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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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    Have a look at http://physics.stackexchange.com/q/92925/50583 – ACuriousMind May 30 '16 at 13:42
  • Look at the limit definition of $v$: $v=\lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t}=\frac{dx}{dt}$. In Physics we treat $dx$ and $dt$ as real numbers. – Gert May 30 '16 at 13:53
  • Also related: http://physics.stackexchange.com/q/70376/2451 , http://physics.stackexchange.com/q/255709/2451 and links therein. – Qmechanic May 30 '16 at 13:58
  • Thanks I'll check those links out, I did a stack search before asking this question. I didn't get any answers so I went on with the question sorry for the inconvenience. – user3621843 May 30 '16 at 14:06

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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  • This doesn't answer the question. You don't explain what $dx$ and $dt$ are and so why this cross multiplication is allowed. Also, velocity is important in describing motion: $v=\frac{dx}{dt}$, $a=\frac{dv}{dt}$, $a=\frac{d^2x}{dt^2}$ (in one dimension only). – Gert May 30 '16 at 14:00
  • dx is a small distance travelled in dt time(An infinitesimally small interval of time). I meant that to derive a=dv/dt is more significant than v=dx/dt. – Sid May 30 '16 at 14:02
  • I meant that to derive a=dv/dt is more significant than v=dx/dt. I think you mean that Newtonian equation of motion are based on $a$. But to solve them, very often we need $v=\frac{dx}{dt}$. – Gert May 30 '16 at 14:06
  • Yes, I agree with that. But, Kinematic equations of motion are usually based on constant acceleration. But, to calculate that, v=dx/dt is required. – Sid May 30 '16 at 14:17
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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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  • Could you please tell me what u is here. – user3621843 May 30 '16 at 16:03
  • @user3621843 have you learned u-substitutions in calculus yet? If not, look them up on wikipedia. If you understand the chain rule, you should be able to understand u-substitution. – Jahan Claes May 30 '16 at 17:17
  • Yeah just looked up a Khan Academy video. Since I'm relatively new to this I couldn't figure out what you took as the variable u – user3621843 May 30 '16 at 17:18
  • Oh, I see. In this case, $x$ is some function of $t$. So I took my variable $u$ to be $x$. Then $dx=\frac{dx}{dt}dt$. – Jahan Claes May 30 '16 at 17:23