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?

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.

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.

To determine when the particle is moving upward or downward, we need to analyze the sign of its velocity. The particle's velocity is the derivative of its position function.

Given that the particle's coordinate at time t is given by y = t − 12t + 3, we can find its velocity function by taking the derivative with respect to time, which is dy/dt.

dy/dt = 1 - 12 = -11

The velocity function is therefore -11.

If the velocity is positive, the particle is moving upward. If the velocity is negative, the particle is moving downward.

Since the velocity is constant at -11, it is always negative, indicating that the particle is moving downward at all times.

Now, to find the distance the particle travels in the first 3 seconds, we need to calculate the total displacement of the particle in that time period.

The displacement can be found by evaluating the position function at the starting and ending times and taking the absolute difference.

y(0) = 0 - 0 + 3 = 3 (initial position)
y(3) = 3 - 12(3) + 3 = -33 (position after 3 seconds)

The absolute difference between y(0) and y(3) gives us the displacement:

|y(3) - y(0)| = |-33 - 3| = 36 units

Therefore, the particle travels a distance of 36 units in the first 3 seconds.

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.