<p>I have been absent to class for the last two days(Double Blocked), so I am a little lost on what the class is learning right now. I have most of it down, but I just can't figure out how to find the acceleration at time, t and at t=4 for the following problem.
f(t)=(t^3)-(6t^2)+9t
I also have been having problems on the question; When is the particle moving to the right? To the left?</p>
<p>The acceleration is the second derivative of f(t), so take f"(t) and plug in 4 for t</p>
<p>Assuming right is positive, the particle is moving right when velocity (first deriv of f) is positive and left when velocity is negative.</p>
<p>f(t)=(t^3)-(6t^2)+9t
f'(t)=(3t^2)-12t+9
f''(t)=6t-12
6(4)-12=12
Is this right?</p>
<p>Looks good to me :)</p>
<p>Thanks you guys! :)</p>