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

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The average acceleration of an object described by the position function x=4t^2-3t^3 from t=0 to t=2 seconds needs to be calculated using the definition of average acceleration as the change in velocity over time. The second derivative, which provides instantaneous acceleration, is not directly applicable for this calculation. Instead, the correct approach involves determining the velocities at t=0 and t=2, then applying the average acceleration formula. Acknowledgment of the misstep in focusing on instantaneous acceleration rather than average acceleration was noted, leading to a clearer understanding of the problem. The discussion concludes with a light-hearted note wishing good luck for an upcoming midterm.
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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 teh 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 derivitive, 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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