🎯⭐ INTERACTIVE LESSON

Motion Graphs

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Motion Graphs - Complete Interactive Lesson

Part 1: Position-Time Graphs

📈 Position-Time Graphs

Part 1 of 7 — Motion Graphs

Graphs are one of the most powerful tools in physics. A single position-time graph can tell you everything about an object's motion — where it is, how fast it's going, and which direction it's moving — all at a glance.

Reading Position-Time ( $x$ - $t$ ) Graphs

On a position-time graph:

The horizontal axis is time ( $t$ )
The vertical axis is position ( $x$ )

What the Graph Tells You

Feature	Meaning
Point on the curve	Position at that time
Slope	Velocity
Positive slope	Moving in + direction
Negative slope	Moving in − direction
Zero slope (flat)	Object is at rest
Straight line	Constant velocity
Curved line	Changing velocity (acceleration)

The Key Relationship

$\text{slope of } x\text{-}t \text{ graph} = \frac{\Delta x}{\Delta t} = v$

The steeper the line, the faster the object moves.

Constant Velocity on $x$ - $t$ Graphs

When an object moves at constant velocity, the $x$ - $t$ graph is a straight line.

Examples

Steep positive slope: fast motion in + direction
Gentle positive slope: slow motion in + direction
Negative slope: motion in − direction
Horizontal line: object at rest ( $v = 0$ )

Curved Lines: Non-Constant Velocity

When the $x$ - $t$ graph is curved, the velocity is changing — the object is accelerating.

Reading Acceleration from Curvature

Curve Shape	Motion
Curving upward (concave up)	Velocity is increasing (accelerating in + direction)
Curving downward (concave down)	Velocity is decreasing (decelerating)

Instantaneous Velocity

For a curved graph, the instantaneous velocity at any moment is the slope of the tangent line at that point.

Average velocity = slope of the secant line (line connecting two points) Instantaneous velocity = slope of the tangent line (at a single point)

Concept Check — Position-Time Graphs 🎯

Reading Graphs — Calculations 🧮

A position-time graph shows a straight line passing through the points $(t=0, x=5$ m $)$ and $(t=10$ s m.

Graph Interpretation 🔍

Exit Quiz — Position-Time Graphs ✅

Part 2: Velocity-Time Graphs

📊 Velocity-Time Graphs

Part 2 of 7 — Motion Graphs

Velocity-time ( $v$ - $t$ ) graphs are arguably the most information-rich graphs in kinematics. From a single $v$ - $t$ graph, you can determine velocity, acceleration, AND displacement.

Slope = Acceleration

On a $v$ - graph:

Part 3: Acceleration-Time Graphs

📉 Acceleration-Time Graphs

Part 3 of 7 — Motion Graphs

The third type of motion graph completes the picture. Acceleration-time ( $a$ - $t$ ) graphs show how acceleration varies over time. Just like with $v$ - $t$ graphs, the area under the curve carries physical meaning.

Reading $a$ - Graphs

Part 4: Slopes & Areas Under Curves

🔄 Converting Between Graph Types

Part 4 of 7 — Motion Graphs

One of the most important skills in AP Physics is translating between $x$ - $t$ , $v$ - $t$ , and $a$ - $t$ graphs. Each graph type contains enough information to construct the others (given initial conditions).

Part 5: Translating Between Graphs

🌀 Non-Uniform Motion on Graphs

Part 5 of 7 — Motion Graphs

So far we've focused mostly on constant velocity and constant acceleration. Now let's tackle more complex scenarios: objects that speed up, slow down, stop, and reverse — all visible on their graphs.

Recognizing Direction Changes

On $x$ - $t$ Graphs

A direction reversal appears as a turning point (local max or min):

The slope changes sign
At the turning point, the slope is zero (momentary rest)
Before: positive slope → after: negative slope (or vice versa)

On $v$ - $t$ Graphs

Part 6: Problem-Solving Workshop

🛠️ Problem-Solving Workshop

Part 6 of 7 — Motion Graphs

Time to sharpen your graph skills with a variety of practice problems. These problems mirror the style and difficulty of AP exam questions involving motion graphs.

Graph Reading Warm-Up 🎯

Graph Conversion Practice 🧮

An $a$ - $t$ graph shows:

$a = +6$ m/s² from $t = 0$ to s

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

Part 7 of 7 — Motion Graphs

Let's consolidate everything about motion graphs. This review covers all three graph types, their relationships, and AP-level interpretation skills.

Complete Graph Relationships

The Derivative/Integral Chain

Operation	From → To	What It Gives
Slope	$x$ - $t$ → $v$ - $t$

Calculating Velocity from the Graph

Pick any two points $(t_1, x_1)$ and $(t_2, x_2)$ on the line:

$v = \frac{x_2 - x_1}{t_2 - t_1} = \frac{\Delta x}{\Delta t}$

A straight line passes through $(0, 2)$ and $(4, 14)$ :

$v = \frac{14 - 2}{4 - 0} = \frac{12}{4} = 3 \text{ m/s}$

What is the velocity? (in m/s)
What is the position at $t = 6$ s? (in meters)
At what time does the object reach $x = 20$ m? (in seconds)

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

Feature	Meaning
Positive slope	Positive acceleration
Negative slope	Negative acceleration
Zero slope (horizontal)	Constant velocity ( $a = 0$ )
Straight line	Constant acceleration
Curved line	Changing acceleration

A $v$ - $t$ graph shows a straight line from $(0, 4)$ to $(6, 16)$ :

$a = \frac{16 - 4}{6 - 0} = \frac{12}{6} = 2 \text{ m/s}^2$

Area Under the Curve = Displacement

The area between the $v$ - $t$ curve and the time axis gives the displacement:

$\Delta x = \int v \, dt = \text{area under } v\text{-}t \text{ graph}$

Sign Convention

Area above the time axis → positive displacement
Area below the time axis → negative displacement
Net displacement = total area (with signs)

Common Shapes

Shape	Area Formula
Rectangle	$\text{base} \times \text{height}$
Triangle	$\frac{1}{2} \times \text{base} \times \text{height}$

Example

A constant velocity of $v = 5$ m/s for $t = 4$ s:

Area = rectangle = $5 \times 4 = 20$ m
This matches $\Delta x = v \cdot t = 20$ m ✓

Putting It Together

From a single $v$ - $t$ graph you can extract:

Velocity at any time → read directly from the graph
Acceleration → calculate the slope
Displacement → calculate the area under the curve
Speed → absolute value of velocity
Direction of motion → sign of velocity (above or below axis)

Special Case: Crossing the Time Axis

When the $v$ - $t$ graph crosses the time axis ( $v = 0$ ):

The object momentarily stops
It then reverses direction
The displacement before and after may partially cancel

Concept Check — Velocity-Time Graphs 🎯

$v$ - $t$ Graph Calculations 🧮

A velocity-time graph shows:

From $t = 0$ to $t = 4$ s: velocity increases linearly from 0 to 12 m/s
From $t = 4$ to $t = 8$ s: velocity stays constant at 12 m/s

What is the acceleration during the first 4 seconds? (in m/s²)
What is the displacement from $t = 0$ to $t = 4$ s? (in meters)
What is the total displacement from $t = 0$ to $t = 8$ s? (in meters)

$v$ - $t$ Graph Interpretation 🔍

Exit Quiz — Velocity-Time Graphs ✅

On an acceleration-time graph:

Feature	Meaning
Value at any time	Acceleration at that instant
Horizontal line	Constant acceleration
Line at $a = 0$	Zero acceleration (constant velocity)
Positive value	Acceleration in + direction
Negative value	Acceleration in − direction

Area Under the Curve

$\text{Area under } a\text{-}t \text{ graph} = \int a \, dt = \Delta v$

The area under the $a$ - $t$ graph gives the change in velocity, NOT the velocity itself.

To find the velocity at any time: $v(t) = v_0 + \text{area under } a\text{-}t \text{ from } 0 \text{ to } t$

Constant Acceleration Case

When $a$ is constant, the $a$ - $t$ graph is a horizontal line.

Finding $\Delta v$

The area is simply a rectangle:

$\Delta v = a \times \Delta t$

Example

An $a$ - $t$ graph shows $a = 3$ m/s² from $t = 0$ to $t = 5$ s. If m/s:

$\Delta v = (3)(5) = 15 \text{ m/s}$ $v_f = v_0 + \Delta v = 2 + 15 = 17 \text{ m/s}$

Free Fall on an $a$ - $t$ Graph

For an object in free fall (taking up as positive):

The $a$ - $t$ graph is a horizontal line at $a = -9.8$ m/s²
It stays constant the entire time the object is in the air

Non-Constant Acceleration

When acceleration changes with time:

The $a$ - $t$ graph is no longer a horizontal line
The area under the curve must be calculated using geometry (triangles, trapezoids) or calculus
The kinematic equations ( $v = v_0 + at$ , etc.) no longer apply because they assume constant $a$

Example

If acceleration increases linearly from 0 to 6 m/s² over 4 s:

$\Delta v = \text{area of triangle} = \frac{1}{2}(4)(6) = 12 \text{ m/s}$

Key Chain: $a$ - $t$ area → $\Delta v$ → use with $v_0$ to get $v$

Concept Check — Acceleration-Time Graphs 🎯

$a$ - $t$ Graph Calculations 🧮

An $a$ - $t$ graph shows acceleration increasing linearly from $a = 0$ at $t = 0$ to $a = 8$ m/s² at $t = 4$ s. The initial velocity is $v_0 = 5$ m/s.

What is the change in velocity from $t = 0$ to $t = 4$ s? (in m/s)
What is the velocity at $t = 4$ s? (in m/s)
If the acceleration then remains constant at 8 m/s² for another 3 s (from $t = 4$ to s), what is during that interval? (in m/s)

$a$ - $t$ Graph Review 🔍

Exit Quiz — Acceleration-Time Graphs ✅

The Graph Conversion Chain

$x\text{-}t \xrightarrow{\text{slope}} v\text{-}t \xrightarrow{\text{slope}} a\text{-}t$

$a\text{-}t \xrightarrow{\text{area}} v\text{-}t \xrightarrow{\text{area}} x\text{-}t$

Going "Down" (Differentiation)

From	To	Method
$x$ - $t$	$v$ - $t$	Take the slope of the $x$ - graph

Going "Up" (Integration)

From	To	Method	Need
$a$ - $t$	$v$ - $t$	Find the area under $a$ -

Important: Going "up" requires initial conditions. Going "down" does not.

Example Conversions

From $x$ - $t$ to $v$ - $t$

If $x$ - $t$ is a straight line with slope 5 m/s: → $v$ - $t$ is a horizontal line at $v = 5$ m/s

If $x$ - $t$ is a parabola (concave up): → $v$ - $t$ is a straight line with positive slope (velocity increasing linearly)

From $v$ - $t$ to $a$ - $t$

If $v$ - $t$ is a horizontal line at 8 m/s: → $a$ - $t$ is a horizontal line at $a = 0$

If $v$ - $t$ is a straight line with slope $-3$ m/s²: → $a$ - $t$ is a horizontal line at $a = -3$ m/s²

Constant Acceleration Summary

$x$ - $t$	$v$ - $t$	$a$ - $t$

Concept Check — Graph Conversions 🎯

Match the Graph Relationships 🔍

Graph Conversion Practice 🧮

A $v$ - $t$ graph shows a straight line from $v = 0$ at $t = 0$ to $v = 20$ m/s at $t = 5$ s, then constant at $v = 20$ m/s from $t = 5$ to $t = 10$ s.

What is the acceleration from $t = 0$ to $t = 5$ s? (in m/s²)
What is the displacement from $t = 0$ to $t = 5$ s? (in meters)

Exit Quiz — Graph Conversions ✅

A direction reversal appears when the graph crosses the time axis:

Velocity changes sign
At $v = 0$ : the object momentarily stops
The object then moves in the opposite direction

Example: Thrown Ball

A ball thrown upward:

$v$ - $t$ : straight line from $+v_0$ to $-v_0$ , crossing zero at the top
$x$ - $t$ : parabola opening downward, with vertex at the highest point
$a$ - $t$ : horizontal line at $-g$

Multi-Phase Motion

Real motion often involves several phases. Each phase may have different acceleration.

Example: A car trip

Accelerate (0–10 s): $a = 3$ m/s², $v$ increases from 0 to 30 m/s
Cruise (10–30 s): $a = 0$ , $v$ stays at 30 m/s
Brake (30–40 s): $a = -3$ m/s², $v$ decreases from 30 to 0 m/s

Phase	$v$ - $t$ Graph	$a$ - $t$ Graph	$x$ - $t$ Graph

Total displacement = sum of areas under each phase of the $v$ - $t$ graph.

Concept Check — Non-Uniform Motion 🎯

Multi-Phase Graph Problems 🧮

A $v$ - $t$ graph shows:

Phase 1 ( $t = 0$ to $t = 4$ s): $v$ increases from 0 to 8 m/s (straight line)
Phase 2 ( $t = 4$ to $t = 10$ s): $v$ is constant at 8 m/s
Phase 3 ( $t = 10$ to $t = 14$ s): $v$ decreases from 8 to 0 m/s (straight line)

What is the displacement during Phase 1? (in meters)
What is the total displacement for the entire trip? (in meters)
What is the acceleration during Phase 3? (in m/s²)

Non-Uniform Motion Review 🔍

Exit Quiz — Non-Uniform Motion ✅

a = 0

from

t = 3

t = 7

a = -4

m/s² from

t = 7

t = 10

The initial velocity is $v_0 = 2$ m/s.

What is the velocity at $t = 3$ s? (in m/s)
What is the velocity at $t = 7$ s? (in m/s)
What is the velocity at $t = 10$ s? (in m/s)

AP-Style Graph Problems 🎯

Challenge: Complete Analysis 🏆

A car starts from rest. Its $v$ - $t$ graph is a straight line reaching 20 m/s at $t = 5$ s, then stays at 20 m/s until $t = 15$ s, then decelerates uniformly to rest at $t = 20$ s.

What is the total displacement? (in meters)
What is the average velocity for the entire trip? (in m/s)
What is the deceleration magnitude in the braking phase? (in m/s²)

Exit Quiz — Graph Problem Solving ✅

Shape Relationships (Constant Acceleration)

$a$ - $t$	$v$ - $t$	$x$ - $t$
Horizontal line	Straight line	Parabola

Slope of $x$ - $t$ = velocity
Slope of $v$ - $t$ = acceleration
Area under $v$ - $t$ = displacement
Area under $a$ - $t$ = $\Delta v$
$v = 0$ on $v$ - $t$ → turning point on $x$ - $t$

AP-Style Questions — Set 1 🎯

AP-Style Calculations 🧮

A $v$ - $t$ graph shows the following:

$t = 0$ to $t = 2$ s: $v$ increases linearly from 0 to 6 m/s
$t = 2$ to $t = 6$ s: $v$ is constant at 6 m/s
$t = 6$ to $t = 10$ s: $v$ decreases linearly from 6 m/s to $-2$ m/s

What is the acceleration during the interval $t = 6$ to $t = 10$ s? (in m/s²)
At what time does the object reverse direction? (in seconds)
What is the total displacement from $t = 0$ to $t = 10$ s? (in meters)

Conceptual Mastery Check 🔍

Final AP Review ✅

What is the total displacement from $t = 0$ to $t = 10$ s? (in meters)

Accelerate	Rising line	$a = +3$	Concave up parabola
Cruise	Horizontal at 30	$a = 0$	Straight line
Brake	Falling line	$a = -3$	Concave down parabola

Motion Graphs

Motion Graphs - Complete Interactive Lesson

Part 1: Position-Time Graphs

📈 Position-Time Graphs

Reading Position-Time (xxx-ttt) Graphs

What the Graph Tells You

The Key Relationship

Constant Velocity on xxx-ttt Graphs

Examples

Curved Lines: Non-Constant Velocity

Reading Acceleration from Curvature

Instantaneous Velocity

Part 2: Velocity-Time Graphs

📊 Velocity-Time Graphs

Slope = Acceleration

Part 3: Acceleration-Time Graphs

📉 Acceleration-Time Graphs

Reading aaa- Graphs

Part 4: Slopes & Areas Under Curves

🔄 Converting Between Graph Types

Part 5: Translating Between Graphs

🌀 Non-Uniform Motion on Graphs

Recognizing Direction Changes

On xxx-ttt Graphs

On vvv-ttt Graphs

Part 6: Problem-Solving Workshop

🛠️ Problem-Solving Workshop

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

Complete Graph Relationships

The Derivative/Integral Chain

Calculating Velocity from the Graph

Example

Example

Area Under the Curve = Displacement

Sign Convention

Common Shapes

Example

Putting It Together

Special Case: Crossing the Time Axis

Area Under the Curve

Constant Acceleration Case

Finding Δv\Delta vΔv

Example

Free Fall on an aaa-ttt Graph

Non-Constant Acceleration

Example

The Graph Conversion Chain

Going "Down" (Differentiation)

Going "Up" (Integration)

Example Conversions

From xxx-ttt to vvv-ttt

From vvv-ttt to aaa-ttt

Constant Acceleration Summary

Example: Thrown Ball

Multi-Phase Motion

Example: A car trip

Shape Relationships (Constant Acceleration)

Quick Reference

Reading Position-Time ( $x$ - $t$ ) Graphs

Constant Velocity on $x$ - $t$ Graphs

Reading $a$ - Graphs

On $x$ - $t$ Graphs

On $v$ - $t$ Graphs

Finding $\Delta v$

Free Fall on an $a$ - $t$ Graph

From $x$ - $t$ to $v$ - $t$

From $v$ - $t$ to $a$ - $t$