Easy question, I'm sure its something simple I'm overlooking

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Homework Help Overview

The problem involves determining the average acceleration of an object described by the position function x=4t^2-3t^3 over the time interval from t=0 to t=2 seconds. The context is kinematics, specifically focusing on acceleration derived from a position function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of average acceleration, contrasting it with instantaneous acceleration derived from the second derivative of the position function. Some participants question the method of finding average acceleration and consider the definition of average acceleration as the change in velocity over time.

Discussion Status

The discussion is active, with participants exploring different interpretations of average acceleration and its calculation. Some guidance has been offered regarding the definition of average acceleration, and there is acknowledgment of the confusion surrounding the use of instantaneous values.

Contextual Notes

There is a mention of homework constraints, particularly the need to calculate average acceleration rather than instantaneous values. Participants express uncertainty about the appropriate methods to use in this context.

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Homework Statement


The coordinate of an object is given as a function of time by x=4t^2-3t^3, where x is in meters and t is in seconds. Its average acceleration from t=0 to t=2s is:

A) -4m/s^2
B) 4m/s^2
C) -10m/s^2
D) 10 m/s^2
E) -13 m/s^2

Homework Equations





The Attempt at a Solution



I found the second derivative of the position function, which is acceleration, to be a(x)=-18t+8. I plugged 2 and 0 into the equation and subtracted the answer for 0 from the answer for 2.

-18t+8
[-18(2)+8]-[-18(0)+8]
-28-8
-36 ?
 
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Instead of working with the instantaneous acceleration (which is what the second derivative gives you), consider the definition of average acceleration. (What is it?)
 
Animal said:

Homework Statement


The coordinate of an object is given as a function of time by x=4t^2-3t^3, where x is in meters and t is in seconds. Its average acceleration from t=0 to t=2s is:

A) -4m/s^2
B) 4m/s^2
C) -10m/s^2
D) 10 m/s^2
E) -13 m/s^2

Homework Equations





The Attempt at a Solution



I found the second derivative of the position function, which is acceleration, to be a(x)=-18t+8. I plugged 2 and 0 into the equation and subtracted the answer for 0 from the answer for 2.

-18t+8
[-18(2)+8]-[-18(0)+8]
-28-8
-36 ?
You are not looking for the difference or change in acceleration during the 2 second interval; you are looking for the average acceleration in that period. Since you know the acceleration at t=0 and t=2, and the acceleration varies linearly with time, then how would you calculate the average between those 2 extremes?
 
i can't think of anything to use that doesn't involve subtracting one instantaneous thing from another...area under the curve is the antiderivative, right? so that won't work...is there some kind of equation i should use?
 
Since you've gone to the trouble of finding the second derivative, you can use it to find average acceleration as PhanthomJay suggests. But even easier is to just consider the definition of average acceleration = (change in velocity)/(time).
 
Ah...thank you much. I feel like a fool lol. Wish me luck on the midterm tomorrow guys :D
 
Animal said:
Ah...thank you much. I feel like a fool lol. Wish me luck on the midterm tomorrow guys :D

Good luck!
 

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