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 = 1s? Ok, so I have an answer: v = dx/dt = -12 + 6t At t = 1, v = -12 + 6(1) = -6 m/s but my problem is I want to see the steps of using the formula v = dx/dt in order to achieve -12 + 6t... I am in phx with calc and calc is only a co-requisite for this class so im taking it while im taking physics but as you can see calc is a little behind. Where just now learning limits in calc and I was hoping someone could help me figure this out.