# A clarification on acceleration and velocity

This is one of those questions which require an answer that does not take practical limitations into account. It is a theoretical physics question, perhaps. If there are any loopholes used, please explicitly state them.

If the position is known as $$x(t)$$ from t=0 to t=1 second, how do I get the velocity at the initial and end points, since velocity at the end point will require $$x(1-(\Delta t)/2)$$ and $$x(1+ (\Delta t)/2)$$, which are added and divided by $$\Delta t$$ ?

It gets worse if I want to know the acceleration at the end point, which requires the $$v(1+(\Delta t)/2)$$ which in turn requires $$x(1+(\Delta t))$$, which is simply not available.

Is this an order thing or is it just neglected in calculus?

Strictly speaking, the velocity at the end points is not defined, since you cannot determine either the left-hand or the right-hand limits to the change in position at those times as the time interval gets arbitrarily smaller.

Since velocity is the time-derivative of the position, $$v(t) = \frac{dx(t)}{dt}.$$ For this derivative to be defined at $$t$$, we must accordingly have $$v(t_+) = \lim_{\Delta t \to 0} \frac{x(t + \Delta t) - x(t)}{\Delta t} = \lim_{\Delta t \to 0} \frac{x(t - \Delta t) - x(t)}{-\Delta t} = v(t_-).$$ That is, both the right-hand as well as the left-hand derivatives must be defined, and they must be equal. In the interval $$t \in (t_0, t_1)$$, the right-hand derivative is not defined at $$t = t_1$$, whereas the left-hand derivative is not defined at $$t = t_0$$. Therefore, mathematically, the function $$x(t)$$ is not differentiable at the end-points: the velocity is not defined at those points.

The velocity, however, still exists on the interval $$t \in (t_0, t_1)$$. That is because, by definition of differentiability on an interval, the function $$x(t)$$ just needs to have a right-hand derivative at $$t= t_0$$ and a left-hand derivative at $$t = t_1$$ to be considered differentiable. This is assuming, of course, $$x(t)$$ is sufficiently nice and smooth everywhere in between. See this for more information on differentiability.

• Very nice. You have addressed all my concerns. – Sidarth Jun 3 at 12:55

This question is more interesting if you refer to numerical calculations. For analytical tratement, it is simple: velocity is the derivative of position.

If you say you know $$x(t)$$ for $$t\in[0,1]$$, then velocity is $$v(t)=dx/dt$$. But yes, the endpoints can be more problematic. However, real numbers are so dense that you can easily extrapolate the velocity function. In other words, if you know $$v(t)$$ for $$t\in (0,1)$$, then you can extend it to close the interval.

$$v(t=0)=\lim_{t\rightarrow 0} v(t)$$

On the other hand, if you work with discrete time for numerical calculations, then this extension is more problematic if the interval size is big enough. The initial value is also given sometimes.

• Your answer is actually summarizing my own question in a better way. I should have included the problem of calculating the derivative at the end points (hence a doubt in analysis itself). Could you give me an elaboration about the problem on the end points? Since that is the whole point of my question? Why is v(t=0)=limt→0v(t) allowed? Is this an approximation? (I will add the numerical methods tag as you say.) @FGSUZ – Sidarth Jun 3 at 9:22
• To expand on this, because the real numbers are continuous, you can pick a real number $t$ in the range $(0,1)$ arbitrarily close to $0$ or $1$. If your position is continuous and differentiable over $t\in (0,1)$, then your choice of $t$ arbitrarily close to $0$ or $1$ with have a velocity arbitrarily close to that at $t=0$ and $t=1$, respectively. The same holds for acceleration. You can use a $t$ just greater than $0$ or just less than $1$ and it'll have both velocity and acceleration close enough for any reasonable purpose. The same holds for higher derivatives (e.g. jerk). – TheEnvironmentalist Jun 3 at 9:23
• @TheEnvironmentalist I think you mean the infinitesimal time can be made smaller and smaller to get the velocity closer and closer to the value at t=1 (and also the acceleration). I can finally define an epsilon and when the $1-t<\epsilon$, I can conclude I have the velocity at t=1. I hope I am right. But am I also right in saying that we never actually get the value of velocity at t=1, without using the concept of the limit? (I know that this sounds silly, given a derivative is itself a limit) – Sidarth Jun 3 at 9:30
• @Sidarth It sounds like you're taking a limit without calling it a limit and then asking whether you can treat it like a limit – TheEnvironmentalist Jun 3 at 10:07
• Actually, it's an answer, but my attempt to answer did not work so well, it seems Yejus did better, but I meant exactly the same. What I'm adding in my answer is that, since it is not strictly defined, whet we often do is EXTENDING the function, so that we define v(0)=limit when t→0. One thing is the value at t=0, and another thing is the limit. The limit exists, the function does not. You can either maintain it with no function at the origin, you you can EXTEND the function and DEFINE it to be coincident with the limit. Why? Because we asume all functions are continuous and differentiable – FGSUZ Jun 3 at 17:00