation or a sentence, and good physics moves between them.

A motion diagram marks where an object is at equal time steps, so each gap is how far it moved in one step. Equal gaps mean a steady speed, growing gaps mean speeding up, and shrinking gaps mean slowing down.

A position-time graph puts time across and position up. Its height at a moment is where the object is, measured from an origin you choose along a direction you call positive. It is not a picture of the path. A cart rolling along a flat floor at a steady speed has a position graph that climbs in a straight line, though the floor never climbs.

A sentence and an equation can say the same thing too. "The cart starts 2 m east of the sensor and rolls east at 3 m/s" is $x = 2 + 3 t$ , with east positive, $x$ in meters and $t$ in seconds. At $t = 2$ s the sentence puts the cart 6 m farther east, at 8 m, and so does the equation.

Quick check · Match

Each motion diagram marks a cart's position with a dot once every second. Every cart starts at the left end, and east is to the right. Tap a description of a cart's motion, then the dot spacing its motion diagram would show.

2b. Slope: how fast, and which way

An object's instantaneous velocity is the rate of change of its position: the slope of the line tangent to its position-time graph at that moment. On a curve, draw the tangent at the moment you care about and find its slope, as the diagram does at 2.5 s.

Steep means fast. A rising graph means motion in the positive direction, a falling one means the negative direction, and a level stretch means the object is stopped. The height says where; the slope says how fast and which way.

One level down, instantaneous acceleration is the rate of change of velocity, so it is the slope of the velocity-time graph. Its sign is a direction: an object speeds up when its velocity and acceleration have the same sign.

A ball tossed straight up shows why slope and height must stay separate. Take up as positive. Its velocity falls by 10 m/s every second, through zero at the top, then grows more negative on the way down. The velocity graph is one straight line with slope $- 10 {m/s}^{2}$ . At the top the velocity is zero but the slope is not, so the acceleration is still $10 {m/s}^{2}$ downward.

Quick check · Sketch the graph

A ball is tossed straight up at 20 m/s and caught 4 s later at the height it left your hand. Air resistance is negligible. Take up as positive. Sketch the ball's velocity against time from 0 to 4 s. The first point is fixed at 20 m/s.

Drag a point up or down, or tap it and use the arrow keys. The shape is checked, and the graph must pass through the values the question gives.

2c. Area: how far, and how much faster

Slopes take you down from position to velocity to acceleration; areas take you back up.

The displacement during a time interval is the area between the velocity-time graph and the time axis over that interval. A cart moving at a steady 3 m/s for 4 s has a rectangle of area $(3) (4) = 12$ m under its velocity graph, so it ends 12 m from where it started. The car in the diagram speeds up from rest to 10 m/s in 5 s, a triangle of area $\frac{1}{2} (5) (10) = 25$ m.

Area on the positive side of the axis counts as positive, and area on the negative side counts as negative. A cart that rolls forward and then back can end with no displacement at all, though it covered plenty of ground.

The change in velocity during an interval is the area under the acceleration-time graph, with the same sign rule. The diagram's car has $a = 2 {m/s}^{2}$ for 5 s, a rectangle of area 10 m/s: it goes from 0 to 10 m/s.

An area gives a change, not a starting value. Add the change in velocity to the velocity the object started with.

Quick check · Calculate

A cart on a level track is moving east at 1 m/s at $t = 0$ . Its acceleration-time graph is a level line at $+ 2 {m/s}^{2}$ from 0 to 3 s, then a level line at $- 1 {m/s}^{2}$ from 3 s to 5 s. Take east as positive. What is the cart's velocity at $t = 5$ s?

Your answer m/s

2d. Three equations, for constant acceleration only

When the acceleration is constant, three equations describe motion along one line:

v_{x} = v_{x 0} + a_{x} t

x = x_{0} + v_{x 0} t + \frac{1}{2} a_{x} t^{2}

v_{x}^{2} = v_{x 0}^{2} + 2 a_{x} (x - x_{0})

Each one leaves something out: the first has no position, the third has no time. Pick the one that leaves out what you neither know nor need. They work along any single axis; for vertical motion, write $y$ for $x$ .

Near Earth's surface, the acceleration caused by gravity is downward, constant and about $a_{g} = g \approx 10 {m/s}^{2}$ . Choose a positive direction before you write a sign. With up as positive, $a_{y} = - 10 {m/s}^{2}$ whether the object is rising, at the top or falling. Every question here uses 10, and a student who uses 9.8 or 9.81 m/s² is not penalized. The equation sheet you get on the exam prints $g = 9.8 {m/s}^{2}$ , while College Board's course description tells question writers to use 10. Both are official, and either value is accepted.

The equations need the acceleration to stay constant. A skydiver's acceleration shrinks as air resistance grows, so you cannot put her motion into them. Motion like that you describe and sketch from slopes and areas instead of calculating.

Quick check · Calculate

A phone slips out of someone's hand and falls from rest from a balcony 45 m above the ground. Air resistance is negligible. Take down as positive. How fast is the phone moving just before it hits the ground? Give your answer to the nearest 0.1 m/s.

Your answer m/s

Part 3 Watch it done A cart up a ramp and back

Watch each step, then draw the graph yourself.

You push a cart at the bottom of a long, smooth ramp, and it leaves your hand moving up the ramp at 3 m/s. Friction is negligible, so its acceleration is 1.5 m/s² down the ramp the whole time, going up and coming back down. Take up the ramp as positive and the bottom as $x = 0$ . Describe its motion for the first 4 s, then sketch its position graph.

Step 1 · Write the acceleration with its sign
Down the ramp is the negative direction, so $a_{x} = - 1.5 {m/s}^{2}$ . It keeps that value on the way up, at the top and on the way down.
Step 2 · Find when the cart stops
At the highest point $v_{x} = 0$ . From $v_{x} = v_{x 0} + a_{x} t$ : $0 = 3 + (- 1.5) t$ , so $t = 2$ s. After that the velocity is negative and the cart rolls back down.
Why this step?
Why is the acceleration still $- 1.5 {m/s}^{2}$ at $t = 2$ s, when the cart is not moving?
Because its velocity is still changing. Just before 2 s the cart moves up the ramp, and just after it moves down. Acceleration is the rate of change of velocity, and the velocity graph keeps its slope as it passes through zero.
Step 3 · Find how far up the ramp it gets
Use $x = x_{0} + v_{x 0} t + \frac{1}{2} a_{x} t^{2}$ at $t = 2$ s: $x = 0 + 3 (2) + \frac{1}{2} (- 1.5) (2)^{2} = 6 - 3 = 3$ m. The area under the velocity graph agrees: a triangle of $\frac{1}{2} (2) (3) = 3$ m.
Your step · Sketch the graph
Finish it. A cart leaves the bottom of a smooth ramp moving up the ramp at 3 m/s, and its acceleration is 1.5 m/s² down the ramp the whole time. Up the ramp is positive, and the bottom is position 0. Sketch the cart's position against time from 0 to 4 s. The first point is fixed at 0.

Drag a point up or down, or tap it and use the arrow keys. The shape is checked, and the graph must pass through the values the question gives.

Part 4 Try it Launch it straight up

Lab

Projectile Lab

Set Launch angle above horizontal to 90 so the ball flies straight up, and Launch height to 0. Switch the Graph menu to Velocity components against time. Launch at 10 m/s, then at 20 m/s. The lab redraws its axes to fit each launch, so the two lines look alike; read the numbers on the axes instead. Note where the vertical velocity starts, when it reaches zero and where it ends, and work out the slope each time.

Open the full Projectile Lab lab

Try it · Decide and justify

In the Projectile Lab, a ball is launched straight up. A second ball is then launched straight up from the same height at twice the launch speed. Air resistance is ignored, and up is positive. How do the slopes of their vertical velocity graphs compare?

Which reason supports your answer?

Now go the other way, from a graph to a graph. A motion sensor records a cart on a straight track, with east as positive. Its position graph is three straight pieces.

Position of the cart

Three straight pieces: up to 6 m by 3 s, level until 5 s, back to 0 m at 7 s.

Try it · Sketch the graph

Use the cart's position graph above. Sketch its velocity against time. Each point is the velocity halfway through a second: at 0.5 s, 1.5 s, and so on up to 6.5 s.

Drag a point up or down, or tap it and use the arrow keys. The shape is checked, and the graph must pass through the values the question gives.

Practice

Practice until it is automatic

Constant-acceleration problems with new numbers each time. List what you know and what you want, then pick the equation that leaves out the rest.

Motion with constant acceleration practice page

Part 5 Explain it Graph, equation, and why they agree

Nothing you type leaves this page.

Free response is 50% of the exam score. This one is shaped like Question 2, Translation Between Representations.

Free-response practice · Question 2 style

A cyclist riding east at 12 m/s brakes with a constant acceleration and stops 4 s later. Take east as positive and the point where she starts braking as $x = 0$ .

(a) Describe her velocity-time graph from the start of braking until she stops: where it starts, where it ends, its shape and its slope.

(b) A classmate writes her position during braking as $x = 12 t + 1.5 t^{2}$ , with $x$ in meters and $t$ in seconds. Use your graph to explain why this equation cannot describe her motion, and write one that can.

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

Model answer

(a) A straight line from 12 m/s at $t = 0$ down to 0 at $t = 4$ s. Its slope is $\frac{0 - 12}{4} = - 3 {m/s}^{2}$ , which is her acceleration: constant, and pointing west. (b) In $x = x_{0} + v_{x 0} t + \frac{1}{2} a_{x} t^{2}$ the number in front of $t^{2}$ is half the acceleration. My graph's slope is negative, so that term must be negative too. The classmate's $+ 1.5 t^{2}$ means an acceleration of $+ 3 {m/s}^{2}$ , east, which would speed her up. The equation is $x = 12 t - 1.5 t^{2}$ . (c) The area under the graph is a triangle, $\frac{1}{2} (4) (12) = 24$ m. The equation at $t = 4$ s gives $12 (4) - 1.5 (4)^{2} = 48 - 24 = 24$ m, the same stopping distance.

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

Describes a straight line from 12 m/s at 0 s down to 0 at 4 s.
Gives the slope as $- 3 {m/s}^{2}$ and says the slope is the acceleration.
Explains that the $t^{2}$ term is half the acceleration, so it must be negative, and writes $x = 12 t - 1.5 t^{2}$ .
Finds the stopping distance, 24 m, from the area of the triangle under the graph.
Shows that the equation at 4 s also gives 24 m.

Part 6 Check yourself A toy car on a track

Both case questions use the toy car's velocity table.

A student tracks a toy car on a straight track with a motion sensor, taking east as positive. The car starts at rest, drives west, and comes to rest again. Within each 2-second stretch (0 to 2 s, 2 to 4 s and 4 to 6 s) its acceleration is constant, so its velocity changes steadily between readings.

Velocity of the toy car
Time (s)	Velocity (m/s)
0	0
1	−2
2	−4
3	−4
4	−4
5	−2
6	0

Case question 1 · Calculate

Using the toy car's velocity table, with east as positive, what is the car's displacement from 0 to 6 s?

Your answer m

Case question 2 · Multiple choice

Using the toy car's velocity table, with east as positive, what is the car's acceleration from 0 to 2 s?

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

Check yourself · Spot the error

A runner's position-time graph has three straight segments: from 0 m at 0 s up to 20 m at 4 s, level at 20 m until 6 s, then down to 0 m at 10 s. Position is measured from the starting line, positive away from it. A student describes the run. Tap the line with the mistake.

Check yourself · Sort it

Tap each task, then whether you would find the answer from the slope of the graph named or from the area under it.

From a slope

From an area

Check yourself · Sketch the graph

A skydiver steps out of a hovering helicopter. Air resistance on her grows as she speeds up, so her downward acceleration starts near 10 m/s² and shrinks toward zero. For the whole first 10 s she is still speeding up, by less each second. Take down as positive. Sketch her velocity against time for those 10 s. The first point is fixed at 0.

Drag a point up or down, or tap it and use the arrow keys. Only the shape is checked.

Part 7 Keep it Bring back earlier ideas

Pulling an older idea back up is what makes it last.

From 1.2 Displacement, Velocity, and Acceleration · Calculate

An elevator moves in a vertical shaft. Take up as positive, with positions measured from the ground floor. When a clock reads 4 s the elevator is at 30 m, and when the clock reads 12 s it is at 6 m. What is its average velocity over that interval?

Your answer m/s

From 1.1 Scalars and Vectors in One Dimension · Decide and justify

Take right as positive. A cart rolls to the left at 4 m/s. Lena writes its velocity as $v_{x} = - 4$ m/s. Jordan says she must also put an arrow over the $v$ , or her answer has no direction. Who is right?

Which reason supports your answer?

In three lines

Velocity is the slope of a position-time graph; acceleration is the slope of a velocity-time graph.
Displacement is the area under a velocity-time graph; change in velocity is the area under an acceleration-time graph.
The three kinematic equations need constant acceleration, and near Earth gravity's is $g \approx 10 {m/s}^{2}$ downward.

Vocabulary

Motion diagram: A picture that marks an object's position at equal time intervals, so the spacing between marks shows how fast it moves.
Instantaneous velocity: The rate of change of an object's position, equal to the slope of the line tangent to its position-time graph at that moment.
Instantaneous acceleration: The rate of change of an object's velocity, equal to the slope of the line tangent to its velocity-time graph at that moment.
Displacement: The change in an object's position during a time interval, equal to the area between its velocity-time graph and the time axis over that interval.
Change in velocity: How much an object's velocity changes during a time interval, equal to the area under its acceleration-time graph over that interval.
Kinematic equations: The three equations that describe motion in one dimension when the acceleration is constant, and only then.
Acceleration caused by gravity: The vertical acceleration of an object near Earth's surface due to gravity, downward and constant, $a_{g} = g \approx 10 {m/s}^{2}$ .
Nonuniform acceleration: Acceleration that changes with time, which you describe and sketch from graphs rather than calculate with the kinematic equations.

Course alignment, for teachers

AP Physics 1: Algebra-Based topic 1.3, Unit 1: Kinematics.

Skill 1.C: Create qualitative sketches of graphs that represent features of a model or the behavior of a physical system.
Skill 2.A: Derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway.
Skill 2.C: Compare physical quantities between two or more scenarios or at different times and locations in a single scenario.
Skill 3.B: Apply an appropriate law, definition, theoretical relationship, or model to make a claim.

This lesson maps all 8 essential knowledge statements of the topic to the part that teaches each one: 8 are checked by a scored question. The map is in the lesson file.

Part 1 Think first Where did the cyclist go?

Part 2 Learn it Four short ideas, each checked as you go

2a. One motion, many representations

2b. Slope: how fast, and which way

2c. Area: how far, and how much faster

2d. Three equations, for constant acceleration only

Part 3 Watch it done A cart up a ramp and back

Part 4 Try it Launch it straight up

Projectile Lab

Practice until it is automatic

Part 5 Explain it Graph, equation, and why they agree

Part 6 Check yourself A toy car on a track

Part 7 Keep it Bring back earlier ideas

Vocabulary

Past exam questions on this topic

Labs that use this topic