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This question already has an answer here:

why is the distance of a body equal to the area of its speed-time graph?

the general formula of speed(v) is v=distance(s)÷time taken(t) so the formula of distance(s) should be s=v×t so if the speed-time graph of a body is a triangle, the distance covered by the body should be equal to twice of the area of the graph because area of a triangle is - 1/2×base×height where base=t and height =v so it becomes 1/2×t×v and the formula of "s" is v×t so if v×t is distance, 1/2×v×t should be equal to 1/2 of the distance and not the distance?

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marked as duplicate by Qmechanic Jul 6 '18 at 15:36

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    $\begingroup$ Check out and understand "The Fundamental Theorem of Calculus". $\endgroup$ – JEB Jul 6 '18 at 15:04
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Your reflection here is good.

Just remember that for such a triangular area on the $v-t$ graph, the speed $v$ is not constant. It changes all the time, it is not the same in 1 second as it is in 2 seconds.

So which value of $v$ should you plug into the formula $s=vt$?

The answer in this particular case is that you take the average, which would be half of the highest value $v$ will reach. And then your formulas fit!

Were $v$ constant, then you would have a rectangular graph, and width-times-height would again give the right result.

In cases where you cannot figure out the corresponding value of $v$, what do we then do? We have to take each individual value for itself, calculate the distance for each and then sum it up. This why the formula in general is written as an integral:

$$s=\int v \;d t$$


Let's split the area into billions of tiny columns. The top of each column is almost perfectly horizontal since it is so thin. And so you find the area of this column from its base $b$ and height $h$:

$$A_{column} =h_{column}b$$

And then you add up all the columns:

$$A=\sum A_{column} =\sum h_{column}b$$

The height is the speed that fits the specific column, and $b$ is the change in time $\Delta t$.

$$A=\sum v \;\Delta t$$

When we have basically infinitely many such infinitely thin columns, then we can invent a special symbol. For example the integral symbol $\int$ to mean a sum of infinitely many infinitely tiny things:

$$A=\int v \;d t$$

And people also then use $d$ instead of $\Delta$ when the difference is infinitely small.

This is it. This is why an integral corresponds to an area. This is its geometric definition.

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    $\begingroup$ I think you accidentally a whole sentence in the end. $\endgroup$ – knzhou Jul 6 '18 at 14:35
  • $\begingroup$ @knzhou He he, indeed. $\endgroup$ – Steeven Jul 6 '18 at 16:12

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