# So confused by this!

1. Sep 19, 2011

### einsteinphysi

1. The problem statement, all variables and given/known data
A particle's position is given by x = 3.00 - 12.00t + 3t2, in which x is in meters and t is in seconds. Where is the particle when it momentarily stops?

2. Relevant equations

3. The attempt at a solution
I tried plugging in m and s, like m= 3 - 12s +3s^2 and tried to solve for s but hit a dead end

2. Sep 19, 2011

### einsteinphysi

I know I'm probably doing it dead-wrong, though.

3. Sep 19, 2011

### SteamKing

Staff Emeritus
What is the velocity of the particle when it stops?

4. Sep 19, 2011

### SteamKing

Staff Emeritus
To reply to your attempted solution, why would you substitute 'm' for x and 's' for t?
'm' and 's' are the units of the time variable t and the position variable x.

5. Sep 19, 2011

### einsteinphysi

I don't want to answer but can you give me a hint as to how I would find that?
Well, if it's stopped...0?

6. Sep 19, 2011

### einsteinphysi

x = 3.00 - 12.00t + 3t2
v= 12 +6t?

7. Sep 19, 2011

### einsteinphysi

a=6?

8. Sep 19, 2011

### raymo39

ok so you have a function that describes the velocity (v = 12 + 6t) and you have a value for the velocity that you want to know what time it occurs at...

9. Sep 19, 2011

### einsteinphysi

So plug in 0 for v? 0=12 + 6t
-2=t?

10. Sep 19, 2011

### einsteinphysi

When I then plug -2 into original? 39?

11. Sep 19, 2011

### einsteinphysi

HAHA, I see a mistake (when I took the derivative).
v=-12 + 6t
0=-12 + 6t
2=t

12. Sep 19, 2011

### einsteinphysi

Now what?

13. Sep 19, 2011

### einsteinphysi

Got it!

14. Sep 19, 2011

### raymo39

well done! :)