# Taxicab Geometry-it's against Euclidean geometry, so what?

In taxicab geometry, the shortest distance between two points is determined by many lines. But in Euclidean geometry, the shortest distance between 2 points is determined by only one line. So, while applying the same for physics too, isn't displacement not supposed only one path?

Or is it this way round? Finding the shortest possible ways through 2 given points would be adding the difference of the coordinates -->the distance between points L(3,2) and M(4,6): it becomes |3-4| + |6-2| = 1 + 4 = 5. Does the same go for finding the distance in a velocity-time graph?

Where am I going wrong?

### 3 Answers

- MorewoodLv 71 decade agoFavourite answer
There are many geometries that are useful in different areas of Physics: Especially relativity and gauge theories.

However, when looking at a velocity time graph you probably want the area between the curve and the x-axis.

The vertical is velocity, the horizontal is time. Velocity x Time = Distance. Height x Width = Area.

For example, if an object is going 2 m/s after 3 second {the point L(3,2)} and is going 6 m/s after 4 seconds {the point M(4,6)}, then the average speed is about (3+6)/2=4.5 m/s (if you want more accuracy give me more points) for 1 second. The distance traveled is about 4.5 m.

That is suspiciously close to 5, so lets look at another example: M(4,6) and N(6,6). Taxicab gives 2. Vecocity x Time gives 6x2=12>>2.

- santmann2002Lv 71 decade ago
The distance between to points L and M is given by d=sqrt((4-3)^2+(6-2)^2))

- 5 years ago
Taxicab geometry only follows paths like those on city streets. You only go forward, backward, left, or right. C is the only one that would use taxicab geometry.