Why displacement equals (average a) x (t)

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    Displacement
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Discussion Overview

The discussion revolves around the relationship between displacement, average acceleration, and time, particularly in scenarios involving constant acceleration and constant velocity. Participants explore the conditions under which displacement can be equated to the product of average acceleration and time, examining both theoretical and conceptual aspects.

Discussion Character

  • Exploratory
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants express confusion about why a body with constant acceleration of (1m/s)/s displaces the same distance as one with a constant velocity of 0.5m/s, suggesting this equivalence holds only for specific time intervals.
  • One participant clarifies that the equivalence is true at 0 or 1 second, noting that at 1 second, the average velocity for the constant acceleration equals 0.5m/s.
  • Another participant speculates that the body accelerates towards and away from the velocity of 0.5m/s, proposing that this might explain the observed displacement equivalence.
  • A later reply indicates that when comparing velocity (v) to displacement (at²/2), cancelling out time (t) leads to a comparison of v to at/2, suggesting that if initial velocity is zero, at/2 represents the average speed over the time interval.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the underlying reasons for the displacement equivalence, with multiple interpretations and explanations being offered. The discussion remains unresolved regarding the conceptual understanding of the relationship between acceleration, velocity, and displacement.

Contextual Notes

Participants highlight that the relationship holds under specific conditions, and there is an emphasis on the time intervals considered. The discussion does not resolve the broader implications of these relationships in different contexts.

V0ODO0CH1LD
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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?
 
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V0ODO0CH1LD said:
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.
 
rcgldr said:
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?
 
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.
 

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