Representing motion

Ways to Represent Motion

Three equations let you predict exactly where an object will be and how fast it will move, as long as the acceleration stays constant. Before diving into those equations, though, it helps to see that physics describes the same motion in many ways — each revealing something the others might hide.

Motion can be represented by motion diagrams, figures, graphs, equations, and narrative descriptions. A motion diagram uses dots at equal time intervals along the path of an object. Dots spaced far apart mean the object moves fast; dots bunched together mean it moves slowly. A narrative description tells the same story in words — “the cart speeds up as it rolls down the ramp.”

Figures show a snapshot of a physical setup: an object on an incline, a ball in midair. Graphs encode the same information quantitatively — the shape and slope of a curve tell you how position, velocity, or acceleration change over time. Equations compress all of that into algebra you can solve. Each representation has strengths, so the AP exam expects you to move freely among all five.

Kinematic Equations for Constant Acceleration

Key Equations

Velocity–time equation:

$${v}_{x}={v}_{x0}+{a}_{x}t$$

Variables: ${v}_{x}$ = final velocity (m/s), ${v}_{x0}$ = initial velocity (m/s), ${a}_{x}$ = constant acceleration (m/s²), $t$ = time elapsed (s)

SI unit: m/s

Rearrangements:

$${a}_{x}=\frac{{v}_{x}-{v}_{x0}}{t}$$

$$t=\frac{{v}_{x}-{v}_{x0}}{{a}_{x}}$$

ProportionalityFinal velocity changes linearly with time when acceleration is constant.

Reference sheet status: On the reference sheet.

Position–time equation:

$$x={x}_{0}+{v}_{x0}t+\frac{1}{2}{a}_{x}{t}^{2}$$

Variables: $x$ = final position (m), ${x}_{0}$ = initial position (m), ${v}_{x0}$ = initial velocity (m/s), ${a}_{x}$ = constant acceleration (m/s²), $t$ = time elapsed (s)

SI unit: m

Rearrangements:

$$x-{x}_{0}={v}_{x0}t+\frac{1}{2}{a}_{x}{t}^{2}$$

ProportionalityDisplacement grows linearly with $t$ when ${a}_{x}=0$, but grows with ${t}^{2}$ when starting from rest with constant acceleration. Doubling the time quadruples the displacement contributed by the acceleration term.

Reference sheet status: On the reference sheet.

Velocity–displacement equation:

$${v}_{x}^{2}={v}_{x0}^{2}+2{a}_{x}(x-{x}_{0})$$

Variables: ${v}_{x}$ = final velocity (m/s), ${v}_{x0}$ = initial velocity (m/s), ${a}_{x}$ = constant acceleration (m/s²), $(x-{x}_{0})$ = displacement (m)

SI unit: m²/s² (take the square root for m/s)

Rearrangements:

$${a}_{x}=\frac{{v}_{x}^{2}-{v}_{x0}^{2}}{2(x-{x}_{0})}$$

$$x-{x}_{0}=\frac{{v}_{x}^{2}-{v}_{x0}^{2}}{2{a}_{x}}$$

ProportionalityThe change in ${v}_{x}^{2}$ is directly proportional to displacement when acceleration is constant.

Reference sheet status: On the reference sheet.

These three kinematic equations apply only when acceleration is constant. Each equation connects four of the five kinematic variables — $x-{x}_{0}$, ${v}_{x0}$, ${v}_{x}$, ${a}_{x}$, and $t$ — and leaves out one. Pick the equation that includes the three quantities you know and the one you want.

Reading the kinematic equations in terms of symbols:

The subscript $x$ indicates the direction of motion. These equations work in any single dimension: replace $x$ with $y$ for vertical motion or any other axis label.

Before solving, always define which direction is positive. A negative result for velocity means the object moves opposite to your chosen positive direction. A negative acceleration does not automatically mean “slowing down” — it means the acceleration points in the negative direction. An object slows down only when acceleration and velocity point in opposite directions.

MisconceptionStudents often assume negative acceleration always means “decelerating.” A negative ${a}_{x}$ simply means the acceleration points in the negative direction. If velocity is also negative, the object speeds up.

Exam TipState your positive direction, then check whether acceleration and velocity have the same or opposite signs to decide if the object speeds up or slows down.

Worked Example: Finding Displacement with Constant Acceleration

Scenario

A cyclist starts from rest and accelerates at a constant $2.0{\text{ m/s}}^{2}$ for $8.0\text{ s}$. Determine the displacement during this interval.

Choosing the equation — The position–time equation links ${v}_{x0}$, ${a}_{x}$, $t$, and displacement. Time is known, so this equation fits.

Equation used

$$x-{x}_{0}={v}_{x0}t+\frac{1}{2}{a}_{x}{t}^{2}$$

Given

$${v}_{x0}=0\text{ m/s}$$

$${a}_{x}=2.0{\text{ m/s}}^{2}$$

$$t=8.0\text{ s}$$

Working — substitution

$$x-{x}_{0}=(0)(8.0)+\frac{1}{2}(2.0)(8.0{)}^{2}$$

$$x-{x}_{0}=0+\frac{1}{2}(2.0)(64)$$

$$x-{x}_{0}=64\text{ m}$$

Answer

$$x-{x}_{0}=64\text{ m}$$

Interpretation

The cyclist covers 64 m in 8.0 s. Because the displacement term depends on ${t}^{2}$, most of that distance is covered in the second half of the interval, when the cyclist is moving fastest.

Free Fall Near Earth’s Surface

Objects in free fall near the surface of Earth experience a vertical acceleration caused by the force of gravity. This acceleration is downward, constant, and approximately equal to ${a}_{g}=g\approx 10{\text{ m/s}}^{2}$.

Because the acceleration is constant, the same three kinematic equations apply. Replace the horizontal variables with vertical ones: use $y$ for position and ${a}_{y}=\pm g$ for acceleration. The sign of $g$ depends on the chosen positive direction. If upward is positive, then ${a}_{y}=-g=-10{\text{ m/s}}^{2}$. If downward is positive, then ${a}_{y}=+g=+10{\text{ m/s}}^{2}$.

For a ball thrown straight up, the velocity is positive on the way up and negative on the way down — but the acceleration is $-10{\text{ m/s}}^{2}$ throughout the entire flight. At the very top, the velocity is momentarily zero while the acceleration remains $-10{\text{ m/s}}^{2}$. This is a key distinction: zero velocity does not mean zero acceleration.

MisconceptionStudents often believe that at the highest point of a trajectory the acceleration is zero because the object is momentarily at rest. The gravitational acceleration is $10{\text{ m/s}}^{2}$ downward at every instant, including at the peak.

Exam TipIf asked “what is the acceleration at the top?” the answer is $g$ downward — never zero.

BoundaryFor all situations in which a numerical quantity is required for $g$, the value $g\approx 10{\text{ m/s}}^{2}$ will be used.

This is outside the scope of the AP exam.

Worked Example: Free-Fall Drop Time

Scenario

A stone is dropped from rest off the edge of a cliff $45\text{ m}$ above the ground. Determine how long the stone takes to reach the ground.

Taking downward as positive:

Equation used

$$y={y}_{0}+{v}_{y0}t+\frac{1}{2}{a}_{y}{t}^{2}$$

Rearranging — the stone starts from rest (${v}_{y0}=0$) and the displacement downward is $y-{y}_{0}=45\text{ m}$:

$$y-{y}_{0}=\frac{1}{2}{a}_{y}{t}^{2}$$

$${t}^{2}=\frac{2(y-{y}_{0})}{{a}_{y}}$$

$$t=\sqrt{\frac{2(y-{y}_{0})}{{a}_{y}}}$$

Given

$$y-{y}_{0}=45\text{ m}$$

$${a}_{y}=10{\text{ m/s}}^{2}$$

Working — substitution

$$t=\sqrt{\frac{2(45)}{10}}$$

$$t=\sqrt{\frac{90}{10}}$$

$$t=\sqrt{9.0}$$

$$t=3.0\text{ s}$$

Answer

$$t=3.0\text{ s}$$

Interpretation

The stone takes 3.0 s to fall 45 m. Because displacement grows with ${t}^{2}$, the stone covers most of the 45 m in the final second of the fall.

Graphs of Motion

Position–time, velocity–time, and acceleration–time graphs encode an object’s full kinematic story. Each graph connects to the others through slopes and areas.

Instantaneous velocity is the rate of change of position. On a position–time graph, the instantaneous velocity at any moment equals the slope of the line tangent to the curve at that point. A straight line means constant velocity. A curve means changing velocity — so the object accelerates.

Instantaneous acceleration is the rate of change of velocity. On a velocity–time graph, the instantaneous acceleration at any moment equals the slope of the tangent line at that point. A straight line means constant acceleration. A horizontal line means zero acceleration (constant velocity).

The graphs also work in reverse through areas under curves. The displacement during a time interval equals the area under the velocity–time curve over that interval. The change in velocity during a time interval equals the area under the acceleration–time curve over that interval. Area below the time axis counts as negative.

A quick guide to reading motion graphs:

When the position–time graph is a parabola, the velocity–time graph is a straight line, and the acceleration–time graph is a horizontal line. This set of shapes describes constant acceleration — exactly the situation the kinematic equations handle.

BoundaryAP Physics 1 does not expect students to quantitatively analyze nonuniform acceleration. However, students will be expected to qualitatively analyze, sketch appropriate graphs of, and discuss situations in which acceleration is nonuniform.

This is outside the scope of the AP exam.

Examiner InsightFRQs frequently ask students to sketch one graph given another. Practice extracting slope from a position–time graph to plot velocity, and extracting area from a velocity–time graph to plot displacement.

Exam TipAlways label axes with quantity and unit, and mark key values (intercepts, slopes, area values) on your sketched graph.