📈 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

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 ($v = 0$)

Calculating Velocity from the Graph

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

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

Example

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

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

Curved Lines: Non-Constant Velocity

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

Reading Acceleration from Curvature

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$, x=35$ m$)$.

1. What is the velocity? (in m/s)

2. What is the position at $t = 6$ s? (in meters)

3. At what time does the object reach $x = 20$ m? (in seconds)

Graph Interpretation 🔍

Exit Quiz — Position-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 $v$-$t$ graph:

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

Example

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

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:

1. Velocity at any time → read directly from the graph
2. Acceleration → calculate the slope
3. Displacement → calculate the area under the curve
4. Speed → absolute value of velocity
5. 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

1. What is the acceleration during the first 4 seconds? (in m/s²)

2. What is the displacement from $t = 0$ to $t = 4$ s? (in meters)

3. What is the total displacement from $t = 0$ to $t = 8$ s? (in meters)

$v$-$t$ Graph Interpretation 🔍

Exit Quiz — Velocity-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 $a$-$t$ Graphs

On an acceleration-time graph:

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 $v_0 = 2$ 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.

1. What is the change in velocity from $t = 0$ to $t = 4$ s? (in m/s)

2. What is the velocity at $t = 4$ s? (in m/s)

3. If the acceleration then remains constant at 8 m/s² for another 3 s (from $t = 4$ to $t = 7$ s), what is $\Delta v$ during that interval? (in m/s)

$a$-$t$ Graph Review 🔍

Exit Quiz — Acceleration-Time Graphs ✅

🔄 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).

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)

Going "Up" (Integration)

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

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.

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

2. What is the displacement from $t = 0$ to $t = 5$ s? (in meters)

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

Exit Quiz — Graph Conversions ✅

🌀 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

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

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

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)

1. What is the displacement during Phase 1? (in meters)

2. What is the total displacement for the entire trip? (in meters)

3. What is the acceleration during Phase 3? (in m/s²)

Non-Uniform Motion Review 🔍

Exit Quiz — Non-Uniform Motion ✅

🛠️ 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 $t = 3$ s
• $a = 0$ from $t = 3$ to $t = 7$ s
• $a = -4$ m/s² from $t = 7$ to $t = 10$ s

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

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

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

3. 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.

1. What is the total displacement? (in meters)

2. What is the average velocity for the entire trip? (in m/s)

3. What is the deceleration magnitude in the braking phase? (in m/s²)

Exit Quiz — Graph Problem Solving ✅

🎓 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

Shape Relationships (Constant Acceleration)

Quick Reference

• 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

1. What is the acceleration during the interval $t = 6$ to $t = 10$ s? (in m/s²)

2. At what time does the object reverse direction? (in seconds)

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

Conceptual Mastery Check 🔍

Final AP Review ✅