The position function x(t) of a particle moving along an x axis is x 4.0 6.0t 2 , with x in meters and t in seconds. (a) at what time and (b) where does the particle (momentarily) stop? at what (c) negative time and (d) positive time does the particle pass through the origin?

a.) Differentiate your position function (the one they gave you) with respect to time. That is, take dx/dt. dx/dt is your velocity function, right? So, when you set that function equal to zero, that's when your velocity equals zero. The particle momentarily stops when velocity equals zero. Makes sense. Solve for t to get your time.

b.) Next step is to figure out where it stops. Since you figured out what TIME it stops (from part "a"), now you can determine WHERE it stops. How? Since position and velocity are related by the variable "t", you can just plug the "t" you solved for into the position function.

c.) and d.) are worded interestingly. What is a "negative time" anyway? Well, we consider t=0 to be our starting point in time, so we assume anything that happened before t=0 to be in "negative time". Semantics aside, just set your position function to be equal to zero. You'll get two answers, one positive and one negative. (remember that squaring a negative number provides the same result as squaring a positive one)

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