A clarification on acceleration and velocity

Question

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?

Answer 1

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.

Answer 2

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.