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I just started on learning how to derive the equations of motion using calculus. This is my first time and I have a basic doubt.

So

$v = \large{\frac{dx}{dt}}$

Then,

$dx = v dt $

What does this mean? How can we even do this isn't this mathematical operator?

My brain starts to get more annoyed when both sides are integrated.

Could any one please explain the meaning of the above steps.

Thanks for taking on the trouble.

Gert
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user3621843
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3 Answers3

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v=dx/dt. It simply means that v is the slope of the x-t graph. v=dx/dt. dx=vdt. This is simple cross multiplication. Although I am not too sure, v=dx/dt has anything to do with deriving the equations of motion as it is acceleration that is the main factor in the equations of motion not velocity.

Sid
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The term ${\rm d}t$ means an infinitesimal quantity of time. Like a slice of time so small, but still not zero.

When something is moving with speed $v$ the infinitesimal distance it travels during the infitesimal time is designated ${\rm d}x = v \, {\rm d}t$. Speed just linearly scales time into distance.

Now to get the total time passed, you sum of all the little slices of time $t = \int {\rm d}t$.

If you scale this with the speed you will get the total displacement $x = \int {\rm d}x = \int v {\rm d}t$.

The advantage is that speed $v$ doesn't have to be a constant since at each time $t$ it contributes only ${\rm d}x = v(t)\,{\rm d}t$ of distance. It is not obvious, but I hope it is intuitive that each ${\rm d}x$ depends on time. When speed $v(t)$ is small ${\rm d}x$ is small(er) and when speed is large $v(t)$ is large ${\rm d}x$ is larger(er).

The idea of calculus is that of slicing a problem into infinitesimal slices that behave linearly with each other, solving for the slice of interest and then summing up all the effects using integration.

John Alexiou
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Treating $dx$ as a quantity you can multiply/divide both sides by is really just a shorthand for the chain rule, or equivalently u-substitution. Here's what's going on.

You have

$$ v=\frac{dx}{dt} $$

Thus, if you integrate both sides against time, they should (up to a constant) be equal

$$ \int v dt = \int \frac{dx}{dt}dt $$

Doing a u-substitution on the right side gives

$$ \int v dt = \int dx $$

This looks exactly the same as "multiplying" both sides by $dt$, and taking the integral. The reason cancelling a $dt$ on the top and bottom works is simply because of u-substitution. Frequently, physicists omit these details, and just treat $dt$ as something you can multiply by.

Jahan Claes
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