# Distance traveled equals area under velocity time graph [duplicate]

I know that for the particular case where the velocity of an object along a straight line is constant, the area beneath the velocity as a function of the time passed curve equals to the distance traveled. That really makes sense, it derives from the fact that the distance traveled equals to the product of the constant velocity and the time, which derives immediately from the physical definition of velocity. But why is that true in any case? Why if I'll take any curve of velocity as a function of time and compute the area beneath it I'll get the distance that was traveled? Any intuitive and rigorous proof for that? Thanks in advance :)

• You answered your own question : distance traveled equals to the product of the constant velocity and the time. – sammy gerbil Dec 26 '17 at 1:22
• Why did I answer it? Every point on any velocity- time graph has to be the product of the velocity and the time? – Ozk Dec 26 '17 at 4:16
• If you take a small enough time interval, the velocity is approximately constant. (As the interval becomes smaller, the approximation becomes better.) The formula $\text{distance = velocity x time}$ applies to each small interval, and the areas add up. – sammy gerbil Dec 26 '17 at 13:55

The relationship you are referring to is

$$v = \frac{\Delta x}{\Delta t},$$ which gives you the area under the velocity curve (the displacement) when you rewrite it as $\Delta x = v\Delta t$. This is the discrete form of this relationship; it only works for large time intervals. If you wanted to know the velocity at any instant, you have to take the limit of this equation:

$$\lim_{\Delta t \rightarrow 0} v = \lim_{\Delta t \rightarrow 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}.$$ Separating variables, you get $dx = vdt.$ If we solve this with bounds, we get a numerical answer for the displacement: $$dx = vdt$$ $$\int_{x_0}^x dx = \int_0^t v(t) dt$$ $$x = x_0 + \int_{0}^t v(t)dt.$$ However, if we were to solve it without bounds, we would get a function that describes the position as a function of time: $$dx = vdt$$ $$\int dx = \int v(t)dt$$ $$x(t) = \int v(t)dt$$ In each case, you a numerical displacement or a formula for the displacement, respectively.

• Your comment is appointed to be the most helpful good Sir. – Ozk Dec 24 '17 at 17:36
• Although I'd be happy if you'll elaborate more on the transition between dx=vdt and the integral itself. And I have another question regarding notations , V stands for velocity and T for time, why does the displacement is being notated with X? – Ozk Dec 24 '17 at 17:38
• I've edited my response to provide more detail. $x$ is a common symbol for displacement because you are often dealing with motion along the $x$-axis. For similar reasons, you will also see it notated with $y$, $z$, and also $r$. We tend to not use $d$ for displacement because it is ambiguous, i.e. v = dd/dt is confusing. – zhutchens1 Dec 24 '17 at 21:41

To convince yourself that this works even when the speed varies widely, break up time into lots of small intervals. Over any one interval, you can approximate the speed as linear. If that approximation isn't good enough, then you haven't broken time into small enough pieces.

The distance traveled during any one little slice of time is the average speed times the length of the time slice. For linearly varying speed, the average speed is the middle speed, which is also the average of the two end speeds.

The total distance traveled is the sum of all the little distances traveled. Now realize that the distance traveled during any one slice has nothing to do with the average speed during other slices. Another way of saying that is that the speed can vary up and down, and this method still works.

Congratulations. You have just re-discovered the integral.

I think there is a rigorous proof, yes, and it's essentially the derivation of the Riemann integral, which is something that is covered in a course on real analysis.

A very rough (and very non-rigorous) sketch of how this works is to take the function $v(t)$ and claim that it can be approximated as a sequence of constant values:

$$v(t) \approx \begin{cases} v_0 &t\in[t_0,t_1)\\ v_1 &t\in[t_1, t_2)\\ \cdots \\ v_{n-1} &t\in[t_{n-1}, t_n) \end{cases}$$

Where $(t_0, t_1, \ldots t_n)$ are a monotonically-increasing sequence of values of $t$.

Now, obviously our approximation to $v(t)$ is constant during each of these little intervals, so we can simply use the high-school definition of the distance travelled for each of those intervals. And we can sum these things to get something we claim is an approximation of the total distance travelled:

$$d \approx \sum\limits_{i=0}^{n-1} v_i(t_{i+1} - t_i)$$

So that's the easy bit: now what you need to do is the bit that needs some mathematical sophistication: you need to show that my claim that this expression for an approximate value for $v(t)$ actually is an approximate value for it, and that you can, in a well-defined way, increase the number of steps (and thus decrease the length of the intervals $[t_i, t_{i+1})$) and show that this sum converges in a well-defined way so long as $v(t)$ is well-behaved, and thus that, in the limit, $d$ is the distance travelled. Doing this properly is not completely trivial (and the construction you need to use is a bit cleverer than my very slapdash one), but it can be done properly: any introductory text on real analysis should contain the details.

As to whether this is intuitive: I don't know. I find that it is, but I'm also aware that intuitions vary.