Question

A particle moves on a vertical line so that its coordinate at time t is
3
y = t − 12t+ 3, t≥ 0 .

When is the particle moving upward and when is it moving downward?
Find the distance that the particle travels in the first 3 seconds.

I got that t=2 and t=-2

and its moving upward at (-infitinty,-2) and (-2,2)
and downward and (2, positive infinity)

am I right? If so, how do i find the distance in the first 3 seconds?

Answer 1

I'll assume you meant
y = t^3 - 12t + 3

the particle is moving downward when y is decreasing; that means y' < 0

y' = 3y^2-12
y' < 0 when 3y^2 < 12, or |y| < 2
So, since t>=0, y is decreasing when 0 <= y < 2.
upward when y > 2
So, you are right there, if you remember that t>=0.

The distance traveled in the first 3 seconds is the difference between where it ended and where it started: y(3)-y(0)
y(3) = -6
y(0) = 3

So, it moved down 9 units in the 1st 3 seconds.

Answer 2

do you mean the other way around, when 0<y<2 it is upward, and downward at y>2

When I graph it on a number line (how I was taught) that's what I got, but of course I could be wrong.

Answer 3

You are indeed wrong. Here is the graph of y:

http://www.wolframalpha.com/input/?i=t^3+-+12t+%2B+3

You can see that y is decreasing for t between 0 and 2.