The equivilence of Power and Kinetic Energy

[Unto]

This isn't homework, just be going over a few concepts.

I'm trying to show that the power delivered by a force equals the rate at which the particle is changing.

Now P = \vec{F} \bullet \vec{v}
= m\vec{a} \bullet \vec{v}
= m\vec{v} \stackrel{\delta}{\delta t}V^2

This book is now telling me that the above line = 2\vec{a} \bullet \vec{v}whyyyyy?

 

[Feldoh]

I can't really read what you wrote...

P = \vec{F} \cdot \vec{v}

P = m \vec{a} \cdot \vec{v}

P = m \frac{d \vec{v}}{dt} \cdot \vec{v}

P = m \frac{d}{dt}\frac{||{v}||^2}{2}

But after that you get that

2 \vec{a} \cdot \vec{v} = \frac{d ||{v}||^2}{dt}

 

[Unto]

I can't really read what you wrote...

P = \vec{F} \cdot \vec{v}

P = m \vec{a} \cdot \vec{v}

P = m \frac{d \vec{v}}{dt} \cdot \vec{v}

P = m \frac{d}{dt}\frac{|{v}|^2}{2}

But after that you get that

2 \vec{a} \cdot \vec{v} = \frac{d (|{v}|^2)}{dt}

In your 4th line, where does that 2 come from? I still don't understand the jump from v to a, shouldn't it be only 2a? Not 2a x V?

 

[Feldoh]

In the fourth line comes from the chain rule, which is also the reason you get 2a dot v.

\frac{d}{dt} \frac{v(t)^2}{2} = (2 v(t)) \frac{d}{dt}\frac{v(t)}{2}

 

[Unto]

Oh snap, totally forgot about that. Thank you.