Finding the Minimum Speed of a Particle with Given Position Function

[user8899]

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?

 

[jbunniii]

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 $c'(t)$.

 

[user8899]

Ok, so I would just get <2,3,-2>. Then what would I do?

 

[Ray Vickson]

Ok, so I would just get <2,3,-2>. Then what would I do?

Speed = magnitude of velocity.

 

[jbunniii]

Ok, so I would just get <2,3,-2>. Then what would I do?

The second component should not be 3.

 

[user8899]

The second component should not be 3.

Yea sorry it's <2,0,-2>

 

[user8899]

Speed = magnitude of velocity.

So the speed is 0? How do we find the time then? That's what I am stuck on.

 

[Fredrik]

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.

 

[user8899]

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

 

[Fredrik]

Yes, that's right.

 

[user8899]

Yes, that's right.

Thanks, so for time what would I do since when you find the derivative t is no longer there?

 

[jbunniii]

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.

 

[user8899]

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 :)