# So confused by this!

## Homework Statement

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?

## 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

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

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

SteamKing
Staff Emeritus
Homework Helper
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.

What is the velocity of the particle when it stops?

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?

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

a=6?

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...

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...

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

When I then plug -2 into original? 39?

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

Now what?

Got it!

well done! :)