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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.

  • 1
    It's too bad English composition isn't a prereq...2012-09-20
  • 0
    it's nice that you want to ascribe units to your stuff, but those units are obviously wrong.2012-09-21

3 Answers 3