AP Physics 1 Kinematics: Building One-Dimensional Motion from First Principles

Scalars and Vectors in One Dimension

What a “one-dimensional” model really means

In one-dimensional motion, you describe an object’s position using a single axis (often the x-axis). That does not mean motion is “simple” or slow—it means you are choosing to track motion along one line, such as a car moving along a straight road, an elevator moving up and down, or a cart on a track.

The big idea is that once you commit to a single axis, direction can be represented with a sign (positive or negative). This is the main conceptual bridge between everyday language (“to the left,” “down,” “backward”) and physics language (negative values).

Scalars vs vectors (and why you should care)

A scalar is a quantity that has magnitude only. A vector has magnitude and direction. In 1D, “direction” is captured by whether the number is positive or negative.

This matters because many AP Physics 1 errors come from treating a vector like a scalar—especially when adding, subtracting, or interpreting negative signs.

Common 1D scalars

• Time: how long something takes (no direction)
• Distance: how much ground was covered (no direction)
• Speed: how fast (no direction)

Common 1D vectors

• Displacement: change in position with direction
• Velocity: rate of change of position with direction
• Acceleration: rate of change of velocity with direction

Choosing a coordinate system and sign convention

Before you compute anything, you must decide:

1. Where is the origin (what position counts as zero)?
2. Which direction is positive?

There is no universally “correct” choice. The physics will work out as long as you stay consistent.

Example choices:

• For an elevator: choose “up” as positive.
• For a car moving along a road: choose “east” as positive.
• For a falling object: many students choose “up” positive (so gravity becomes negative), but you can choose “down” positive if you want.

A powerful habit: write the sign convention in words before solving: “Let right be positive” or “Let upward be positive.” This prevents sign mistakes later.

Vector representation in 1D: magnitude plus sign

In one dimension, you typically don’t draw arrows for every vector; instead you represent direction by sign.

• If you choose right as positive, then a displacement of 3 meters to the left is -3\,\text{m}.
• If you choose up as positive, the acceleration due to gravity near Earth’s surface is approximately -9.8\,\text{m/s}^2 (because it points downward).

A key interpretation: the magnitude of a 1D vector is the absolute value.

• Magnitude of -7\,\text{m/s} is 7\,\text{m/s}.

Adding and subtracting vectors in 1D

In 1D, vector addition is “algebra with signs.” This is simpler than 2D vectors because you don’t need components—your vectors already are signed components.

If you walk +5\,\text{m} (right) and then -2\,\text{m} (left), your net displacement is +3\,\text{m}.

What commonly goes wrong is mixing up distance and displacement:

• Distance adds total path length (always nonnegative).
• Displacement adds signed changes in position (can cancel).

Worked example: distance vs displacement

You move along a hallway.

• First: +8\,\text{m}
• Then: -3\,\text{m}

Displacement:
\Delta x = +8 + (-3) = +5\,\text{m}

Distance traveled:
\text{distance} = 8 + 3 = 11\,\text{m}

Notice how the distance is larger because it ignores “canceling.”

Worked example: picking a sign convention wisely

A ball is thrown straight upward. Choose up as positive.

• Initially it moves upward, so velocity is positive.
• Gravity points downward, so acceleration is negative.

That single decision makes it much easier to reason: the velocity decreases over time because acceleration is negative.

Exam Focus

Typical question patterns

• You’re given a verbal description (“moves left,” “turns around”) and asked to assign signs to displacement/velocity/acceleration.
• A multi-part problem asks for both distance and displacement to test whether you confuse scalars and vectors.
• A setup question asks you to choose a coordinate system and then interpret negative values.

Common mistakes

• Treating distance and displacement as interchangeable (they are not).
• Thinking a negative sign means “slowing down” (negative means direction relative to your chosen axis).
• Switching sign conventions mid-problem (choose once, stick to it).

Displacement, Velocity, and Acceleration

Position and displacement: the foundation

Position x tells you where an object is located on your chosen axis, relative to an origin. Position can be positive, negative, or zero.

Displacement is the change in position:
\Delta x = x_f - x_i
Here:

• x_i is initial position
• x_f is final position
• \Delta x is displacement

Displacement is a vector in 1D—its sign tells you direction.

Why this matters: almost all kinematics relationships are ultimately about how position changes over time. If you misunderstand displacement, everything built on top of it becomes shaky.

Time intervals and “delta” notation

A time interval is:
\Delta t = t_f - t_i
In AP Physics 1 problems, \Delta t is almost always positive because final time is later than initial time.

A subtle but important point: negative velocities and accelerations come from changes in position and velocity, not from negative time.

Velocity: how fast position changes

Average velocity is defined as displacement per time:
v_{avg} = \frac{\Delta x}{\Delta t}
Average velocity depends only on where you started and ended, not on the path in between.

Speed is different: average speed is distance traveled per time, and it cannot be negative.

Why velocity matters: velocity tells you both “how fast” and “which way.” It connects directly to graphs and to the kinematics equations for constant acceleration.

Instantaneous vs average velocity (conceptual)

Even in algebra-based physics, you should understand the idea that velocity can change from moment to moment. The instantaneous velocity is the velocity at a specific time (like the speedometer reading at an instant), while the average velocity is over an interval.

Graphically, instantaneous velocity corresponds to the slope of the position-time graph at a point (more on this in the Representing Motion section).

Acceleration: how fast velocity changes

Average acceleration is defined as change in velocity per time:
a_{avg} = \frac{\Delta v}{\Delta t}
where
\Delta v = v_f - v_i

Acceleration is a vector. In 1D, the sign tells you direction along your axis.

Why acceleration matters: it tells you how velocity evolves. Many real motions (cars speeding up/slowing down, objects under gravity, carts pushed by constant force) are well-modeled by constant acceleration.

Clearing up a major misconception: acceleration vs “slowing down”

Students often say “acceleration means speeding up.” In physics, acceleration means changing velocity, which can happen in two ways in 1D:

• Velocity magnitude increases (speeding up)
• Velocity magnitude decreases (slowing down)

The rule that actually works is about relative signs:

• If velocity and acceleration have the same sign, speed increases.
• If velocity and acceleration have opposite signs, speed decreases.

Example: choose right as positive.

• v = +6\,\text{m/s} and a = -2\,\text{m/s}^2 means the object is moving right but accelerating left, so it slows down.
• v = -6\,\text{m/s} and a = -2\,\text{m/s}^2 means moving left and accelerating left, so it speeds up (in the left direction).

Constant-acceleration kinematics (the core AP Physics 1 toolkit)

A huge fraction of 1D motion problems in AP Physics 1 assume constant acceleration. Under that assumption, several very useful relationships hold.

Notation you’ll see

Different teachers use slightly different symbols. Here’s a common set:

Typically, v_0 means the velocity at t=0, and x_0 means the position at t=0.

The constant-acceleration equations (and when to use them)

If acceleration is constant (same value for the entire interval), then:

1) Velocity as a function of time:
v = v_0 + at
Use this when you know a and time and want final velocity, or vice versa.

2) Position as a function of time:
x = x_0 + v_0 t + \frac{1}{2}at^2
Use this when you want position after time t.

3) Velocity-position relationship (eliminates time):
v^2 = v_0^2 + 2a(x - x_0)
Use this when time is not given or not needed.

4) Displacement from average velocity (valid for constant acceleration):
\Delta x = \frac{(v_0 + v)}{2}t
This works because under constant acceleration, velocity changes linearly, so the average of initial and final velocities equals the average velocity.

A practical strategy: pick the equation that includes the variables you know and the one you want, and excludes the one you don’t.

Worked problem: braking car (signs matter)

A car moving to the right at 20\,\text{m/s} brakes with constant acceleration -4\,\text{m/s}^2. How long until it stops, and how far does it travel while braking?

Step 1: Interpret the signs
Right is positive. The car is moving right, so v_0 = +20\,\text{m/s}. Braking means acceleration is opposite the motion, so a = -4\,\text{m/s}^2. Stopping means v = 0\,\text{m/s}.

Step 2: Find time using v = v_0 + at
0 = 20 + (-4)t
4t = 20
t = 5\,\text{s}

Step 3: Find displacement (distance here equals displacement since it never reverses)
Use \Delta x = \frac{(v_0 + v)}{2}t:
\Delta x = \frac{(20 + 0)}{2}(5)
\Delta x = 10 \cdot 5 = 50\,\text{m}

A common check: the time is reasonable (5 seconds), and the distance is plausible for highway braking.

Worked problem: tossed upward (negative acceleration does not mean “going down”)

A ball is thrown upward with v_0 = +12\,\text{m/s} from position x_0 = 0\,\text{m}. Take up as positive and use a = -9.8\,\text{m/s}^2. How long until it reaches its highest point?

At the highest point, the velocity is zero (it changes direction there):
v = 0
Use v = v_0 + at:
0 = 12 + (-9.8)t
9.8t = 12
t \approx 1.22\,\text{s}

Important interpretation: the ball is still moving upward during this time, but its velocity is shrinking because acceleration is downward.

Exam Focus

Typical question patterns

• “Object slows down” or “speeds up” prompts: you must determine the sign of a relative to v.
• Constant-acceleration word problems (cars braking, objects tossed up/down) requiring the kinematics equations.
• Multi-representation questions: compute a value and then interpret its sign physically.

Common mistakes

• Using \Delta x when the problem asks for distance traveled (especially if the object reverses direction).
• Plugging in g = 9.8\,\text{m/s}^2 without a sign (you must choose a direction and stick to it).
• Assuming “at the top” means acceleration is zero (it is not; only velocity is zero there for vertical motion under gravity).

Representing Motion

Why representations are as important as equations

In AP Physics 1, you are often asked to translate between words, graphs, and diagrams. This isn’t “extra”—it’s a core skill because representations reveal relationships quickly:

• Graphs show how quantities change.
• Motion diagrams make direction and relative speed visible.
• Equations let you compute exact values.

Strong problem-solvers constantly cross-check: does the graph match the story, and do the numbers match the graph?

Motion diagrams (a visual story of movement)

A motion diagram represents an object’s position at equally spaced time intervals.

• Each dot is the position at a specific time.
• The spacing between dots shows speed: larger spacing means greater speed.
• You can add velocity arrows: longer arrows mean greater speed; arrow direction shows direction.
• You can infer acceleration: if the velocity arrows grow, acceleration points in the direction of motion; if they shrink, acceleration points opposite.

Why this matters: motion diagrams help you reason without equations, especially on conceptual questions.

Example: interpreting dot spacing

If dots get farther apart as time progresses, the object is speeding up. If dots get closer, it’s slowing down. If the spacing is constant, velocity is constant and acceleration is zero.

A common misconception is to treat the dot diagram like a “path drawing.” In 1D, it’s not about shape—it’s about spacing over equal time steps.

Position-time graphs: slope means velocity

A position-time graph plots position x versus time t.

Key idea: the slope of the position-time graph is velocity.

• A steeper slope means greater speed.
• A positive slope means positive velocity.
• A negative slope means negative velocity.

For a straight line on an x vs t graph, the velocity is constant.

Average velocity over an interval corresponds to the slope of a secant line:
v_{avg} = \frac{\Delta x}{\Delta t}

Instantaneous velocity corresponds to the slope of the tangent line at a point (conceptual understanding is enough for AP Physics 1).

Example: what “turning around” looks like on an x vs t graph

If an object reverses direction, the slope must change sign. That usually appears as a curve that reaches a maximum or minimum position (slope zero at the turning point).

Common error: thinking a graph “crossing the axis” means turning around. Crossing x = 0 only means passing the origin, not reversing direction.

Velocity-time graphs: slope means acceleration, area means displacement

A velocity-time graph plots velocity v versus time t.

Two powerful interpretations:

1) Slope is acceleration:
a = \frac{\Delta v}{\Delta t}
On a v vs t graph, a straight line means constant acceleration.

2) Area under the curve is displacement.
For a constant velocity (horizontal line), displacement is rectangle area:
\Delta x = vt

For changing velocity, you can break the area into shapes (rectangles, triangles, trapezoids). The sign of the area matters:

• Area above the time axis (positive v) contributes positive displacement.
• Area below the time axis (negative v) contributes negative displacement.

This “area gives displacement” idea is one of the most tested skills because it connects graph-reading to physical motion.

Worked example: displacement from a v vs t graph (piecewise)

An object has:

• v = +4\,\text{m/s} for the first 3\,\text{s}
• then velocity decreases linearly from +4\,\text{m/s} to 0\,\text{m/s} over the next 2\,\text{s}

Compute displacement.

First interval (rectangle):
\Delta x_1 = (4)(3) = 12\,\text{m}

Second interval (triangle/trapezoid): velocity goes from 4 to 0 over 2 s, so the average velocity in that interval is:
v_{avg} = \frac{4 + 0}{2} = 2\,\text{m/s}
Then:
\Delta x_2 = v_{avg}t = (2)(2) = 4\,\text{m}

Total displacement:
\Delta x = 12 + 4 = 16\,\text{m}

Notice how this matches the constant-acceleration idea: linearly changing v corresponds to constant a.

Acceleration-time graphs: area means change in velocity

An acceleration-time graph plots acceleration a versus time t.

Key interpretation: the area under an a vs t graph is change in velocity:
\Delta v = a\Delta t
for constant acceleration (rectangle area). For varying acceleration, the total change in velocity is the signed area under the curve.

This representation is especially useful when acceleration is piecewise constant (common in AP questions).

Connecting all three graphs (and the constant-acceleration story)

When acceleration is constant:

• a vs t is a horizontal line.
• v vs t is a straight line (since slope is constant).
• x vs t is a parabola-like curve (because position changes faster and faster when speeding up).

You don’t need calculus to understand the shape: if velocity increases each second, the object covers more distance each second, so the position curve gets steeper over time.

Example: matching a verbal description to graphs

“An object moves to the right, slows to a stop, then moves left speeding up.”

Translate to signs (right positive):

• Initially v > 0.
• It slows down, so acceleration is negative while it’s still moving right.
• At the turning point, v = 0.
• Then it moves left, so v < 0.
• It speeds up to the left, so acceleration remains negative (same sign as velocity).

On a v vs t graph, you’d expect a line or curve that starts above zero, crosses to zero, and continues below zero, with a negative slope during the slowing-down and speeding-up phases if acceleration stays negative.

Common graph-reading pitfalls (and how to avoid them)

1) Confusing slope with height. On a v vs t graph:

• Height is velocity.
• Slope is acceleration.
  Students often see a high point and call it “high acceleration,” which is incorrect.

2) Ignoring sign of area. If velocity is negative, the area contributes negative displacement.

3) Mixing up “steep” with “fast” on the wrong graph. Steep on an x vs t graph means large velocity. Steep on a v vs t graph means large acceleration.

Worked problem: infer acceleration and displacement from a v vs t graph

A velocity-time graph is a straight line from v = -2\,\text{m/s} at t = 0\,\text{s} to v = +6\,\text{m/s} at t = 4\,\text{s}.

1) Acceleration (slope):
a = \frac{\Delta v}{\Delta t} = \frac{6 - (-2)}{4} = \frac{8}{4} = 2\,\text{m/s}^2

2) Displacement (area):
Since v changes linearly, the average velocity over the interval is:
v_{avg} = \frac{v_0 + v}{2} = \frac{-2 + 6}{2} = 2\,\text{m/s}
Then:
\Delta x = v_{avg}t = (2)(4) = 8\,\text{m}

Even though velocity is negative at first, it becomes positive and the overall displacement ends up positive because the positive velocities dominate the area.

Exam Focus

Typical question patterns

• Given a graph (often v vs t), find displacement using area and acceleration using slope.
• Match a verbal motion description to the correct set of graphs.
• Determine when an object changes direction (identify when v = 0) and distinguish that from when it passes the origin (when x = 0).

Common mistakes

• Calculating “distance” by taking signed area (signed area gives displacement; distance requires adding magnitudes of areas without cancellation).
• Using the wrong graph feature (using height instead of slope, or slope instead of area).
• Concluding acceleration is zero at the top of a position curve without checking the velocity and the physical situation (curved x vs t generally indicates nonzero acceleration).