Why displacement equals (average a) x (t)

[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?

 

[rcgldr]

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

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.

 

[V0ODO0CH1LD]

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.

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 accelerating 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?

 

[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.