Average Acceleration of a Accel vs Speed Graph

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Discussion Overview

The discussion revolves around determining the average acceleration from a given acceleration versus speed graph, specifically for a scenario where speed increases from 0 to 80 mph. Participants explore the relationship between acceleration, speed, and distance using various mathematical approaches and graphical representations.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks to find the average acceleration that would yield the same distance traveled as an acceleration from 0 to 80 mph, using the formula Distance = (Vfinal^2)/(2*Acceleration).
  • Another participant suggests that plotting 1/acceleration versus speed relates to the time spent at a specific velocity, while speed/acceleration versus speed relates to the distance traveled.
  • A participant expresses confusion regarding unit consistency and the results obtained from plotting data in Excel, questioning the validity of the area under the curve.
  • There is a discussion about whether the area under a velocity versus time graph gives total distance and if this holds for an (a/v)/v graph, raising the question of a constant needed after integration.
  • One participant provides a simple example of constant acceleration and derives the distance formula, showing how integration of v/a leads to the same result.
  • Another participant mentions that the calculations become less accessible under the influence of alcohol but finds the algebraic approach helpful, suggesting piecewise integration for non-constant acceleration.
  • A later reply emphasizes that the area under the speed/acceleration versus speed graph is equal to the distance traveled, not just proportional to it, and proposes an alternative method using 1/(2a) as a function of v^2.
  • Concerns are raised about the prefactor in the equations, with a participant questioning its validity and applicability to non-constant acceleration scenarios.

Areas of Agreement / Disagreement

Participants express differing views on the relationships between acceleration, speed, and distance, with some agreeing on the mathematical approaches while others raise questions about unit consistency and the validity of the methods discussed. The discussion remains unresolved regarding the best approach to calculate average acceleration from the given graph.

Contextual Notes

Participants note limitations regarding the assumptions made in their calculations, particularly in relation to constant versus non-constant acceleration and the need for constants in integration. There is also uncertainty about the correctness of the area calculations derived from the graphs.

lboucher
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Hi All

So I have an Accel vs Speed graph and data.
Goes from 0 to 80 mph.
This is all the info I have.

I would like to figure out what Average Acceleration would give me the same distance, of an acceleration from 0 MPH to 80 MPH.
AKA, using the formula Distance = (Vfinale^2)/(2*Acceleration)

Any ideas on how to figure this out?
 
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If you plot 1/acceleration versus speed, your function value is proportional to the time spent at a specific velocity (assuming your acceleration is always positive).
If you plot speed/acceleration versus speed, the area below the function is proportional to the distance travelled.
 
Sorry, but that still confuses me as the units don't add up. Also when I attempt to plot this out in excel and add up the area under the curve, the resulting numbers cannot be right
 
The area under the v/t graph should give total distance, whatever the instantaneous velocity and accelerations are. Does that hold for an (a/v)/v graph too? Is there not a constant needed after the integration?
 
A simple example: v(t)=at+c with constant a and c.
Distance after time T is ##\frac{1}{2}aT^2 + cT##.

v/a = t + c/a
t=(v-c)/a
Therefore, v/a = (v-c)/a + c/a = v/a (trivial in this example, as a is constant)
Speed increases from c to c+aT. If we integrate v/a from c to c+aT, we get ##\int \frac{v}{a}dv = \frac{1}{2a}((c+aT)^2-c^2) = cT+\frac{1}{2}aT^2## - the same as above.
 
Wine and gin make this less accessible but the sums seem right in this case. It that also OK for non constant acceleration? I guess it could be - piecewise. Good one. I love it when someone else does the algebra and I can follow it.
 
mfb said:
If you plot 1/acceleration versus speed, your function value is proportional to the time spent at a specific velocity (assuming your acceleration is always positive).
If you plot speed/acceleration versus speed, the area below the function is proportional to the distance travelled.
Not just proportional to the distance traveled. It is equal to the distance traveled. Another way to do this is to plot 1/(2a) as a function of v2. This will also be equal to the distance traveled.
 
I did not check the prefactor when I wrote my first post here, but it is 1, right.

It that also OK for non constant acceleration?
It works there as well, but the integration can get more difficult.
 

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