#### notes

Physical experiments have confirmed that the body dropped from a tall cliff covers a distance of 4.9`t^2` metres in t seconds, i.e., distance s in metres covered by the body as a function of time t in seconds is given by

s = 4.9`t^2`.

In the following table the distance travelled in metres at various intervals of time in seconds of a body dropped from a tall cliff.

The objective is to find the veloctiy of the body at time t = 2 seconds from this data. One way to approach this problem is to find the average velocity for various intervals of time ending at t = 2 seconds and hope that these throw some light on the velocity at t = 2 seconds.

Average velocity between t = `t_1` and t = `t_2` equals distance travelled between t = `t_1` and t = `t_2` seconds divided by `(t_2 – t_1)`. Hence the average velocity in the first two seconds.

t | s |

0 | 0 |

1 | 4.9 |

1.5 | 11.025 |

1.8 | 15.876 |

1.9 | 17.689 |

1.95 | 18.63225 |

2 | 19.6 |

2.05 | 20.59225 |

2.1 | 21.609 |

2.2 | 23.716 |

2.5 | 30.625 |

3 | 44.1 |

4 | 78.4 |

= `("Distance travelled between" t_2 = 2 and t_1 = 0)/("Time interval" (t_2-t_1))`

= `((19.6 - 0)m)/((2-0)s)` = 9.8 m/s

Similarly, the average velocity between t = 1 and t = 2 is

`((19.6 - 4.9)m)/((2-1)s)` = 14.7 m/s

we compute the average velocitiy between t = `t_1` and t = 2 for various `t_1`. gives the average velocity (v), t = `t_1` seconds and t = 2 seconds

`t_1` | 0 | 1 | 1.5 | 1.8 | 1.9 | 1.95 | 1.99 |

v | 9.8 | 14.7 | 17.15 | 18.62 | 19.11 | 19.355 | 19.551 |

From the above table, The average velocity is gradually increasing. the time intervals ending at t = 2 smaller, we see that we get a better idea of the velocity at t = 2.Hoping that nothing really dramatic happens between 1.99 seconds and 2 seconds, we conclude that the average velocity at t = 2 seconds is just above 19.551 m/s.