# Why displacement equals (average a) x (t)

1. Mar 12, 2012

### V0ODO0CH1LD

I get the proof with the velocity as a function of time graphic.

I don't get why that is...

E.G.

Why does something with a constant acceleration of (1m/s)/s gets displaced the same as something with a constant velocity of .5m/s?

2. Mar 12, 2012

### rcgldr

That's only true for a specific amount of time, in this case 0 or 1 seconds, zero because no movement occurs, and 1 second because average velocity for the constant acceleration will be .5 m/s. At any other time, the average velocity and displacement will not be the same.

3. Mar 12, 2012

### V0ODO0CH1LD

Yeah, I realized that only after I posted...

But it still doens't answer the question of why that is..

Is it because the body spends the same amount of time accelerating towards the velocity of .5m/s as it spends accelarating away from it? And all of that at the same rate?

So it kind of compansates and adds up to the velocity which the body accelarates around?

4. Mar 12, 2012

### Ken G

Yes. You are comparing vt with at2/2, so cancelling out one t means we are comparing v to at/2. Note that at is the final v, so if v starts out zero, then at/2 is the same thing as the average speed over the time interval t. So you will always have the same average v if you set t such that v = at/2.