V dot (dv/dt) = (0.5)*(d/dt)*(v^2) ?

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In summary, the conversation discusses a derivation involving the equation F=ma and the power equation. The conversation also addresses the question of where the \frac{1}{2} comes from in Step 4 of the derivation. The answer is that it comes from simplifying the equation d/dt(v2)=2v*dv/dt.
  • #1
wavingerwin
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This is not a homework question, but a derivation in my class which I can't get around.

Homework Statement


Step1 [tex]F=ma[/tex]
Step2 [tex]\vec{F} = m\frac{d\vec{v}}{dt}[/tex]
Step3: Multiply both side by v [tex]\vec{F}.\vec{v} = m\vec{v}.\frac{d\vec{v}}{dt}[/tex]
Step4 [tex]Power = \frac{d}{dt}\frac{1}{2}m\vec{v}.\vec{v}[/tex]
Step5 [tex]Power = \frac{d}{dt}\frac{1}{2}mv^{2} = \frac{dK}{dt}
[/tex]


Homework Equations



The Attempt at a Solution


I just can't figure out where did the [tex]\frac{1}{2}[/tex] come from in step 4.

Please help & thanks in advance!
 
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  • #2
v2=v.v

So that d/dt(v2)=2v*dv/dt

They just wrote v2 as v.v
 
  • #3
rock.freak667 said:
So that d/dt(v2)=2v*dv/dt

Aha, I see.
Or more elaborately:

[tex]\frac{d}{dt}v^{2} = \frac{dv}{dt}\frac{d}{dv}v^{2} = \frac{dv}{dt}2v[/tex]

Thanks rock!
 

FAQ: V dot (dv/dt) = (0.5)*(d/dt)*(v^2) ?

What is the meaning of "V dot (dv/dt) = (0.5)*(d/dt)*(v^2)"?

"V dot (dv/dt) = (0.5)*(d/dt)*(v^2)" is a mathematical equation that represents the acceleration of an object as it changes over time. The "V dot" term refers to the derivative of velocity, while the "(dv/dt)" term represents the change in velocity over time. The other terms in the equation, "(0.5)*(d/dt)*(v^2)", represent the change in kinetic energy over time.

What is the significance of this equation in physics?

This equation is known as the "power equation" and is used to calculate the power or rate at which a force is doing work on an object. It is also used in the field of fluid dynamics to describe the motion of fluids.

How is this equation derived?

This equation can be derived from Newton's second law of motion, which states that the net force on an object is equal to its mass multiplied by its acceleration. By substituting the derivative of velocity for acceleration, the above equation is obtained.

What are some practical applications of this equation?

This equation has many practical applications in physics and engineering. It is used in the design and analysis of engines and turbines, as well as in the study of fluid flow and aerodynamics. It is also used in the field of robotics to calculate the power needed to move a robot or other machinery.

Are there any limitations to this equation?

This equation is only applicable to objects moving in a straight line with constant mass. It also does not take into account external forces such as friction or air resistance, which can affect the acceleration of an object. Additionally, it assumes that the mass of the object is constant, which may not always be the case in real-world scenarios.

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