Net Work Calculus: Understanding mΔx(Δv/Δt)

[Nile3]

In the equation:

W_net=F_netΔX=[m*Δv/Δt]Δx=m*Δv*Δx/Δt

I just don't understand how the 3rd part of the equation is changed into the 4rth (Dt going under the Dx). If I can get some help with this, that would be great.

The way I think about it is [m*Δv/Δt]Δx = mΔx(Δv/Δt)

I have no idea how Dt went under the Dx...

 

[Mark44]

In the equation:

W_net=F_netΔX=[m*Δv/Δt]Δx=m*Δv*Δx/Δt

I just don't understand how the 3rd part of the equation is changed into the 4rth (Dt going under the Dx). If I can get some help with this, that would be great.

The way I think about it is [m*Δv/Δt]Δx = mΔx(Δv/Δt)

I have no idea how Dt went under the Dx...

It's no different from writing
$$\frac{3}{4}\cdot 5 = 3\cdot \frac{5}{4}$$

 

[HallsofIvy]

Note that $\Delta x$, $\Delta v$, and $\Delta t$ are numeric approximations to the differentials, not the differentials. They are simply numbers and what is going on is simply manipulation of numbers.
$$a\frac{b}{c}= \frac{ab}{c}= \frac{ba}{c}= b\frac{a}{c}$$

 

[Nile3]

Ah thank you. I couldn't see it for some reason...