Calculus and Physics Help! If a particle's position is given by $x = 4-12t+3t^2$ (where $t$ is in seconds and $x$ is in meters):
a) What is the velocity at $t = 1$ s?
Ok, so I have an answer:
$v = \frac{dx}{dt} = -12 + 6t$
At $t = 1$, $v = -12 + 6(1) = -6$  m/s
But my problem is that I want to see the steps of using the formula $v = \frac{dx}{dt}$ in order to achieve $-12 + 6t$...
I am in physics with calc, and calc is only a co-requisite for this class, so I'm taking it while I'm taking physics. As you can see calc is a little behind. We're just now learning limits in calc, and I was hoping someone could help me figure this out.

Think it this way. On a $x$ versus $t$ plot. The slope of the line is the rate of change of $x$ with respect to $t$ which is given by, 
$lim \ \Delta t\rightarrow0 \ \frac{\Delta x}{\Delta t}=\frac{dx}{dt}=v$
For more information, check this out
http://en.wikipedia.org/wiki/Derivative
