Conceptual question about the Kinetic Energy Equation

[Xyius]

So as you all know, kinetic energy is..
K=\frac{1}{2}mv^2

The question I have is the followng.

The "v" in this formula isn't really velocity, its speed is it not? Which is just the magnitude of the velocity vector with no direction information. Kinetic energy is a scalar quantity so with this formula am I allowed to write..

v=\frac{dx}{dt}

Since this is just speed, which is the magnitude of the velocity vector, how does this make sense? Couldn't it just as easily be dy/dt or dz/dt? When deriving the kinetic energy formula from Newtons Second Law, you end up with the integral of "Fv" on one side and manipulate v to "dx/dt" to get the work integral.

I guess the main point of my question is, how can v=dx/dt when "v" isn't specifying any particular direction?

 

[rcgldr]

v² is the dot product of (v)(v) which is a scalar equal to the square of the magnitude of v. So velocity² = speed², but velocity in the expression 1/2 m v² is a vector and does have a direction.

 

[Xyius]

v² is the dot product of (v)(v) which is a scalar equal to the square of the magnitude of v. So velocity² = speed², but velocity in the expression 1/2 m v² is a vector and does have a direction.

So basically, only the square of v is scalar? But velocity itself is a vector? Makes sense! Thanks :D

 

[jtbell]

Consider this:

K = \frac{1}{2}mv^2 = \frac{1}{2}m (v_x^2 + v_y^2 + v_z^2)<br /> = \frac{1}{2}m \left[ {\left( \frac{dx}{dt} \right)}^2 + {\left( \frac{dy}{dt} \right)}^2<br /> + {\left( \frac{dz}{dt} \right)}^2 \right]