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I wanted to ask in general what area under the graph means. Also which physical quantity is highlighted by area under distance vs time graph. I'm confused that area is a 2 dimensional concept and it may indicate distance or displacement, which are 1 dimensional quantities.

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The interpretation of the area under a curve, depending upon the curve, will vary. If it is a Velocity v. Time Graph, the area from a given time to another time, will be the distance traveled between those times. If it is an Acceleration v. Time Graph, the area from one time to another, will be the change in velocity of the object between those two times. For other dimensions, the quantity's significance will vary. In the case of a y v. x graph, if the unit(s)/dimension(s) of y is/are a, and the unit(s)/dimension(s) of x is/are b, then the units/dimensions of the area quantity will be 'a times b'.

For a distance v time graph, the area is indeed two dimensions, but not necessarily with both being a length. The Area under a graph is the geometrical meaning, not the physically significant one. For your situation, the area-quantity would be two dimensions: Length times Time. A quantity of this dimension is known as absement. An absement is a measure of a distance and how long something has been at that distance. Visit here for more info: https://en.wikipedia.org/wiki/Absement .

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There is no relation between the area being two-dimensional in your graph and what it means. For example, consider you make the $x$-axis as an indicator of the temperature, so what is the difference between $x= 5 K$ and $x = 8 K$? of course it's $ \Delta x = 3K $. Now, isn't $x$ a one-dimensional quantity?

So the graph gains its meaning from you not from itself as a draw.

What does the area under graph mean? well, it means the area under a graph, any graph. (provided some conditions which are satisfied in nearly all physics) but what is the area refers to depends on how you interpret the axes.

let $A$ be the area under the graph of $f(x)$,

$A =\int_{x_{0}}^{x_{1}} f(x) \space dx $

you can always interpret this as a product of $f(x)$ and $dx$ (a small quantity of $x$).

when each of them is a distance so the dimensions of $A$ are L$^2$. However, in the velocity-time relation, i.e. $f(x)$ measures velocity and $x$ measures time (that's how we usually write it $t$ and $ f(t)$ instead of $x $ and$ f(x)$ ), we get the dimensions of $A$ as $[L T^{-1} . T = L$ so it's a distance.

I hope this helps.

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If we consider the velocity-time graph area under the whole line is the distance. For example, enter image description here

Now look at the second part of the diagram (rectangular, dark blue) $$v=\frac{dx}{dt}\rightarrow\int{vdt}=\int{dx}\rightarrow x=vt$$

But the first part (triangle), we should consider acceleration

$$a=\frac{d^2x}{d^2t}=\frac{dv}{dt}\rightarrow\int{adt}=\int{dv}\rightarrow v=at\rightarrow\frac{dx}{dt}=at\rightarrow x=\frac{1}{2}at^2$$

As you can see, all the meaning is behind the meaning of integral.

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I'm confused that area is a 2 dimensional concept and it may indicate distance or displacement , which are 1 dimensional quantities.

You are right. Area is a 2 dimensional quantity but what you have missed out is that you didn't use dimensional analysis properly. The dimensions of velocity is $[LT^{-1}]$. So if you multiply velocity with time, you get the dimensions of the resultant quantity as $[L]$ which is the dimensions of displacement/distance.

The whole concept of area comes from integration. As you must be aware, $v=\dfrac{dx}{dt}$. So the displacement is $\int vdt$ which when used with proper limits is area under the curve of displacement vs time.

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protected by Qmechanic Jun 9 '16 at 8:53

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