# Acceleration, and displacement

1. Mar 1, 2012

### jaja1990

I think we can't have a negative displacement. Is that right?

Acceleration: the change in displacement / change in time squared
Since displacement can't be negative, and time can't be negative, acceleration also can't be negative.
But I know (it's in the books) that when we have decreasing acceleration, it's negative. How is that?

2. Mar 1, 2012

### Staff: Mentor

Why do you think this? Realize that whether a vector is negative or not just depends on the sign convention used when specifying its components.
Better to think of acceleration as Δv/Δt.
Again, the sign of a vector is rather arbitrary.
Acceleration is a vector. A negative acceleration just means that the acceleration vector points in the negative direction. For example, if you take up as positive, the acceleration of a falling body will be negative.

3. Mar 1, 2012

### jaja1990

I understand now, it's the direction that is negative, not the magnitude.

Why is it better to think of acceleration as Δv/Δt?
I guess it just a matter of which is more intuitive/elegant, but maybe there is another reason; is there?

4. Mar 2, 2012

### SergioPL

Acceleration is the derivative of the velocity so it's dv/dt.

The sign of the acceleration can have different criteria, I think an acceleration is negative if the scalar product (dv/dt)•v < 0 and positive if (dv/dt)•v > 0. If the scalar product it's zero then the acceleration is perpendicular to the motion. This is the case in circular motion, for example.

5. Mar 2, 2012

### Staff: Mentor

Well, Δv/Δt is the definition of acceleration (at least average acceleration).

Blindly using Δx/(Δt)2 can lead to silly results. Imagine something moving at a constant velocity of 10 m/s for 1 second. Δx = 10, Δt = 1. Obviously the acceleration is zero here, so that formula fails.