Particle Position and Momentary Stop: Solving a Motion Equation

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The particle's position is defined by the equation x = 3.00 - 12.00t + 3t². To find where the particle momentarily stops, the velocity equation, derived from the position equation, is v = -12 + 6t. Setting the velocity to zero to find the stopping time results in t = 2 seconds. Substituting t back into the position equation gives the particle's position as x = 39 meters when it stops. The discussion highlights the importance of correctly deriving and solving the equations of motion.
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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?




Homework Equations





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
 
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I know I'm probably doing it dead-wrong, though.
 
What is the velocity of the particle when it stops?
 
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.
 
SteamKing said:
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...
 
raymo39 said:
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?
 
  • #10
When I then plug -2 into original? 39?
 
  • #11
HAHA, I see a mistake (when I took the derivative).
v=-12 + 6t
0=-12 + 6t
2=t
 
  • #12
Now what?
 
  • #13
Got it!
 
  • #14
well done! :)
 
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