Particle Position and Momentary Stop: Solving a Motion Equation

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Homework Help Overview

The discussion revolves around a particle's motion described by the equation x = 3.00 - 12.00t + 3t², focusing on determining the position of the particle when it momentarily stops.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between position and velocity, with attempts to derive the velocity function from the position equation. Questions arise regarding the substitution of variables and the implications of setting velocity to zero to find when the particle stops.

Discussion Status

The discussion has seen various attempts to clarify the relationship between position, velocity, and time. Some participants have identified errors in their calculations and are revisiting their approaches. There is an ongoing exploration of how to correctly apply the derived velocity function.

Contextual Notes

Participants are navigating through potential misunderstandings regarding variable substitution and the implications of the motion equation. There is a recognition of the need to accurately derive and interpret the velocity function in the context of the problem.

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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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