Q: If we use reasonably high constant
accelerations, how long does it take to travel across the country, say,
or across the planet?

[24 Jan 2002]

A: If we accelerate and decelerate uniformly
with acceleration ±`a` over a total distance `s`
then

time = `t` = sqrt(4`s` / `a`),
midpoint speed = `v` = sqrt(`as`)

so for a few sample accelerations we have

s | a |
t | v |
||
---|---|---|---|---|---|

100km | 1g | 202s | 3.4min | 990m/s | 2210mph |

4g | 101s | 1.7min | 1980m/s | 4431mph | |

10g | 64s | 1.1min | 3130m/s | 7000mph | |

1000km | 1g | 639s | 10.7min | 3130m/s | 7000mph |

4g | 319s | 5.3min | 6260m/s | 14000mph | |

10g | 202s | 3.4min | 9900m/s | 22100mph | |

Earth's diameter | 1g | 2340s | 39min | 11459m/s | 25600mph |

4g | 1170s | 19.5min | 22900m/s | 51300mph | |

10g | 739s | 12.3min | 36300m/s | 81100mph |

This page is maintained by Thomas Bending,
and was last modified on 7 March 2017.

Comments, criticisms and suggestions are welcome.
Copyright © Thomas Bending 2017.