Motion Graphs

Position-time, velocity-time, and acceleration-time graphs

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Motion Graphs

Introduction

Graphs are powerful tools for visualizing and analyzing motion. The three main types are:

  1. Position-time graphs (xx vs tt)
  2. Velocity-time graphs (vv vs tt)
  3. Acceleration-time graphs (aa vs tt)

Position-Time Graphs

Interpreting Slope

The slope of a position-time graph gives the velocity.

slope=ΔxΔt=v\text{slope} = \frac{\Delta x}{\Delta t} = v

Key features:

  • Horizontal line → zero velocity (at rest)
  • Positive slope → positive velocity (moving forward)
  • Negative slope → negative velocity (moving backward)
  • Steeper slope → greater speed
  • Curved line → changing velocity (acceleration)

Interpreting Curvature

  • Straight line: constant velocity (a=0a = 0)
  • Curve upward: increasing velocity (positive acceleration)
  • Curve downward: decreasing velocity (negative acceleration)

Reading the Graph

  • y-value: position at that time
  • Slope: velocity at that time
  • Change in slope: acceleration

Velocity-Time Graphs

Interpreting Slope

The slope of a velocity-time graph gives the acceleration.

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

Key features:

  • Horizontal line → zero acceleration (constant velocity)
  • Positive slope → positive acceleration
  • Negative slope → negative acceleration (deceleration)
  • Steeper slope → greater acceleration

Interpreting Area

The area under a velocity-time graph gives the displacement.

area=∫v dt=Ī”x\text{area} = \int v \, dt = \Delta x

For constant velocity (rectangle): Δx=v⋅t\Delta x = v \cdot t

For constant acceleration (trapezoid or triangle): Δx=12(v0+v)t\Delta x = \frac{1}{2}(v_0 + v)t

Sign matters:

  • Area above time axis → positive displacement
  • Area below time axis → negative displacement

Reading the Graph

  • y-value: velocity at that time
  • Slope: acceleration at that time
  • Area under curve: displacement

Acceleration-Time Graphs

Interpreting Area

The area under an acceleration-time graph gives the change in velocity.

area=∫a dt=Ī”v\text{area} = \int a \, dt = \Delta v

For constant acceleration: Δv=a⋅t\Delta v = a \cdot t

Reading the Graph

  • y-value: acceleration at that time
  • Area under curve: change in velocity
  • Horizontal line: constant acceleration

Relationships Between Graphs

| If you have... | To get... | Operation | |----------------|-----------|-----------| | Position-time | Velocity-time | Take slope (derivative) | | Velocity-time | Acceleration-time | Take slope (derivative) | | Velocity-time | Position-time | Find area (integral) | | Acceleration-time | Velocity-time | Find area (integral) |

Common Motion Patterns

Constant Velocity

  • Position-time: straight line (slope = velocity)
  • Velocity-time: horizontal line
  • Acceleration-time: zero (horizontal at a=0a = 0)

Constant Acceleration

  • Position-time: parabola (curved)
  • Velocity-time: straight line (slope = acceleration)
  • Acceleration-time: horizontal line

Speeding Up vs. Slowing Down

Speeding up: velocity and acceleration have the same sign

  • Moving right and accelerating right: v>0v > 0, a>0a > 0
  • Moving left and accelerating left: v<0v < 0, a<0a < 0

Slowing down: velocity and acceleration have opposite signs

  • Moving right and accelerating left: v>0v > 0, a<0a < 0
  • Moving left and accelerating right: v<0v < 0, a>0a > 0

Problem-Solving Tips

  1. Identify the type of graph (position, velocity, or acceleration)
  2. Look at the y-value for the quantity itself
  3. Calculate slope to find the derivative quantity
  4. Calculate area to find the integral quantity
  5. Check signs carefully (positive/negative)
  6. Draw the graph if only given data

šŸ“š Practice Problems

1Problem 1easy

ā“ Question:

A position-time graph is a straight line passing through (0,5)(0, 5) m and (10,25)(10, 25) m. What is the velocity?

šŸ’” Show Solution

Given information:

  • Position at t=0t = 0 s: x0=5x_0 = 5 m
  • Position at t=10t = 10 s: x=25x = 25 m
  • Graph is a straight line (constant velocity)

Find: Velocity

Solution: The velocity is the slope of the position-time graph.

v=Ī”xĪ”t=xāˆ’x0tāˆ’t0v = \frac{\Delta x}{\Delta t} = \frac{x - x_0}{t - t_0}

v=25āˆ’510āˆ’0=2010=2Ā m/sv = \frac{25 - 5}{10 - 0} = \frac{20}{10} = 2 \text{ m/s}

Answer: The velocity is 2 m/s (constant).

Interpretation: The object moves in the positive direction at a steady 2 m/s. Since the slope is constant (straight line), there is no acceleration.

2Problem 2easy

ā“ Question:

A velocity-time graph shows a horizontal line at v=15v = 15 m/s from t=0t = 0 to t=8t = 8 s. Find the displacement and acceleration.

šŸ’” Show Solution

Given information:

  • Velocity: v=15v = 15 m/s (constant)
  • Time interval: 00 to 88 s
  • Graph is horizontal line

Find:

  1. Displacement
  2. Acceleration

Part 1: Displacement Displacement = area under velocity-time graph

Since velocity is constant, the area is a rectangle: Area=baseƗheight\text{Area} = \text{base} \times \text{height} Ī”x=tƗv=8Ɨ15=120Ā m\Delta x = t \times v = 8 \times 15 = 120 \text{ m}

Part 2: Acceleration Acceleration = slope of velocity-time graph

Since the line is horizontal: a=ΔvΔt=08=0 m/s2a = \frac{\Delta v}{\Delta t} = \frac{0}{8} = 0 \text{ m/s}^2

Answers:

  • Displacement: 120 m
  • Acceleration: 0 m/s² (moving at constant velocity)

Key insight: Horizontal line on vv-tt graph means constant velocity (zero acceleration).

3Problem 3medium

ā“ Question:

A velocity-time graph shows a straight line from (0,0)(0, 0) to (5,20)(5, 20) m/s. Find: (a) the acceleration, (b) the displacement during this time.

šŸ’” Show Solution

Given information:

  • Initial velocity at t=0t = 0: v0=0v_0 = 0 m/s
  • Final velocity at t=5t = 5 s: v=20v = 20 m/s
  • Graph is a straight line (constant acceleration)

Part (a): Find acceleration Acceleration = slope of velocity-time graph

a=Ī”vĪ”t=vāˆ’v0tāˆ’0a = \frac{\Delta v}{\Delta t} = \frac{v - v_0}{t - 0}

a=20āˆ’05āˆ’0=205=4Ā m/s2a = \frac{20 - 0}{5 - 0} = \frac{20}{5} = 4 \text{ m/s}^2

Part (b): Find displacement Displacement = area under velocity-time graph

The graph forms a triangle with:

  • Base = 55 s
  • Height = 2020 m/s

Area=12ƗbaseƗheight\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

Ī”x=12Ɨ5Ɨ20=12Ɨ100=50Ā m\Delta x = \frac{1}{2} \times 5 \times 20 = \frac{1}{2} \times 100 = 50 \text{ m}

Alternative method for (b): Use kinematic equation: Δx=12(v0+v)t\Delta x = \frac{1}{2}(v_0 + v)t

Δx=12(0+20)(5)=12(100)=50 m\Delta x = \frac{1}{2}(0 + 20)(5) = \frac{1}{2}(100) = 50 \text{ m}

Answers:

  • (a) Acceleration: 4 m/s²
  • (b) Displacement: 50 m

Check: Using v=v0+atv = v_0 + at: 20=0+4(5)=2020 = 0 + 4(5) = 20 āœ“