Why Is Mass Halved in the Work-Energy Theorem Derivation?

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Scalise
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Hello,

Someone could explain me why in the derivation below the mass m is divided by 2 in the last step?:

##\int\vec{F}\cdot d\vec{s}=m\int\frac{d\vec{v}}{dt}\cdot\vec{v}dt=\frac{m}{2}\int \frac{d}{dt}(v^{2})dt##
 
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Scalise said:
Hello,

Someone could explain me why in the derivation below the mass m is divided by 2 in the last step?:

##\int\vec{F}\cdot d\vec{s}=m\int\frac{d\vec{v}}{dt}\cdot\vec{v}dt=\frac{m}{2}\int \frac{d}{dt}(v^{2})dt##

The derivative ##\frac{d}{dt}(v^{2})## is equal to ##2\vec{v}\cdot\frac{d\vec{v}}{dt}##