# How area under Velocity-Time graph represents magnitude of displacement?

I understand that if the velocity is constant (acceleration=$0$) throughout the course of motion (where graph shows a rectangle) then it would simply be like playing with equation:

(1). velocity=displacement/time

(2). displacement=velocity $\times$ time =area of rectangle =Velocity-axis $\times$ time-axis,

putting the value of velocity and time, then getting out the displacement.

But what if the acceleration is constant instead of $0$ (where graph shows a triangle) ? I know it would be $\frac{1}{2}$ velocity $\times$ time, but how we can put a value of velocity in (1) displacement=velocity $\times$ time When velocity is changing all time due to uniformly accelerated motion?

The velocity function is an infinite set of values(corresponding to time) that describe an infinite set of velocities(points). In this situation, that set of points is perfectly constant in its rate of change. An example range is [ 0, 1, 2, 3, 4 ]. Now, a single value that perfectly represents all of these values is an average since the rate of change is constant. This means that you can represent this set of values perfectly with an average, that average being ( 4 - 0 ) / 2. Now, with a single value that perfectly represents the entire range of values (the velocity function) you can calculate the area! Imaging that instead of a graph with increasing values, making a positively sloped line between time 0 and time x (time on the x axis), you represent the same value set with the constant single value average(2 in our example) and have a line with a slope of zero as your new value-representative function. Simply multiply the width of the rectangle(delta x) by the height of the rectangle (the average we calculated, 2) and you have your area. This is effectively the formula for the area of a triangle.

I gave this answer because I assume the question implies a limited or no experience with calculus and I see that all other answers are derived from concepts from calculus.

I suggest looking into the Fundamental Theorem of Calculus for a more intuitive understanding of this situation, or alternatively looking into Riemann rectangular summation which was probably described by someone on this question.

Imagine dividing your graph of velocity vs. time into a bunch of extremely thin vertical rectangles. It's reasonable to say that, over such a short time, velocity is constant in any given rectangle. So we can say that the distance traveled in each tiny instant of time is equal to the velocity in that interval times the length of the interval (i.e. the area of the rectangle). Add up the distance traveled over each thin rectangle and you get the total distance traveled.

Well this is the point, you can't use (1) except if the interval is very small such that your "distance / time" will be a good approximation for your instantaneous velocity. So to get the total distance or displacement, (depending on how you define it, but let's assume that we're talking about distance), you need to find the distance travelled during the whole time, but the velocity you were going with was changing all the time, so instead, you divide the whole time into very small time intervals such that your velocity can be approximated to be constant on those intervals, and then use the value of the velocity $\frac{dx}{dt}$ at the beginning of each small interval and multiply by the value representing the time interval to get the distance, that is, integrate $\frac{dx}{dt}$ over time. But if you represent the velocity on the y-axis and the time on the x-axis, you'll find yourself, while summing up all the $\frac{dx}{dt} * \Delta t$, actually summing up areas of small rectanges whose heights are $\frac{dx}{dt}$ and whose widths are $\Delta x$. In other words, the whole "area under the graph" thing is just a consequence of your representation of velocity and time using rectangular coordinates.

Equation (1) as you provide it, gives the Average Velocity. It just so happens that when the instantaneous velocity (the velocity at a single time value) is constant, as is the case when the acceleration is 0, the average velocity equals the instantaneous velocity.

In the case of constant non-zero acceleration, as you pointed out, the velocity changes. The average velocity $v_{av}$ in this case is $$v_{av} = \frac{v_0 + v_1}{2}$$ where $v_0$ is the velocity at the moment you start counting the time and $v_1$ is the velocity at the moment you stop counting the time i.e. the time interval for which you want to calculate the displacement. If your velocity at the moment you start counting the time (at $t_0$) is 0 ($v_0 = 0$) then the average velocity is $$v_{av} = \frac{v_0 + v_1}{2} = \frac{0 + v_1}{2} = \frac{v_1}{2}$$ or simply $$v_{av}=\frac{v}{2}$$

Note that the above equations hold only for constant acceleration.

Therefore I would change your equations to

(1) Average Velocity = Displacement / Time

(2) Displacement = Average Velocity x Time

Now in both the zero acceleration case and the constant (non-zero) acceleration case the "velocity" value, given a fixed time interval, is a single value, the Average Velocity.

The key point here is exactly the idea of the constant speed (rectangle).

Let it be an arbitrarily changing velocity. Since at every instant in time the speed changes, it will be almost impossible to calculate the displacement. At this point, the best you can do is to give away of the actual value and accept an approximated value. To do so, at certain moments in time, let’s say at every 2 seconds you pick the corresponding value of speed and calculate the displacement as if speed was constant. At this point, what you will end up with are small rectangles representing small amounts of the displacement. The total approximated displacement will be the sum of the small displacements.

To better show this, let’s use the triangle (constant acceleration) of your example. In the first figure the plot of displacement vs time for a particle moving with a constant acceleration of $\rm a=1m/s^2$ is shown. After $\rm t=10 s$ it forms the triangle with sides of 10 by 10 units, whose area is $\rm \frac{1}{2}×10×10$ represented in red.

After Appling the 2 by 2 seconds sampling process you get the green triangles of the second figure. Has you can see, the total approximated displacement is an underestimation of the actual displacement, since for example, in $\rm t=3s$ you still consider that velocity equals $\rm 2m/s$ but in reality it is higher ($\rm 3m/s$).

However, if you make thinner and thinner rectangles and then sum it up, your approximation will get closer and closer to the actual value. For example, if one uses a 1 second sampling, the error shown before for for $\rm t=3s$ will disappear, but not the error for $\rm t=2.5s$ or $\rm t=3.5s$.

In the end, if you use instant sampling (infinitely thin rectangles) and then sum the resulting infinite number of rectangles your approximated value will match the actual value and the green area (which we shown to be equal to the approximated displacement) will become equal to the red area, i.e, the actual value for the displacement equals the area under the speed vs time curve.

For further reading search for the following keywords.

1. Geometric representation of integration.

2. Integration as summation.

PS: This doesn’t mean displacement is a sort of “area”. It is just the result of representing graphs or plots of functions in a “piece of paper”. This means that when one represents any quantity as a graph the magnitude of such a quantity will be represented as a length, therefore, any quantity that results from the multiplication of the axes of such a graph will be represented as an area. For example, work will be equal to the area under the force vs displacement curve, and electric charge will be equal to the area under the current intensity vs time curve.

## protected by Qmechanic♦Jul 11 '17 at 7:40

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