### Linearisation of Experimential Data

Given the following Data we wish to understand the relationship between Distance and Time

Time (s) | Distance (m) |
---|---|

0.01 | 0.00 |

0.10 | 0.05 |

0.20 | 0.11 |

0.30 | 0.43 |

0.40 | 0.75 |

0.50 | 1.14 |

0.60 | 1.84 |

0.70 | 2.53 |

0.80 | 3.45 |

0.90 | 3.67 |

1.00 | 4.87 |

## Step 1) Plot the data

#### Plot the Data Using Computer Software

## Step 2) Identify the General Relationship

#### Given Our Data Plot and the Plots of our Basic Relationships We need to choose the one that is the best match when compared to our data

Plot | Basic Relationship | Our Data Plot |
---|---|---|

y = kx | ||

y = k x^{2} |
||

y = sqrt(x) | ||

y=1/x | ||

y = 1/x^{2} |

#### In Our Case our data is best matched by the Quadratic Plot y = k x^{2}

## Step 3) Linearise the Data

#### Determine How we Need to Modify Our Data to get a Linear Plot

#### So as we can see we will need to square the time and then create a new plot

## Step 4) Plot the Linearised Data

#### Data after Modification

Time (s) | Time^{2} (s^{2}) |
Distance (m) |
---|---|---|

0.01 | 0.0001 | 0.00 |

0.10 | 0.010 | 0.05 |

0.20 | 0.04 | 0.11 |

0.30 | 0.09 | 0.43 |

0.40 | 0.16 | 0.75 |

0.50 | 0.25 | 1.14 |

0.60 | 0.36 | 1.84 |

0.70 | 0.49 | 2.53 |

0.80 | 0.64 | 3.45 |

0.90 | 0.81 | 3.67 |

1.00 | 1.00 | 4.87 |

## We will plot Distance (y axis) versus Time^{2} (x axis)

## Step 5) Draw your Conslusions

## Understanding Our Results

## Slope of Our Line 4.898 m/s^{2}

## Y - Intercept effectivly zero

## Compare to Actual Physics Equations

## Equation that Governs a Falling Ball

## Distance = (1/2)a_{g} time^{2}

This means that our slope is equal to (1/2) a

_{g}, where a

_{g}is the acceleration due to gravity

## 4.898 = (1/2) a_{g}

## a_{g} = 9.79 m/s^{2} Which is our measured acceleration due to gravity

The accepted value for the acceleration due to gravity on Earth is 9.81 m/s^{2}so we

are in very good aggrement.