# What does v=dx/dt means?

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#### Explanation

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47
Gió Share
Sep 16, 2015

It represents the instantaneous variation of POSITION with TIME. also known, as velocity.

#### Explanation:

The explanation can be a little bit boring but...
Cosider a car that moves from position ${x}_{1}$ at instant ${t}_{1}$ to position ${x}_{2}$ at instant ${t}_{2}$.
You can represent the variation of position using a number (a kind of position parameter) called average velocity as:
${v}_{a v} = \frac{{x}_{2} - {x}_{1}}{{t}_{2} - {t}_{1}} = \frac{\Delta x}{\Delta t}$

The problem is: "what happened in the middle? Did the car stop...go faster...slower...?"

To "look" inside your interval you can reduce the time interval and try to focus on a specific instant.

This means reducing $\Delta t$ to zero or at least tend to zero!

So, basically, you'll be able to evaluate the velocity at a point (not interval) and have an instantaneous velocity!

It is easy to say but mathematically...you need:
${v}_{\text{inst}} = {\lim}_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{\mathrm{dx}}{\mathrm{dt}}$ which is the "symbol" for an operation done on a function called Derivative.

For example:
consider a car that has a position modelled by the function:
$x \left(t\right) = - 4 {t}^{2} + 3 t - 2$ (I invented it)
So
instantaneous velocity will be given as:
$\frac{\mathrm{dx}}{\mathrm{dt}} = - 8 t + 3$
So at each instant you will get the velocity at exactly that instant: for example at $t = 0$ ${v}_{\text{inst}} = - 8 \cdot 0 + 3 = 3 \frac{m}{s}$

Hope it is not too confusing!

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#### Explanation

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#### Explanation:

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12
Oct 26, 2016

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

This means that the velocity over a certain period of time is the instantaneous change in position $\left(\mathrm{dx}\right)$ over the instantaneous change in time $\left(\mathrm{dt}\right)$. This period of time is intentionally very small, hence "instantaneous".

Imagine looking at a graph of position (x) vs. time (t) and determining the slope of that graph.

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{\Delta x}{\Delta t}$

...when the change, $\Delta$, is very, very small. You are essentially zooming into the graph until it looks linear (try it on your calculator on, let's say, $y = \sqrt{x}$).

The little square represents what you see on your calculator.

When that is the case, you can take two points very, very close to each other, like $\text{1.003 s}$ and $\text{1.006 s}$ for example, calculate the slope using $\frac{{x}_{2} - {x}_{1}}{{t}_{2} - {t}_{1}}$, and you have the velocity at about $\text{1.0045 s}$ in $\text{m/s}$ (halfway between the two times).

You can also shift this small square window around and find velocities at other points on the same curve in a similar fashion, and they would be instantaneous velocities, at instantaneous moments in time.

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