# Average acceleration problem

1. Apr 13, 2007

### newton9point8

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

Okay here's the problem, The coordinate of an object is given as a function of time by x = 4t^2 - 3t^3, where v is in m/s and t is in seconds. Its average acceleration over the interval form t = 0 to t = 2

I wast told the answer is -10m/s^2 but i can't figure out how to get the same anwer.

2. Relevant equations

Aavg = V2-V1/time

3. The attempt at a solution

ok i've already tryed to plug in 4*2^2 - 3*2^3 and 4*0^2 - 3*0^3 but it doesn't work....

i know if i can get the right values for the change of velocity over 2seconds (my time) i can get the right answer but i can't figure out how to obtain
the change of velocity with the given function

2. Apr 13, 2007

### nrqed

Welcome to the forum.
First, let's start with a simpler question. Given x(t), do you know how to calculate th evelocity at a given time? (I am talking about the instantaneous velocity, not the average velocity). You need to know that because the first thing you need to do is to calculate the velocity at t=0 and then at t=2 seconds.

3. Apr 13, 2007

### newton9point8

the instantaneus velocitiy formula is V= dx/dt but I still need to know what to do with the given function..... i don't know how to use the given funtion to get the answer. If I was given the v1 and v2 in m/s i would just take the difference of v1 and v2 divide it by the time but that functions is what is holding me back. :(

4. Apr 13, 2007

### nrqed

You just provided the answer yourself! To find the instantaneous velocity at a certain time, you need to calculate the derivative of th efunction x(t) and then evaluate it at that time. So to find v1 (= velocity at t=0). find dx/dt and then set t=0. To find v2, plug t=2 in the derivative dx/dt.

5. Apr 13, 2007

### newton9point8

Got it!!!
so i got the derivative v=8t-9t^2 and then i got -20 after i pluged in my v1 and v2 then divided it by two and that gave me -10m/s^2!!! (right answer)