AP Physics 1 · Algebra-Based · CED Unit 1

Kinematics

How things move — before we ever ask why. Position, velocity, acceleration, graphs, and motion in two dimensions. Every later unit in this course speaks the language you learn here.

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42 SLIDES · 10–15% OF THE EXAM

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2 · Learning Objectives

By the end of this unit, you will be able to…

LO 1 Distinguish scalar quantities from vector quantities and add vectors in one and two dimensions.
LO 2 Define and calculate displacement, average velocity, instantaneous velocity, average acceleration, and instantaneous acceleration.
LO 3 Translate between motion diagrams, position–time, velocity–time, and acceleration–time graphs — and read slopes and areas correctly.
LO 4 Apply the three constant-acceleration kinematic equations, including free fall with g ≈ 10 m/s².
LO 5 Describe motion from different inertial reference frames and compute relative velocities.
LO 6 Resolve vectors into components and solve projectile-motion problems by treating x and y independently.

🎯

Why this unit matters: Kinematics is 10–15% of the multiple-choice section, but it's also the hidden 100% — every force diagram, energy bar chart, and rotation problem in Units 2–8 requires you to describe motion correctly first.

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3 · AP CED Mapping

Unit 1 → the five CED topics

Taken directly from the Unit at a Glance in the 2024 revised AP Physics 1 Course and Exam Description.

Topic	Suggested Skills	Slides
1.1 Scalars and Vectors in One Dimension	1.A, 2.C, 3.B, 3.C	5–7
1.2 Displacement, Velocity, and Acceleration	1.C, 2.B, 2.C, 3.C	8–13
1.3 Representing Motion	1.C, 2.A, 2.C, 3.B	14–25
1.4 Reference Frames and Relative Motion	1.C, 2.A, 2.B, 3.C	26–29
1.5 Vectors and Motion in Two Dimensions	1.B, 2.A, 2.D, 3.A, 3.C	30–37

Skill 1 · Creating Representations

1.A diagrams & schematics · 1.B quantitative graphs with scales and units · 1.C qualitative sketches of graphs

Skill 2 · Mathematical Routines

2.A derive a symbolic expression · 2.B calculate with units · 2.C compare quantities · 2.D predict factors of change

Skill 3 is Scientific Questioning: 3.A create an experimental procedure · 3.B apply a law to make a claim · 3.C justify a claim with evidence.

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4 · Key Terms

The vocabulary of motion

Scalar

A quantity with magnitude only: distance, speed, mass, time, energy. Fully described by a number and a unit.

Vector

A quantity with magnitude and direction: displacement, velocity, acceleration, force, momentum. Written with an arrow, e.g. v.

Position (x)

Where an object is, relative to a chosen origin. Has units of metres and a sign.

Displacement (Δx)

x − x₀. The change in position — a vector. Not the same as the distance travelled.

Velocity (v)

Rate of change of position. A vector: v = Δx/Δt. Its magnitude is speed.

Acceleration (a)

Rate of change of velocity. A vector: a = Δv/Δt. Any change in speed or direction means acceleration.

Free fall

Motion under gravity alone, near Earth's surface, with a = −g ≈ −10 m/s² (taking up as positive).

Projectile

An object moving under gravity alone after being launched — horizontal acceleration zero, vertical acceleration −g.

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5 · Topic 1.1 · Scalars and Vectors in One Dimension

Two kinds of quantity

Definition

A scalar is described completely by a magnitude (number + unit). A vector requires a magnitude and a direction. In one dimension, the sign of the number carries the entire directional message.

Scalar	Vector partner	Key difference
Distance (m)	Displacement (m)	Distance is path length, always ≥ 0. Displacement can be zero after a round trip.
Speed (m/s)	Velocity (m/s)	Speed = \|velocity\|. A car going 20 m/s east and 20 m/s west has the same speed, opposite velocity.
Mass (kg)	Force (N)	Force's vector nature is Unit 2's whole story.
Time (s)	Acceleration (m/s²)	Acceleration points wherever the velocity is heading, not where the object is going.

🍺

Meme break

A scalar walks into a bar. That's it. That's the whole joke — it had no direction. (A vector walked in too, but honestly, you had to be facing the right way to get it.)

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6 · Topic 1.1 · Vector Addition in One Dimension

Same direction adds; opposite direction subtracts

Rule

Same direction: |A + B| = |A| + |B| | Opposite: |A + B| = ||A| − |B||

The resultant points the same way as whichever vector has the larger magnitude.

Worked

Q1. A cart moves +12 m, then −5 m. Find its displacement and the distance travelled.

→ Displacement = (+12) + (−5) = +7 m (7 m to the right).

→ Distance = 12 + 5 = 17 m. Distance ignores sign entirely — that's what makes it a scalar.

Worked

Q2. An object moves +30 m, then −50 m. What is the resultant displacement?

→ Δx = (+30) + (−50) = −20 m. Magnitude 20 m, direction negative (left / down / west — whatever you defined as negative).

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7 · Checkpoint · Topic 1.1

Quick check: scalars and vectors

A student walks 40 m east, then 30 m west. Which of the following correctly gives the distance travelled and the magnitude of the displacement?

ADistance 10 m; displacement 70 m

BDistance 70 m; displacement 10 m

CDistance 70 m; displacement 70 m

DDistance 10 m; displacement 10 m

✅ Answer: B. Distance is the total path length: 40 + 30 = 70 m. Displacement is the net change in position: (+40) + (−30) = +10 m, magnitude 10 m east. Distance and displacement are equal only when the motion never reverses direction.

📝

AP habit: on any FRQ, always define your positive direction in writing ("taking east as positive") before you plug in signs. It costs two seconds and protects every sign that follows.

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8 · Topic 1.2 · Position and Displacement

Position, displacement, and distance

Definition

Δx = x − x₀

Displacement depends only on the endpoints — never on the path taken.

Δx is a vector: sign = direction. Distance is a scalar: always positive or zero.
If an object returns to its start, Δx = 0 but distance > 0.

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9 · Topic 1.2 · Velocity

Average velocity vs average speed

Definitions

v_avg = Δx ⁄ Δt = (x − x₀) ⁄ (t − t₀) | average speed = total distance ⁄ Δt

Velocity is a vector; speed is its magnitude.

Average velocity

Uses displacement. Can be positive, negative, or zero. Units: m/s.

Average speed

Uses distance. Always ≥ 0. Never negative — not even if the object moves backwards.

Worked

Q. A runner goes 200 m east in 25 s, turns around, and jogs 80 m west in 20 s. Find the average velocity and average speed for the whole trip.

1 Total displacement = (+200) + (−80) = +120 m. Total time = 45 s.

2 Average velocity = 120 / 45 = +2.67 m/s (east).

3 Total distance = 200 + 80 = 280 m. Average speed = 280 / 45 = 6.22 m/s.

✓ Note they are very different numbers. That gap is the entire point of the distinction.

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10 · Topic 1.2 · Instantaneous Velocity

Instantaneous velocity: the slope of x vs t

Definition

The instantaneous velocity at time t is the velocity over an infinitesimally small interval around t. On a position–time graph it equals the slope of the tangent line at that instant.

💡

No calculus needed. On the AP Physics 1 exam you will always read the instantaneous velocity from a graph by drawing a tangent line and computing its rise/run — or by being given a data table close enough to estimate it.

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11 · Topic 1.2 · Acceleration

Acceleration: how fast velocity changes

Definition

a_avg = Δv ⁄ Δt = (v − v₀) ⁄ (t − t₀)

Units: m/s². An object accelerates whenever its speed changes or its direction changes — or both.

Speeding up

v and a have the same sign. Moving right and accelerating right → faster. Moving left and accelerating left → faster.

Slowing down

v and a have opposite signs. The acceleration fights the motion.

Changing direction only

Uniform circular motion (Unit 2): speed constant, but velocity keeps turning, so a ≠ 0 and points inward.

Zero acceleration

Constant velocity — straight line, steady speed. Does not mean the object is at rest.

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12 · Topic 1.2 · The Most Tested Distinction

Negative acceleration ≠ slowing down

This single idea is responsible for more lost points in Unit 1 than anything else.

Velocity	Acceleration	What happens	Real example
+ (right)	+ (right)	Speeding up, moving right	Car leaving a stoplight
+ (right)	− (left)	Slowing down, moving right	Car braking while heading east
− (left)	− (left)	Speeding up, moving left	Car reversing out of a driveway faster and faster
− (left)	+ (right)	Slowing down, moving left	Reversing car hitting the brakes
+ (right)	0	Constant velocity	Cruise control on a straight highway

The rule, memorise it

Compare signs, not words. If v and a share a sign → speeding up. If v and a have opposite signs → slowing down. "Negative acceleration" tells you the acceleration points in the negative direction — nothing about whether the object is slowing.

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13 · Topic 1.2 · Worked Examples

Three worked problems

A · average acceleration

Q1. A cyclist goes from 4.0 m/s to 16.0 m/s in 6.0 s. Find the average acceleration.

→ a = Δv/Δt = (16.0 − 4.0)/6.0 = 12.0/6.0 = 2.0 m/s² in the direction of motion.

B · slowing down with signs

Q2. A car moving east at 24 m/s brakes to a stop in 4.0 s. Take east as positive.

→ v₀ = +24 m/s, v = 0, Δt = 4.0 s.

→ a = (0 − 24)/4.0 = −6.0 m/s². The negative sign means the acceleration points west — it opposes the motion, so the car slows. ✓

C · reverse applied

Q3. A cart has velocity −3.0 m/s and acceleration −1.5 m/s². Is it speeding up or slowing down, and which way is it moving?

→ v < 0 so it moves in the negative direction. v and a are both negative → same sign → speeding up while travelling in the negative direction.

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14 · Topic 1.3 · Representing Motion · Setup

The three equations — and their one condition

The condition that matters

These equations are valid only when acceleration is constant. If a is changing, they do not apply. Everything else on this slide assumes a is a fixed number.

Where they come from

a = Δv/Δt (definition) ⟶ v = v₀ + at

Combine that with v_avg = (v₀ + v)/2 (true only for constant a) and you get the other two.

Why v_avg = (v₀ + v)/2 works

For constant acceleration, velocity changes linearly with time, so the average of the endpoints is exactly the average over the whole interval. For non-constant a this is false.

Five variables

x₀, x, v₀, v, a, t — six quantities. Know any three (plus one you're solving for) and you can find the rest.

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15 · Topic 1.3 · The Kinematic Equations

Which equation should I use?

Equation	Missing quantity	Reach for it when…
v = v₀ + a t	Δx (position)	You are not given and not asked for displacement.
Δx = v₀ t + ½ a t²	v (final velocity)	You are not given and not asked for final velocity.
v² = v₀² + 2 a Δx	t (time)	Time is absent from the problem — the classic "how fast at the bottom?" question.
Δx = ½ (v₀ + v) t	a (acceleration)	You don't know a but you have both velocities and the time.

🧭

The three-step pick method: ① list what you know, ② list what you want, ③ choose the equation whose "missing quantity" is the one variable that appears in neither list. It takes five seconds and eliminates almost all wrong starts.

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16 · Topic 1.3 · Worked

Worked: choosing and solving

Worked

Q. A car starts from rest and accelerates uniformly at 3.0 m/s². How fast is it going after it has travelled 150 m?

1 Known: v₀ = 0, a = 3.0 m/s², Δx = 150 m. Want: v.

2 Time t is in neither list → use the equation with t missing.

3 v² = v₀² + 2aΔx = 0 + 2(3.0)(150) = 900.

4 v = √900 = 30 m/s (about 67 mph). ✓

✓ Sanity check: 30 m/s ≈ 108 km/h. Reasonable for a car accelerating hard over a football field and a half.

Same problem, different question

Q. How long did that 150 m take?

→ Now t is wanted, so use v = v₀ + at: 30 = 0 + 3.0t → t = 10 s.

→ Check with Δx = v₀t + ½at² = 0 + ½(3.0)(100) = 150 m. ✓ Consistent.

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17 · Topic 1.3 · Free Fall

Free fall: a = −g, and g ≈ 10 m/s²

On the AP Physics 1 exam

g ≈ 10 m/s² (taking upward as positive: a = −g = −10 m/s²)

All objects in free fall near Earth accelerate downward at the same rate, regardless of mass — neglect air resistance.

At the very top of the flight, velocity is zero — but acceleration is still −10 m/s². This is the single most common free-fall trap.
Going up: v > 0, a < 0 → slowing down. Coming down: v < 0, a < 0 → speeding up.

🎢

Meme break

Physics teachers love free fall because it's the only place you experience 9.8 m/s² of honest acceleration without paying for a rollercoaster ticket.

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18 · Free Fall · Worked A

Worked: dropped from rest

Worked

Q. A ball is dropped from rest from a 45 m tall building. (a) How long to hit the ground? (b) How fast on impact? Use g = 10 m/s², up positive.

1 Known: v₀ = 0, a = −10 m/s², Δx = −45 m (final position is below the start).

2 (a) Use Δx = v₀t + ½at²: −45 = 0 + ½(−10)t² = −5t².

3 t² = 9 → t = 3.0 s.

4 (b) v = v₀ + at = 0 + (−10)(3.0) = −30 m/s. Negative = downward, magnitude 30 m/s.

✓ Cross-check with v² = v₀² + 2aΔx = 0 + 2(−10)(−45) = 900 → v = ±30, and we know it's going down, so −30 m/s. ✓

⚠️

Sign discipline: if up is positive, then a = −10, and a falling object's Δx is negative. Two negatives multiply to give positive v² — which is exactly why the cross-check works. Never quietly drop the signs.

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19 · Free Fall · Worked B

Worked: thrown straight up

Worked

Q. A ball is thrown straight upward at 20 m/s from ground level. Find (a) max height, (b) time to the top, (c) total time in the air. Use g = 10 m/s².

1 Known: v₀ = +20 m/s, a = −10 m/s². At the top, v = 0.

2 (a) v² = v₀² + 2aΔx → 0 = 400 + 2(−10)Δx → 400 = 20Δx → Δx = 20 m.

3 (b) v = v₀ + at → 0 = 20 − 10t → t = 2.0 s to the top.

4 (c) Returning to the launch height is symmetric → total time = 2 × 2.0 = 4.0 s.

5 Check with Δx = v₀t + ½at² at t = 4: Δx = 20(4) + ½(−10)(16) = 80 − 80 = 0. ✓ Back to the start.

6 Impact speed: v = 20 − 10(4) = −20 m/s. Same magnitude as launch, opposite direction. That symmetry is free points on the exam.

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20 · Topic 1.3 · Position–Time Graphs

Reading an x vs t graph

The one rule

slope of x–t graph = velocity | value of the graph = position

Straight line → constant velocity. Curved → accelerating. Horizontal → at rest.

Positive slope ↗

Moving in the positive direction. Steeper = faster.

Negative slope ↘

Moving in the negative direction.

Zero slope →

At rest (position not changing).

Curvature

Concave up (∪) → a > 0. Concave down (∩) → a < 0. Curvature, not slope, tells you the acceleration.

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21 · Topic 1.3 · Velocity–Time Graphs

Reading a v vs t graph — two things at once

The two rules

slope of v–t = acceleration | area under v–t = displacement

Area above the axis counts positive; area below counts negative.

Displacement vs distance from a v–t graph

Add signed areas for displacement; add absolute areas for distance. If the graph crosses the axis, the two answers differ — and both might be asked.

Crossing the t-axis

Means the object changed direction. The moment of crossing is the turnaround point (v = 0).

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22 · Topic 1.3 · Acceleration–Time Graphs

a vs t: area gives Δv

The rule

Δv = area under the a–t graph = a · Δt (when a is constant)

The a–t graph is the least commonly asked of the three, but it's the fastest way to get Δv from a graph.

Worked

Q. An object has v = 4 m/s at t = 0. Its acceleration is +2 m/s² from t = 0 to 3 s, then −3 m/s² from t = 3 s to 7 s. Find v at t = 7 s.

1 Δv over [0, 3] = (2)(3) = +6 m/s → v(3) = 4 + 6 = 10 m/s.

2 Δv over [3, 7] = (−3)(4) = −12 m/s → v(7) = 10 − 12 = −2 m/s.

✓ Negative final velocity, so the object reversed direction somewhere between t = 3 s and t = 7 s — specifically when v hit 0, at t = 3 + 10/3 ≈ 6.33 s.

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23 · Topic 1.3 · Constant-Acceleration Graph Families

The three-graph family you must know cold

Type of motion	x vs t	v vs t	a vs t
At rest	horizontal line	horizontal line at 0	horizontal line at 0
Constant positive velocity	straight line, positive slope	horizontal line above axis	horizontal line at 0
Constant positive acceleration	parabola opening up (∪)	straight line, positive slope	horizontal line above axis
Constant negative acceleration	parabola opening down (∩)	straight line, negative slope	horizontal line below axis
Increasing acceleration	cubic-like, steeper and steeper	curve increasing (∪)	straight line, positive slope

🔗

The chain: slope takes you down one level (x → v → a) and area takes you up one level (a → v → x). Every graph-translation FRQ is just this chain applied twice.

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24 · Topic 1.3 · Motion Diagrams (Dot Diagrams)

Motion diagrams: a picture of the motion itself

A motion diagram is a series of dots showing the object's position at equal time intervals, with velocity arrows drawn between consecutive dots.

Dots are drawn at equal time intervals — the spacing is the velocity.
Velocity arrows (v) connect consecutive dots and grow or shrink with the spacing.
Acceleration arrows come from subtracting consecutive velocity arrows: Δv = v₂ − v₁.

🛹

Meme break

Newton’s First Law is basically "an object at home stays at home" — unless an unbalanced external force (mum) acts on it. (Inertia explains both mechanics and why you’re still in bed. Fnet = 0 is a lifestyle.)

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25 · Checkpoint · Topic 1.3

Reading graphs: AP-style MCQ

The velocity–time graph of a cart is a straight line starting at v = +6 m/s at t = 0 and reaching v = −6 m/s at t = 4 s, crossing zero at t = 2 s. What is the cart's displacement over the full 4 seconds, and how far did it travel?

ADisplacement 0 m; distance 12 m

BDisplacement 12 m; distance 12 m

CDisplacement 0 m; distance 24 m

DDisplacement 24 m; distance 24 m

✅ Answer: A. Sketch it: from 0–2 s the triangle above the axis has area ½(2)(6) = +6 m. From 2–4 s the triangle below has area ½(2)(6) = −6 m. Signed sum = 0 m (it returned to the start). Total distance uses absolute values: 6 + 6 = 12 m. The constant slope is a = Δv/Δt = (−6 − 6)/4 = −3 m/s².

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26 · Topic 1.4 · Reference Frames and Relative Motion

Motion depends on who's watching

Definition

A reference frame is the coordinate system and observer used to measure motion. An inertial reference frame is one that is not accelerating — one in which Newton's first law holds.

Frame-dependent (changes)

Position, velocity, displacement. Ask a person on the train and a person on the platform how fast the coffee cup is moving, and you get two different — both correct — answers.

Frame-independent (same for all inertial observers)

Acceleration, the time interval between two events, and any force. This is why physics works at all: everyone agrees on the acceleration, so everyone agrees on ΣF = ma.

🚂

The canonical example: you sit on a train moving at a constant 30 m/s and toss a ball straight up. To you it goes straight up and down. To someone on the platform it traces a parabola. Same event, two descriptions — both valid, because both frames are inertial.

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27 · Topic 1.4 · Relative Velocity in One Dimension

Adding velocities between frames

The subscript rule

v_AC = v_AB + v_BC

"Velocity of A relative to C = velocity of A relative to B + velocity of B relative to C." The middle letters must match and cancel.

Why the subscripts work

Write v_AC as "A w.r.t. C." Then AB + BC: the two B's are adjacent, so they cancel, leaving A…C. If the letters don't cancel, you've written it wrong.

Reverse a frame

v_AB = −v_BA. If the train moves at +30 m/s relative to the ground, the ground moves at −30 m/s relative to the train.

Worked

Q. A train moves east at 30 m/s. A passenger walks east at 2 m/s relative to the train. What's the passenger's velocity relative to the ground?

→ v_PT = +2, v_TG = +30. v_PG = v_PT + v_TG = 2 + 30 = +32 m/s east.

→ If the passenger walked west: v_PT = −2, so v_PG = −2 + 30 = +28 m/s east.

Original comic · ap-study.com

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28 · Topic 1.4 · Worked

Worked: two observers, one event

Worked

Q. A flatcar moves east at a constant 20 m/s. A student standing on the flatcar throws a ball straight up at 15 m/s relative to herself. Describe the ball's motion in (a) the student's frame and (b) the ground frame.

(a) Student's frame: the ball goes straight up at 15 m/s, decelerates at 10 m/s², stops after 1.5 s at height 11.25 m, and falls back into her hand. Time of flight 3.0 s. Horizontal velocity zero.

(b) Ground frame: the ball has v_x = 20 m/s (inherited from the flatcar) and v_y0 = +15 m/s. It follows a parabola.

→ Time of flight is still 3.0 s — vertical motion is unaffected by the constant horizontal velocity.

→ Horizontal range = (20)(3.0) = 60 m. The ball lands 60 m east of where it was thrown.

✓ Both observers agree the ball lands back in the student's hand — because the student also moved 60 m east in those 3 seconds. Physics is consistent.

🔑

Takeaway: both observers measure the same acceleration (−10 m/s² down) and the same time interval. Only position and velocity differ. That's what makes inertial frames special.

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29 · Topic 1.4 · Summary Table

Switching frames: what changes and what stays

Quantity	Changes between inertial frames?	Notes
Position	✅ Yes	Depends entirely on where you put the origin.
Displacement	✅ Yes	Two observers can disagree on Δx if their frames move relative to each other.
Velocity	✅ Yes	Add the relative velocity of the frames.
Acceleration	❌ No	All inertial observers agree — this is the foundation of Newton's laws.
Time interval	❌ No	In classical (non-relativistic) physics, time is universal.
Mass / force	❌ No	Same for everyone, so ΣF = ma gives the same physics.

Accelerating frames are not inertial. If your frame is speeding up, slowing, or turning, objects appear to accelerate with no force on them (the "fictitious force" you feel when a car brakes). AP Physics 1 stays in inertial frames.

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30 · Topic 1.5 · Vectors and Motion in Two Dimensions

Breaking a vector into components

Component resolution

θ is measured from the nearest positive x-axis. Always sketch first and check the sign of each component.

Components are independent — changing one never affects the other. This is the entire basis of projectile motion.
A vector pointing into quadrant II or III has a negative x-component; quadrant III or IV has a negative y-component. Let the sketch tell you the signs.

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31 · Topic 1.5 · Vector Addition

Head-to-tail, or add the components

Method 1 · Graphical (head-to-tail)

Place the tail of B at the head of A. The resultant runs from A's tail to B's head. Good for estimation and for showing your reasoning on an FRQ diagram.

Method 2 · Component (exact)

Add all x-components to get R_x, all y-components to get R_y, then |R| = √(R_x² + R_y²) and θ = tan⁻¹(R_y/R_x). This is what you use when numbers matter.

Worked

Q. Vector A is 30 m at 0°, vector B is 40 m at 90°. Find the magnitude and direction of R = A + B.

1 A_x = 30, A_y = 0. B_x = 0, B_y = 40.

2 R_x = 30 + 0 = 30; R_y = 0 + 40 = 40.

3 |R| = √(900 + 1600) = √2500 = 50 m.

4 θ = tan⁻¹(40/30) = tan⁻¹(1.333) = 53° above the +x axis.

✓ The classic 3-4-5 triangle. When you see 30 and 40, the answer 50 should feel instant.

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32 · Topic 1.5 · The Independence Principle

Horizontal and vertical motion are independent

Projectile motion — the two halves

Horizontal: a_x = 0, v_x = constant | Vertical: a_y = −g = −10 m/s²

The only thing the two directions share is time.

🎯

The most useful fact on the exam: a bullet fired horizontally and a bullet dropped from the same height hit the ground at the same instant. Horizontal velocity has zero effect on fall time.

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33 · Topic 1.5 · Horizontal Launch

Case 1: launched horizontally

Setup (up positive, launch height h, speed v₀)

v_x = v₀ | v_y0 = 0 | Δy = −h = −½gt² ⟹ t = √(2h/g)

Solve the vertical half for time first, then use that time in the horizontal half.

Find the time from the vertical motion. t = √(2h/g) — it depends only on the height, never on v₀.
Find the range from the horizontal motion. Δx = v₀ · t.
Find the impact velocity from both components. v_x = v₀, v_y = −gt, then |v| = √(v_x² + v_y²) and θ = tan⁻¹(v_y/v_x) below horizontal.

💡

Order matters. Time is the bridge between the two directions. Almost every projectile problem is "vertical → time → horizontal," in that order.

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34 · Horizontal Launch · Worked

Worked: ball rolled off a table

Worked

Q. A ball rolls off a 1.25 m high table at 4.0 m/s horizontally. (a) Time to hit the floor? (b) Horizontal distance? (c) Speed on impact? Use g = 10 m/s².

1 v_x = 4.0 m/s (constant). v_y0 = 0. Δy = −1.25 m.

2 (a) −1.25 = 0 − ½(10)t² = −5t² → t² = 0.25 → t = 0.50 s.

3 (b) Δx = (4.0)(0.50) = 2.0 m from the base of the table.

4 (c) v_y = −gt = −(10)(0.50) = −5.0 m/s. v_x = 4.0 m/s.

5 |v| = √(4.0² + 5.0²) = √(16 + 25) = √41 = 6.4 m/s.

✓ Direction: θ = tan⁻¹(5.0/4.0) = 51° below the horizontal.

Note the answer doesn't depend on the ball's mass at all. If the question adds "a second ball twice as heavy," both land together — a favorite conceptual MCQ.

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35 · Topic 1.5 · Launch at an Angle

Case 2: launched at an angle θ

Resolve first, then treat each direction separately

v_0x = v₀ cos θ | v_0y = v₀ sin θ | t_up = v₀ sin θ ⁄ g

Everything else follows from those three lines.

Maximum height

h_max = (v₀ sin θ)² ⁄ (2g). At the peak v_y = 0 but a_y is still −g.

Total time (level ground)

T = 2 v₀ sin θ ⁄ g — twice the time to the top.

Range (level ground)

R = v₀² sin(2θ) ⁄ g. Maximum range at θ = 45°.

Complementary angles

θ and (90° − θ) give the same range, because sin(2θ) = sin(180° − 2θ). Higher arc = longer flight, shorter arc = faster trip.

⚠️

Range formula conditions: R = v₀²sin(2θ)/g only works when the projectile lands at the same height it launched from. If it lands higher or lower, go back to solving the vertical quadratic for t. Using the shortcut off a cliff is a guaranteed wrong answer.

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36 · Angled Launch · Worked

Worked: kicked football

Worked

Q. A ball is kicked at 20 m/s at 30° above horizontal from level ground. Find (a) time of flight, (b) max height, (c) range. Use g = 10 m/s².

1 v_0x = 20 cos 30° = 20(0.866) = 17.3 m/s. v_0y = 20 sin 30° = 20(0.5) = 10.0 m/s.

2 (a) T = 2v_0y/g = 2(10)/10 = 2.0 s.

3 (b) h_max = v_0y²/(2g) = 100/20 = 5.0 m.

4 (c) R = v_x·T = (17.3)(2.0) = 34.6 m.

5 Check with the range formula: R = v₀²sin(2θ)/g = (400)(sin 60°)/10 = 400(0.866)/10 = 34.6 m. ✓

✓ Sanity: 34.6 m ≈ 38 yards — a believable punt. If you get 346 m, you missed a factor of ten somewhere.

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37 · Checkpoint · Topic 1.5

Projectile MCQ

Two balls are launched from the same height at the same instant. Ball A is dropped from rest. Ball B is fired horizontally at 15 m/s. Which statement is correct? (Neglect air resistance.)

ABall A hits the ground first, because it has no horizontal velocity to "waste."

BBoth hit the ground at the same time, but Ball B lands farther away.

CBall B hits the ground first, because it travels a longer path faster.

DBoth hit at the same time and at the same horizontal distance.

✅ Answer: B. Fall time depends only on the vertical equation Δy = −½gt², and both balls have v_y0 = 0 and the same height — so t = √(2h/g) for both. Ball B's 15 m/s horizontal velocity is constant and carries it a distance Δx = 15t away from the launch point before it lands. This is the independence principle in its purest form.

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38 · FRQ Type ① · Mathematical Routines

FRQ walkthrough: derive, then calculate

Prompt

Q. A cart of mass m starts from rest at the top of a ramp of length L inclined at angle θ. The cart is released and slides down a frictionless track. Derive an expression for the cart's speed at the bottom in terms of g, L, and θ only. Then calculate the speed for L = 4.0 m, θ = 30°.

1 pt Identify the acceleration along the ramp. For a frictionless incline, a = g sin θ. (Full derivation is Unit 2; in a Unit 1 question this would be given or measured.)

1 pt Pick the equation. Time isn't given or wanted → v² = v₀² + 2aΔx.

2 pts Derive symbolically. v² = 0 + 2(g sin θ)(L) → v = √(2gL sin θ). Do not plug numbers until the symbol is isolated.

1 pt Calculate with units. v = √(2 · 10 · 4.0 · sin 30°) = √(80 · 0.5) = √40 = 6.3 m/s.

1 pt Check. Units: √(m/s² · m) = √(m²/s²) = m/s ✓. Does it behave sensibly? θ = 0 → v = 0 ✓; larger L → faster ✓.

✍️

Grader behaviour: the symbolic derivation is worth more than the final number. If you go straight to numbers and get 6.3 with no algebra, you lose the derivation points even though the number is right.

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39 · FRQ Type ③ · Experimental Design & Analysis

FRQ walkthrough: design a lab

Prompt

Q. You are given a cart on a track, a meterstick, a stopwatch, and a phone with slow-motion video. Design an experiment to determine whether the cart's acceleration is constant. List the quantities you'd measure, the equipment for each, and how you'd analyse the data.

1 pt Quantities to measure: position x at a series of known times t. Measure x with the meterstick (or marks on the track), t with the video's timestamp or the stopwatch.

1 pt Procedure: release the cart from rest at x = 0; record video with the meterstick in frame; use frame-by-frame playback to read x at t = 0.0, 0.2, 0.4, 0.6, 0.8, 1.0 s. Repeat three times and average.

2 pts Analysis — linearise. If a is constant, x = ½at², so a graph of x versus t² should be a straight line through the origin with slope ½a. Plot it; if the data fall on a line, a is constant, and a = 2 × slope.

1 pt Alternative check: compute v = Δx/Δt for each interval and plot v versus t — a straight line of slope a confirms constant acceleration.

1 pt Uncertainty: repeat trials, use the spread of the slopes as the uncertainty in a, and state whether the result is consistent with the accepted value within that uncertainty.

📊

The linearisation trick is the highest-value skill on the whole exam. Whenever you're asked to analyse data, ask: "what do I plot against what to get a straight line?" Then read the physics off the slope and intercept.

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40 · Common Mistakes

Five traps that cost the most points in Unit 1

"Negative acceleration means slowing down"

False. Compare the signs of v and a. Same sign = speeding up, opposite signs = slowing. An object moving left and accelerating left is speeding up with a negative acceleration.

"At the top, acceleration is zero"

At the peak of a projectile's path, v_y = 0 — but a_y = −g the whole time, including at the top. If a were zero at the top, the ball would hover there forever.

Using distance and displacement interchangeably

Average speed uses distance; average velocity uses displacement. On a round trip, average velocity is exactly zero while average speed is not.

Using the constant-a equations when a isn't constant

The three kinematic equations require constant acceleration. If a changes (or if the problem gives you a v–t graph that isn't a straight line), use slopes and areas instead.

Mixing x and y quantities in projectiles

v₀ goes into BOTH the horizontal and vertical equations only after you resolve it into components. Putting the full v₀ into a vertical equation (or vice versa) is the #1 projectile error.

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41 · Video Resources

Four videos worth your time

Full lesson · 22 min

AP Physics 1 — Kinematics Review

Flipping Physics · Complete Unit 1 walkthrough with AP-style problems

Graphs · 18 min

Position, Velocity & Acceleration vs Time Graphs

The Organic Chemistry Tutor · Slope/area relationships worked in detail

Projectiles · 25 min

Projectile Motion — Complete Introduction

Professor Leonard · Every projectile case, derived slowly

Visual intuition · 12 min

Vectors and 2D Motion — The Basics

Khan Academy · Clean animations for vector components

AP Classroom also hosts official Unit 1 progress-check videos that use the exact wording of the exam — worth 20 minutes if you have access.

Original comic · ap-study.com

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42 · Coverage Checklist

Unit 1 · complete topic coverage

Every topic and skill in the CED Unit 1 Unit at a Glance, mapped to this deck.

1.1 Scalar vs vector quantities — definitions and the distance/displacement, speed/velocity pairs (slides 4–7)
1.1 Adding vectors in one dimension — same direction and opposite direction (slide 6)
1.2 Position, displacement Δx = x − x₀, and the object model (slide 8)
1.2 Average velocity, average speed, and instantaneous velocity (slides 9–10)
1.2 Average and instantaneous acceleration; speeding up vs slowing down by sign (slides 11–13)
1.3 The three constant-acceleration kinematic equations and how to choose (slides 14–16)
1.3 Free fall with g ≈ 10 m/s²; dropped and thrown-upward cases; top-of-flight behaviour (slides 17–19)
1.3 x–t graphs: slope = velocity; curvature = sign of acceleration (slide 20)
1.3 v–t graphs: slope = acceleration, signed area = displacement (slide 21)
1.3 a–t graphs: area = Δv (slide 22)
1.3 Graph families for constant acceleration; motion (dot) diagrams (slides 23–24)
1.4 Inertial reference frames; what changes and what doesn't between frames (slides 26, 29)
1.4 Relative velocity v_AC = v_AB + v_BC and worked frame changes (slides 27–28)
1.5 Resolving vectors into components; adding vectors head-to-tail and by components (slides 30–31)
1.5 Projectile motion: independence of x and y, horizontal launch, angled launch, range and max height (slides 32–37)
Skills FRQ types ① Mathematical Routines and ③ Experimental Design & Analysis, with linearisation (slides 38–39)

🎓

Unit 1 complete. Next: Unit 2 — Force and Translational Dynamics, the single heaviest unit on the exam at 18–23%. Everything you learned about acceleration now gets a cause: ΣF = ma.

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Unit 1 1 / 42 Kinematics