Unit 1: Kinematics

Describing Motion: Position, Time, and Reference Frames

Kinematics is the study of motion without analyzing what causes it. In Unit 1, the goal is to describe how an object moves using measurable quantities such as position, displacement, velocity, and acceleration, and to translate smoothly among equations, graphs, and verbal descriptions.

The most basic idea is position: where an object is located relative to a chosen origin. In one dimension, position is usually labeled with a coordinate such as

$x$

Kinematics always starts by choosing a reference frame, meaning a coordinate system (origin and positive direction) plus a clock. The choice is up to you, but once you choose it, you must stay consistent.

Position

Position is the location of an object relative to a chosen reference point. A coordinate system is typically used to show where an object is located.

For example, if an object is at

$x = 2\ \text{m}$

that means it is 2 meters from the origin in the positive direction (based on your sign convention).

Position, Displacement, and Distance

Displacement is the change in position from an initial position to a final position:

$\Delta x = x_f - x_i$

Displacement is a vector (direction matters). In 1D, direction is represented with a sign. A positive displacement means the final position is in the positive direction relative to the initial position; a negative displacement means the opposite. Displacement can be represented graphically as an arrow pointing from the initial position to the final position.

Distance is the total length of the path traveled. Distance is a scalar (magnitude only) and can never be negative. It is measured in units like meters, kilometers, or miles. Distance is often denoted by the letter s or x in some contexts, but AP kinematics equations are built around displacement rather than total distance traveled.

Distance and displacement are only the same if the motion is in one direction with no reversals.

Scalars vs. Vectors (and Sign Conventions)

A scalar has magnitude only (examples: mass, temperature, time, speed, distance, energy, power).

A vector has magnitude and direction (examples: displacement, velocity, acceleration, force, momentum). Vectors are often represented graphically as arrows: the arrow length shows magnitude and the arrow direction shows direction. Vector quantities can be added and subtracted using vector algebra that accounts for both magnitude and direction.

In AP Physics 1, 1D vectors are typically handled with signs. If right is positive, then leftward velocities and accelerations are negative.

A key misconception to avoid: “negative” does not automatically mean “slowing down.” Negative usually means “pointing in the negative direction.” Whether something speeds up or slows down depends on the relationship between velocity and acceleration.

Time and Time Intervals

Time is usually represented by

$t$

and a time interval is

$\Delta t = t_f - t_i$

In typical AP problems, $\Delta t$ is positive because time moves forward.

Worked Example: Displacement vs. Distance

You walk from $x = 0\ \text{m}$ to $x = 6\ \text{m}$ , then back to $x = 2\ \text{m}$ .

Displacement:

$\Delta x = x_f - x_i = 2 - 0 = 2\ \text{m}$

Distance traveled: 6 m out and 4 m back, so distance is

$10\ \text{m}$

This is why “how far did you go?” and “where did you end up relative to where you started?” are fundamentally different.

Exam Focus

Typical question patterns:
- Translate a written motion description into displacement, distance, or a sign convention.
- Compare distance and displacement for motion with turnarounds.
- Choose a coordinate system and justify the sign of quantities.
Common mistakes:
- Treating distance as if it can be negative.
- Using displacement and distance interchangeably in equations.
- Changing the positive direction mid-problem.

Velocity and Speed: How Fast and Which Direction

Once you can describe where something is, the next question is how fast its position changes. That leads to velocity and speed.

Speed (Average Speed)

Speed describes how fast an object moves and is a scalar. Average speed is total distance traveled divided by elapsed time:

$speed_{avg} = \frac{\text{distance}}{\Delta t}$

You may also see this written generically as

$S = \frac{D}{t}$

The SI unit of speed is

$\text{m/s}$

Average speed is always nonnegative.

Velocity (Average Velocity)

Velocity describes the rate of change of position and is a vector. Average velocity is displacement divided by elapsed time:

$v_{avg} = \frac{\Delta x}{\Delta t}$

You may also see this written generically as

$V = \frac{x}{t}$

but in physics it is important to interpret this as displacement over time (not just “position divided by time” in every situation).

Average velocity depends only on starting position, ending position, and total time. It ignores “wiggles” in between if they do not change the net displacement.

Instantaneous Velocity (Conceptually)

Instantaneous velocity is the velocity at a specific moment. In AP Physics 1, you interpret it from a position-versus-time graph as the slope of the tangent line at that time. If you zoom in on a curved position-time graph, a small region looks nearly straight, and its slope corresponds to the instantaneous velocity.

Why Sign Matters (Direction vs. Speeding Up)

The sign of velocity indicates direction. Speeding up or slowing down depends on whether the magnitude of velocity increases or decreases.

If velocity and acceleration point in the same direction, the object speeds up.
If velocity and acceleration point in opposite directions, the object slows down.

Worked Example: Average Velocity vs. Average Speed

A runner moves from $x = 0\ \text{m}$ to $x = 100\ \text{m}$ in 20 s, then back to $x = 40\ \text{m}$ in the next 10 s.

Total time:

$\Delta t = 30\ \text{s}$

Displacement:

$\Delta x = 40 - 0 = 40\ \text{m}$

Average velocity:

$v_{avg} = \frac{40}{30} = 1.33\ \text{m/s}$

Distance traveled: $100\ \text{m}$ out plus $60\ \text{m}$ back gives

$160\ \text{m}$

Average speed:

$speed_{avg} = \frac{160}{30} = 5.33\ \text{m/s}$

These are very different because the runner reversed direction.

Exam Focus

Typical question patterns:
- Compute average velocity from initial and final positions (often with a turnaround included).
- Interpret the sign of velocity from motion descriptions or graphs.
- Distinguish average speed from average velocity in conceptual questions.
Common mistakes:
- Using distance instead of displacement in $v_{avg} = \Delta x/\Delta t$ .
- Assuming “average velocity” means averaging a list of speeds without considering direction.
- Confusing “negative velocity” with “slowing down.”

Acceleration: How Velocity Changes

Velocity tells you how position changes. Acceleration tells you how velocity changes.

Calculating Acceleration (Average Acceleration)

Average acceleration is change in velocity over change in time:

$a_{avg} = \frac{\Delta v}{\Delta t}$

where

$\Delta v = v_f - v_i$

This is also commonly written as

$a = \frac{v_f - v_i}{t}$

when the time interval is understood.

Units of Acceleration

The SI unit of acceleration is

$\text{m/s}^2$

Other units you might encounter include

$\text{ft/s}^2$

and

$\text{(km/h)}^2$

Positive and Negative Acceleration (What the Sign Really Means)

Acceleration is a vector, so it can be positive or negative depending on the chosen positive direction.

A common oversimplification is “speeding up means positive acceleration and slowing down means negative acceleration.” The correct rule is:

If acceleration points in the same direction as velocity, the object speeds up.
If acceleration points opposite velocity, the object slows down.

Whether that corresponds to a positive or negative number depends on your coordinate choice.

Instantaneous Acceleration (Conceptually)

Instantaneous acceleration is the acceleration at a specific moment. Graphically, it is the slope of a velocity-versus-time graph at that moment.

Uniform vs. Non-Uniform Acceleration

Uniform (constant) acceleration means acceleration stays constant over time, so velocity changes by the same amount in each equal time interval.

Non-uniform acceleration means acceleration changes over time. Analyzing non-uniform acceleration in full generality involves calculus, but AP Physics 1 often handles it using graphs or by splitting the motion into piecewise-constant intervals.

A useful special-case relationship: if an object starts from rest and accelerates constantly so that its displacement is s in time t, then

$s = \frac{1}{2}at^2$

which can be rearranged to

$a = \frac{2s}{t^2}$

This does not apply to every situation; it relies on starting from rest and having constant acceleration.

Acceleration Does Not Mean “Moving”

You can have acceleration even when velocity is zero at an instant. For example, at the top of a vertical toss, the instantaneous velocity is zero but the acceleration is still downward due to gravity.

Worked Example: Velocity Zero but Acceleration Nonzero

A ball is thrown straight up. At the highest point:

Velocity is momentarily $0\ \text{m/s}$ .
Acceleration is approximately $-9.8\ \text{m/s}^2$ if up is positive.

So the ball is not “stuck.” Its velocity is changing from positive (upward) to negative (downward), and that change is caused by downward acceleration.

Exam Focus

Typical question patterns:
- Use a velocity-time graph to find acceleration from slope.
- Determine whether an object is speeding up or slowing down from signs of $v$ and $a$ .
- Analyze vertical motion at special points (launch, peak, landing).
Common mistakes:
- Thinking acceleration must be zero when velocity is zero.
- Treating acceleration as always opposite motion (it is opposite only in specific cases like slowing down).
- Mixing up “negative acceleration” with “decreasing speed.”

Motion Graphs: Turning Pictures into Physics

AP Physics 1 emphasizes representing motion in multiple ways. Graphs are compact descriptions of relationships between kinematic quantities.

The three core graphs are position vs. time, velocity vs. time, and acceleration vs. time. Two big ideas make these graphs powerful:

Slope tells you a rate of change.
Area under a curve tells you an accumulated change.

Position-Time Graphs

On a position-versus-time graph, the slope represents velocity.

To determine which way an object is moving, look at whether the graph slopes upward (positive velocity) or downward (negative velocity).
A steeper slope means a larger magnitude of velocity.
A straight-line slope means constant velocity.
A curved line means velocity is changing, so acceleration is not zero.
A zero slope (horizontal line) means the object is at rest.
The y-intercept corresponds to the object’s initial position.

A common trap is thinking a curved position-time graph means a curved physical path. In 1D, a curved graph means changing velocity, not a turning trajectory in space.

Velocity-Time Graphs

A velocity-versus-time graph depicts acceleration.

The slope represents acceleration.
The area under the curve represents displacement.

Displacement from $t_i$ to $t_f$ is

$\Delta x = \int_{t_i}^{t_f} v\,dt$

In AP Physics 1 you typically compute this using geometry (rectangles, triangles, trapezoids). This area is signed: negative velocity contributes negative displacement.

Acceleration-Time Graphs

On an acceleration-versus-time graph, the area under the curve represents change in velocity:

$\Delta v = \int_{t_i}^{t_f} a\,dt$

Again, this is often handled with geometric area for constant or piecewise-constant acceleration.

Connecting the Graphs

These representations are linked:

The slope of $x(t)$ gives $v(t)$ .
The slope of $v(t)$ gives $a(t)$ .
The area under $v(t)$ gives change in position.
The area under $a(t)$ gives change in velocity.

If acceleration is constant and positive, you should expect: an acceleration-time graph that is a horizontal line above zero, a velocity-time graph that is a straight line with positive slope, and a position-time graph that curves upward (increasing slope).

Worked Example: Displacement from a Velocity-Time Graph

An object’s velocity increases linearly from $2\ \text{m/s}$ to $10\ \text{m/s}$ over 4 s. Find displacement.

For linear change, the average velocity is

$v_{avg} = \frac{v_i + v_f}{2} = \frac{2 + 10}{2} = 6\ \text{m/s}$

Then displacement is

$\Delta x = v_{avg}\Delta t = 6 \cdot 4 = 24\ \text{m}$

This equals the trapezoid area under the velocity-time graph.

Worked Example: Change in Velocity from an Acceleration-Time Graph

Acceleration is constant at $-3\ \text{m/s}^2$ for 5 s. If $v_i = 12\ \text{m/s}$ , find $v_f$ .

Change in velocity:

$\Delta v = a\Delta t = (-3)(5) = -15\ \text{m/s}$

Final velocity:

$v_f = v_i + \Delta v = 12 - 15 = -3\ \text{m/s}$

The negative sign indicates the object reversed direction during the interval.

Exam Focus

Typical question patterns:
- Determine velocity from the slope of an $x$ vs $t$ graph.
- Determine acceleration from the slope of a $v$ vs $t$ graph.
- Find displacement or change in velocity from the area under $v$ vs $t$ or $a$ vs $t$ graphs.
Common mistakes:
- Forgetting that “area under the curve” is signed (negative velocity gives negative displacement).
- Reading steepness on $x$ vs $t$ as acceleration instead of velocity.
- Confusing “curved graph” with “object moving on a curved path.”

Constant Acceleration in 1D: The Kinematic Model

Many AP Physics 1 problems assume constant (uniform) acceleration, meaning acceleration does not change with time. This model is excellent for situations like free fall (ignoring air resistance) or a cart pulled with roughly constant net force.

The Core Definitions You Build From

Start with

$a = \frac{\Delta v}{\Delta t}$

For constant acceleration:

$v_f = v_i + a\Delta t$

Average velocity is

$v_{avg} = \frac{\Delta x}{\Delta t}$

For constant acceleration in 1D, velocity changes linearly, so

$v_{avg} = \frac{v_i + v_f}{2}$

Combining with $\Delta x = v_{avg}\Delta t$ gives

$\Delta x = \frac{v_i + v_f}{2}\Delta t$

Substituting $v_f = v_i + a\Delta t$ yields

$\Delta x = v_i\Delta t + \frac{1}{2}a(\Delta t)^2$

Eliminating time gives

$v_f^2 = v_i^2 + 2a\Delta x$

The “BIG FIVE” Equations of Motion (Uniform Acceleration)

A common way to organize constant-acceleration relationships is the “BIG FIVE.” Using the common alternative notation

$u$ = initial velocity
$v$ = final velocity
$a$ = acceleration
$s$ = displacement
$t$ = time

the equations are:

$v = u + at$

$s = ut + \frac{1}{2}at^2$

$v^2 = u^2 + 2as$

$s = \frac{1}{2}(u + v)t$

$a = \frac{v - u}{t}$

These are the same relationships as the standard AP forms; they are just notationally different.

When These Equations Are Valid

These equations require that acceleration is constant over the time interval you are analyzing. If acceleration changes (for example, if a velocity-time graph is curved), you must use graphs or piecewise analysis rather than applying a single constant-acceleration equation to the entire motion.

Notation Reference (Common Equivalent Forms)

Quantity	Common symbols used	Meaning
initial position	$x_i$ , $x_0$	position at start
final position	$x_f$	position at end
displacement	$\Delta x$	$x_f - x_i$
initial velocity	$v_i$ , $v_0$	velocity at start
final velocity	$v_f$	velocity at end
time interval	$\Delta t$ , $t$	elapsed time (often start at 0)
acceleration	$a$	constant acceleration in model

Problem-Solving Strategy (How to Choose an Equation)

Instead of memorizing equations as separate tools, treat them as different views of the same constant-acceleration model.

List what you know and what you’re asked to find.
Pick an equation containing your unknown and the known variables.
Watch signs carefully.
Check whether the result makes physical sense.

If time is not needed, the equation

$v_f^2 = v_i^2 + 2a\Delta x$

is often the fastest route.

Worked Example: Car Braking to a Stop

A car traveling at $20\ \text{m/s}$ brakes with constant acceleration of magnitude $4.0\ \text{m/s}^2$ until it stops. Find the stopping distance.

Choose positive in the initial direction of motion.

Given:

$v_i = 20\ \text{m/s}$
$v_f = 0\ \text{m/s}$
$a = -4.0\ \text{m/s}^2$

Use

$v_f^2 = v_i^2 + 2a\Delta x$

Substitute:

$0^2 = 20^2 + 2(-4.0)\Delta x$

$0 = 400 - 8\Delta x$

$\Delta x = 50\ \text{m}$

Worked Example: Finding Time and Final Velocity

A cart starts from rest and accelerates at $2.5\ \text{m/s}^2$ for 6.0 s. Find its final velocity and displacement.

Final velocity:

$v_f = v_i + a\Delta t = 0 + 2.5(6.0) = 15\ \text{m/s}$

Displacement:

$\Delta x = v_i\Delta t + \frac{1}{2}a(\Delta t)^2 = 0 + \frac{1}{2}(2.5)(6.0^2) = 45\ \text{m}$

A quick check: average velocity is halfway from 0 to 15, so 7.5 m/s; over 6 s that gives 45 m.

Worked Example (BIG FIVE): Car Accelerating from Rest

A car starts from rest and accelerates uniformly at $5\ \text{m/s}^2$ for 10 seconds. Find final velocity and displacement.

Using

$v = u + at$

with $u = 0$ gives

$v = 0 + 5(10) = 50\ \text{m/s}$

Using

$s = ut + \frac{1}{2}at^2$

gives

$s = 0(10) + \frac{1}{2}(5)(10^2) = 250\ \text{m}$

Exam Focus

Typical question patterns:
- Use constant-acceleration equations to solve for one missing variable.
- Multi-part problems: find time first, then use it to find displacement (or vice versa).
- Explain which equation is appropriate and why (validity of constant acceleration).
Common mistakes:
- Forgetting that $\Delta x$ is displacement, not distance.
- Plugging in acceleration magnitude without a sign.
- Using a constant-acceleration equation across an interval where acceleration changes.

Free Fall and Vertical Motion: Acceleration Due to Gravity

A special and extremely common case of constant-acceleration motion is free fall near Earth’s surface, where acceleration is approximately constant and downward.

The Acceleration Due to Gravity

Near Earth, the magnitude of gravitational acceleration is approximately

$g = 9.8\ \text{m/s}^2$

If you choose up as positive, then

$a = -g$

If you choose down as positive, then

$a = +g$

The acceleration due to gravity does not change direction just because the object is moving up or down. Even at the top of the motion when velocity is zero, acceleration is still downward (ignoring air resistance).

Free Fall from Rest (Common Model Equation)

If an object is dropped from rest and air resistance is neglected, the distance fallen after time t is

$d = \frac{1}{2}gt^2$

The sign and whether you call the vertical change d or $\Delta y$ depend on your axis choice; the physics is the same.

Vertical Toss: Understanding the Peak

If you throw an object upward (taking up as positive):

Initially, velocity is positive.
Acceleration is negative.
The object slows down until velocity reaches zero at the peak.
Then velocity becomes negative and the object speeds up downward.

At the peak:

$v = 0$

but

$a = -g$

Worked Example: Maximum Height of a Vertical Throw

A ball is thrown upward with initial velocity $12\ \text{m/s}$ . Find the maximum height above the launch point (ignore air resistance). Choose up as positive.

Given:

$v_i = 12\ \text{m/s}$
$a = -9.8\ \text{m/s}^2$
at the top, $v_f = 0$

Use

$v_f^2 = v_i^2 + 2a\Delta x$

Substitute:

$0^2 = 12^2 + 2(-9.8)\Delta x$

$0 = 144 - 19.6\Delta x$

$\Delta x = 7.35\ \text{m}$

Worked Example: Time Up vs. Time Down (Same Height)

If the ball lands back at the same height it was launched (ignoring air resistance), the time going up equals the time coming down. This symmetry comes from constant acceleration and the fact that the speed at a given height is the same on the way up and down (with opposite direction).

Time to the peak comes from

$v_f = v_i + a\Delta t$

At the peak, $v_f = 0$ :

$0 = 12 - 9.8\Delta t$

$\Delta t = 1.22\ \text{s}$

So the total time back to launch height is about

$2.44\ \text{s}$

Exam Focus

Typical question patterns:
- Vertical motion with a toss: find maximum height, time to peak, or speed on return.
- Interpret signs of $v$ and $a$ during upward vs downward motion.
- Explain why acceleration is constant even as direction changes.
Common mistakes:
- Setting acceleration to zero at the top.
- Using $g = -9.8\ \text{m/s}^2$ regardless of coordinate choice (sign must match your axis).
- Confusing “speed” with “velocity” at points where direction changes.

Two-Dimensional Kinematics: Vectors and Components

Many real motions are not confined to a single line. In two dimensions, position is described with two coordinates (often x and y). The key idea is that you can treat each axis with the same kinematics tools you already know.

Position, Velocity, and Acceleration as 2D Vectors

In 2D, it is useful to think of

$\vec{r} = \langle x, y \rangle$

$\vec{v} = \langle v_x, v_y \rangle$

$\vec{a} = \langle a_x, a_y \rangle$

You do not need advanced vector math in AP Physics 1, but you must be comfortable with components and with the idea that vectors can be added and subtracted in a way that respects direction.

Independence of Perpendicular Components

A central idea: motion in the x-direction and motion in the y-direction are independent, except that they share the same time variable.

That means you can write separate constant-acceleration equations for each axis.

Horizontal:

$v_{x,f} = v_{x,i} + a_x\Delta t$

$\Delta x = v_{x,i}\Delta t + \frac{1}{2}a_x(\Delta t)^2$

Vertical:

$v_{y,f} = v_{y,i} + a_y\Delta t$

$\Delta y = v_{y,i}\Delta t + \frac{1}{2}a_y(\Delta t)^2$

Breaking an Initial Velocity into Components

If an object is launched with speed $v_0$ at angle $\theta$ above horizontal:

$v_{x,i} = v_0\cos\theta$

$v_{y,i} = v_0\sin\theta$

A self-check: if $\theta = 0$ , then $v_{x,i}$ should equal $v_0$ and $v_{y,i}$ should be zero.

Worked Example: Component Decomposition

A ball is launched at $v_0 = 18\ \text{m/s}$ at $30^\circ$ above horizontal.

$v_{x,i} = 18\cos 30^\circ = 15.6\ \text{m/s}$

$v_{y,i} = 18\sin 30^\circ = 9.0\ \text{m/s}$

(using $\cos 30^\circ \approx 0.866$ and $\sin 30^\circ = 0.5$ )

Exam Focus

Typical question patterns:
- Resolve a launch velocity into x and y components.
- Use separate constant-acceleration equations in each axis with a shared time.
- Interpret vector directions and signs from a diagram.
Common mistakes:
- Treating speed and velocity as the same in 2D (direction matters).
- Swapping $\sin$ and $\cos$ when finding components.
- Using different times for x and y motion (they must match).

Projectile Motion: A Special Case of 2D Constant Acceleration

Projectile motion describes an object moving in 2D under gravity alone (ignoring air resistance). The path is parabolic.

The Core Model (Assumptions)

In ideal projectile motion near Earth:

$a_x = 0$

and (if up is positive)

$a_y = -g$

So horizontal velocity is constant:

$v_x = v_{x,i}$

and vertical velocity changes linearly:

$v_{y,f} = v_{y,i} - g\Delta t$

The horizontal and vertical components are independent, connected only by the fact that they share the same time.

“Same Time” Is the Bridge Between x and y

A standard method:

Use vertical motion to find time.
Use that same time in horizontal motion to find horizontal displacement.

Projectile Launched Horizontally

If an object is launched horizontally,

$v_{y,i} = 0$

Gravity acts immediately, so the object begins gaining downward velocity right away even though it starts with no vertical velocity.

Worked Example: Horizontal Launch from a Height

A ball rolls off a table of height $1.25\ \text{m}$ with horizontal speed $3.0\ \text{m/s}$ . How far from the table does it land?

Choose up as positive. Then $a_y = -9.8\ \text{m/s}^2$ .

Vertical motion:

$\Delta y = -1.25\ \text{m}$
$v_{y,i} = 0$

Use

$\Delta y = v_{y,i}\Delta t + \frac{1}{2}a_y(\Delta t)^2$

Substitute:

$-1.25 = 0 + \frac{1}{2}(-9.8)(\Delta t)^2$

$-1.25 = -4.9(\Delta t)^2$

$(\Delta t)^2 = 0.255$

$\Delta t = 0.505\ \text{s}$

Horizontal motion:

$\Delta x = v_x\Delta t = 3.0(0.505) = 1.52\ \text{m}$

Projectile Launched at an Angle (Same Launch and Landing Height)

On level ground (same launch and landing height), useful symmetries apply:

Time up equals time down.
The speed at landing equals the speed at launch (directions differ).
At the peak, $v_y = 0$ but $v_x$ is unchanged.

Time to peak:

$0 = v_{y,i} - gt_{up}$

$t_{up} = \frac{v_{y,i}}{g}$

Total time of flight:

$T = 2t_{up} = \frac{2v_{y,i}}{g}$

Range:

$R = v_x T$

Substituting component expressions gives the level-ground range formula:

$R = \frac{v_0^2\sin(2\theta)}{g}$

This formula applies only when launch and landing heights match and air resistance is neglected.

Maximum height (above the launch point) for level-ground projectile motion comes from the vertical component reaching $v_y = 0$ at the peak:

$h = \frac{(v_0\sin\theta)^2}{2g}$

Worked Example: Range on Level Ground

A projectile is launched at $v_0 = 20\ \text{m/s}$ at $40^\circ$ above horizontal. Find the time of flight and range on level ground.

Components:

$v_x = 20\cos 40^\circ = 15.3\ \text{m/s}$

$v_{y,i} = 20\sin 40^\circ = 12.9\ \text{m/s}$

Time of flight:

$T = \frac{2v_{y,i}}{g} = \frac{2(12.9)}{9.8} = 2.63\ \text{s}$

Range:

$R = v_x T = 15.3(2.63) = 40.2\ \text{m}$

Unequal Launch and Landing Heights

If launch and landing heights differ, symmetry shortcuts (and the simple range formula) do not apply. Use

$\Delta y = v_{y,i}\Delta t + \frac{1}{2}a_y(\Delta t)^2$

solve for $\Delta t$ (often a quadratic), then use

$\Delta x = v_x\Delta t$

Exam Focus

Typical question patterns:
- Horizontal launch problems: find time to fall, then horizontal distance.
- Angled launch on level ground: find time of flight, maximum height, or range.
- Conceptual questions about independence of x and y motion (what changes, what stays constant).
Common mistakes:
- Assuming vertical velocity affects horizontal motion (it doesn’t in the ideal model).
- Using the level-ground range formula when heights differ.
- Forgetting that at the top, $v_y = 0$ but $a_y = -g$ still.

Relative Motion and Reference Frames (Common Kinematics Extension)

Kinematics quantities depend on your reference frame. A velocity measured relative to a moving train can differ from one measured relative to the ground, and both can be correct.

Relative Velocity in 1D

A common 1D relationship is

$v_{A/B} = v_{A/E} - v_{B/E}$

where $v_{A/B}$ is “velocity of A relative to B” and E is a chosen Earth (ground) frame.

Worked Example: Walking on a Moving Train

A train moves east at $12\ \text{m/s}$ relative to the ground. A passenger walks west at $2.0\ \text{m/s}$ relative to the train. Take east as positive.

$v_{T/G} = +12\ \text{m/s}$

$v_{P/T} = -2.0\ \text{m/s}$

Passenger relative to ground:

$v_{P/G} = v_{P/T} + v_{T/G} = -2.0 + 12 = +10\ \text{m/s}$

So the passenger still moves east relative to the ground, just slower than the train.

Exam Focus

Typical question patterns:
- Combine velocities for motion on moving platforms (walkway, train, river simplified to 1D).
- Interpret what “relative to” means in words and translate to symbols.
- Choose a frame that simplifies the reasoning.
Common mistakes:
- Adding velocities without consistent sign conventions.
- Mixing up “velocity of A relative to B” with “velocity of B relative to A” (they are negatives).
- Treating relative motion as a new kind of physics instead of a coordinate choice.

Building and Testing Kinematic Models: Using Data and Linearization

AP Physics 1 also connects kinematics to experimental evidence: how to measure acceleration, how to check whether acceleration is constant, and how to linearize data.

What Constant Acceleration Looks Like in Data

If acceleration is constant:

A velocity-time graph should be a straight line.
A position-time graph should be curved (quadratic in time).

If velocity-time data curve rather than forming a line, that suggests acceleration is not constant or measurement issues are present.

Extracting Acceleration from a Velocity-Time Graph

For a straight-line velocity-time graph, acceleration is the slope:

$a = \frac{\Delta v}{\Delta t}$

This is often a clean experimental method because slopes can reduce certain types of noise.

Finding g from Free-Fall Data (Linearization)

If you drop an object from rest and measure speed after time t, the model predicts

$v = gt$

so a plot of v vs t has slope g (with sign depending on axis choice).

If you measure vertical displacement from rest, the model predicts

$\Delta y = \frac{1}{2}gt^2$

so plotting $\Delta y$ versus $t^2$ should produce a line whose slope is

$\frac{1}{2}g$

A common mistake is plotting $\Delta y$ versus t and expecting a line; under constant acceleration, that plot should be curved.

Worked Example: Estimating g from a Linear Plot

A plot of $\Delta y$ vs $t^2$ for a dropped object gives a best-fit slope of $4.85\ \text{m/s}^2$ . Since

$\Delta y = \frac{1}{2}gt^2$

the slope equals $\frac{1}{2}g$ , so

$g = 2(4.85) = 9.70\ \text{m/s}^2$

This is close to $9.8\ \text{m/s}^2$ ; differences can come from timing uncertainty or small air resistance.

Exam Focus

Typical question patterns:
- Identify which graph (or linearized graph) would produce a straight line for constant acceleration.
- Determine acceleration from a best-fit slope and interpret units.
- Explain discrepancies between model and data using experimental limitations.
Common mistakes:
- Expecting position vs time to be linear under constant acceleration.
- Ignoring units in slope interpretations (units often reveal what the slope represents).
- Claiming the model is “wrong” without considering measurement uncertainty or air resistance.