Finding the Minimum Speed of a Particle with Given Position Function

user8899
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Homework Statement



The position function of a particle is given by r(t) = <2t, 3, -2t+1> where t is the time in seconds. When is the speed a minimum? What is the minimum speed?

Homework Equations



v(t) = r'(t)

The Attempt at a Solution



The derivative of r'(t) = 2 + 0 + (-2) = 0. The t is canceled out so how are we supposed to find t and the minimum speed? I am so confused. Help please?
 
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user8899 said:

Homework Statement



The position function of a particle is given by r(t) = <2t, 3, -2t+1> where t is the time in seconds. When is the speed a minimum? What is the minimum speed?

Homework Equations



v(t) = r'(t)

The Attempt at a Solution



The derivative of r'(t) = 2 + 0 + (-2) = 0.

No, the derivative should be a vector. If ##r(t) = \langle a(t), b(t), c(t)\rangle##, then ##r'(t) = \langle a'(t), b'(t), c'(t) \rangle##. You don't sum ##a'(t)##, ##b'(t)##, and ##c'(t)##.
 
Ok, so I would just get <2,3,-2>. Then what would I do?
 
user8899 said:
Ok, so I would just get <2,3,-2>. Then what would I do?

Speed = magnitude of velocity.
 
user8899 said:
Ok, so I would just get <2,3,-2>. Then what would I do?
The second component should not be 3.
 
jbunniii said:
The second component should not be 3.
Yea sorry it's <2,0,-2>
 
Ray Vickson said:
Speed = magnitude of velocity.
So the speed is 0? How do we find the time then? That's what I am stuck on.
 
user8899 said:
So the speed is 0?
No, it's not. He meant the magnitude (i.e. the norm) of the vector, not the magnitude of the sum of its components.
 
Fredrik said:
No, it's not. He meant the magnitude (i.e. the norm) of the vector, not the magnitude of the sum of its components.

I am really confused now. Could you give me some guidance?

So the derivative is velocity right? The magnitude of the vector would be √((2)^2 + (-2)^2)? Then that would be speed which is √8?

Is this right?

Thanks
 
  • #10
Yes, that's right.
 
  • #11
Fredrik said:
Yes, that's right.

Thanks, so for time what would I do since when you find the derivative t is no longer there?
 
  • #12
user8899 said:
Thanks, so for time what would I do since when you find the derivative t is no longer there?
You have shown that the speed is constant, so it achieves both a minimum and a maximum at all times.
 
  • #13
jbunniii said:
You have shown that the speed is constant, so it achieves both a minimum and a maximum at all times.

oh! okay that makes sense! Thank you everybody :)
 
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