Average velocity and acceleration

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Homework Help Overview

The discussion revolves around calculating average acceleration between two specific time points, t = 2.2 s and t = 9.6 s, based on a provided graph. The subject area includes concepts of average velocity and acceleration in the context of kinematics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various methods to calculate average acceleration, including using slopes from a position-vs-time graph and applying the formula for average acceleration. There are attempts to clarify the relationship between average velocity and the graph's slope.

Discussion Status

The discussion has seen multiple attempts to calculate the average acceleration, with some participants expressing confusion about the correct approach. One participant notes a realization that calculus is necessary for the problem, indicating a shift in understanding. There is no explicit consensus on the correct method yet, but some productive insights have been shared.

Contextual Notes

Participants mention a lack of instruction on calculus concepts relevant to the problem, which may affect their understanding and approach to solving it. Additionally, there are references to multiple attempts at finding the correct answer, suggesting constraints on the number of tries allowed.

Kajayacht
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Homework Statement



What is the average acceleration between time t = 2.2 s and time t = 9.6 s?
http://img98.imageshack.us/my.php?image=prob20av2yo4.gif
I've already found the average velocity between the two points to be 1.09×10^-1 m/s


Homework Equations



a= "delta"V / "delta" T
V= d/t

The Attempt at a Solution



v1= 9/2.2 = 4.09
v2= 10/9.6 = 1.04

a = (1.04-4.09) / (9.6-2.2) = -3.05 / 7.4 = -.412 m/s^2
 
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Actually, v is the slope of the position-vs.-time graph.
 
Ok, so I tried with the slope between 2.2 and 9.6

V= (10-9)/(9.6-2.2) = 1/7.4 = .135

a= .135/7.4 = .0183m/s^2

still says I'm wrong.
 
Got it, the main problem here was it involved calculus and my calc professor was sick the day she would've taught us that (that's a big thing)

Anyway, I had to take the slopes as it approached 2.2 and 9.6 which I found to be 5 and -10 plug and chug, shoop dah woop and it came out to -2.03 m/s^2 which the lon-capa thing told me yes.

and I only used up 18 of my 30 tries on it.
 

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