Solving Rectilinear Motion Equations - Deriving at^2+v+h

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Discussion Overview

The discussion revolves around deriving the equation for rectilinear motion, specifically the expression involving acceleration, time, initial velocity, and height. Participants explore the integration process and the correct formulation of the equation, addressing potential discrepancies in the terms used.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant recalls needing to derive the equation but struggles with the integration process and the correct formulation.
  • Another participant suggests starting with the relationships dv/dt = a and dx/dt = v to clarify the derivation.
  • A participant proposes a derived equation x(t) = at² + v₀t + h₀, indicating that h₀ represents the initial height.
  • There is a correction regarding the absence of the 1/2 factor in the derived equation, with one participant noting its importance.
  • A later reply questions the presence of a negative sign in the term -1/2at², suggesting it may relate to an object being thrown upwards, where initial velocity is positive and acceleration is negative.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct formulation of the equation, as there are multiple interpretations and corrections regarding the inclusion of the 1/2 factor and the sign of the acceleration term.

Contextual Notes

Participants express uncertainty about the definitions and implications of terms like "fluxion" and the correct application of integration in the context of rectilinear motion.

BloodyFrozen
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I can't remember how to derive this equation...

x(fluxion)
a=acceleration

x=a

From that, how do we get->

at^2+v+h

I think it had to do with integration, but I can't seem to get it to match the above.
 
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Hi BloodyFrozen! :smile:

(try using the X2 icon just above the Reply box :wink:)

I don't understand what you mean by "fluxion" or "x=a" :confused:

Try starting with dv/dt = a (and dx/dt = v). :smile:
 
BloodyFrozen said:
at^2+vt+h

My bad.

Anyways, I think I got it...

dv/dt = a
v(t)=at+C1
v0=C1

v(t)=at+v0
x(t)=at2+v0t+C2
x0=C2

Therefore,

x(t)=at2+v0t+x0

Since x0 is the position at t=0, he can just replace it as the original height.


x(t)=at2+v0t+h0

Correct?:smile:
 
BloodyFrozen said:
x(t)=at2+v0t+h0

Correct?:smile:

erm :biggrin:

what happened to the 1/2 ? :rolleyes:
 
tiny-tim said:
erm :biggrin:

what happened to the 1/2 ? :rolleyes:

Woops my bad

1/2at2...:wink:

In one of the calculus review books, it says it's -1/2at2... Why is it that?
 
BloodyFrozen said:
In one of the calculus review books, it says it's -1/2at2... Why is it that?

i'll guess it's talking about something being thrown up

so if v0 is positive, then the acceleration is negative
 

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