- #1

Matthew Travers

- 10

- 0

## Homework Statement

Saw a calculation that put differentiation of power in terms of acceleration as follows:

E=Fs

dE/dt=Fv=P

dP/dt=Fa=ma^2

It doesn't make sense to me because if power was changing, acceleration must change. Correct me if I'm wrong, but shouldn't the product rule be applied here?[/B]

## Homework Equations

If z=xy then

dz/dt=x(dy/dt)+y(dx/dt)

F=ma

E=Fs

P=F(ds/dt) with s(dF/dt)=0 so it is discarded[/B]

## The Attempt at a Solution

Instantaneously

P=Fs/t=F(ds/dt) provided ds/dt is smaller than c. Now first application of product rule with second diff of position defined as first diff of velocity

dP/dt=F(dv/dt)+(ds/dt)(dF/dt) since F=ma, subst in and apply product rule again

dP/dt=ma^2+(ds/dt)[m(da/dt)+a(dm/dt)][/B]

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