# Distance < Displacement

1. Jun 25, 2005

### jacyh

Why is the distance never less than the displacement?
I can't seem to find a scientific explanation for it.

2. Jun 25, 2005

### howie222

think about travelling half way around a circle, the distance travelled would be half the circumference, and the displacement would be the diameter.

in this case distance > displacement

distance can be equal to the displacement (if you travel in a straight line), but it can never be less.

3. Jun 25, 2005

### jacyh

distance can be equal to the displacement (if you travel in a straight line), but it can never be less.

I know, but... why? Haha.

4. Jun 25, 2005

### howie222

just think about the example i gave you and think of any other ones you can think of..... the displacement will never be greater than the distance.

Think of distance as "distance travelled"

So if you're going from point A to point B in any situation you can think of (around curves, over mountains etc...) The distance travelled will be greater because you had travel "around" things.
The displacement is a straight line between point A and B, so it is always the shortest possible distance.

Last edited: Jun 25, 2005
5. Jun 25, 2005

### robphy

Strictly speaking, distance is a magnitude but displacement is a vector.
So, yoiu really mean
"why is distance >= magnitude of displacement ?"

Mathematically,
$$\int \left| d\vec s \right| \geq \left| \int d\vec s \right|$$

Essentially, distance [ the arc-length of a curve from A to B ] is the sum of non-negative quantities.
The magntude of displacement [ the magnitude of a vector from A to B ] is the non-negative magnitude of a sum-of-(signed)-vector-quantities.
The proof of the inequality is essentially the triangle inequality.

6. Jun 25, 2005

### seiferseph

distance is total distance traveled, displacement is from the initial location to the final location.

7. Jun 25, 2005

### Knavish

The displacement is the shortest way possible from the start to the finish (that is, it's a line); thus, nothing can be shorter.

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