# Homework Help: Easy question, I'm sure its something simple I'm overlooking

1. Dec 18, 2006

### Animal

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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 ?

2. Dec 18, 2006

### Staff: Mentor

Instead of working with the instantaneous acceleration (which is what the second derivative gives you), consider the definition of average acceleration. (What is it?)

3. Dec 18, 2006

### PhanthomJay

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?

4. Dec 18, 2006

### Animal

i cant think of anything to use that doesnt involve subtracting one instantaneous thing from another...area under the curve is the antiderivative, right? so that wont work...is there some kind of equation i should use?

5. Dec 18, 2006

### Staff: Mentor

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).

6. Dec 18, 2006

### Animal

Ah...thank you much. I feel like a fool lol. Wish me luck on the midterm tomorrow guys :D

7. Dec 18, 2006

### cristo

Staff Emeritus
Good luck!!