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

## 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

## 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 ?

Doc Al
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?)

PhanthomJay
Homework Helper
Gold Member

## 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

## 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 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?

Doc Al
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).

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

cristo
Staff Emeritus