Average Velocity and Instantaneous Velocity — Question 5

Question 5

A position can be continuous while velocity fails to exist

A particle follows $\vec r(t)=[|t/(1\ \mathrm s)|\hat\imath+(t/(1\ \mathrm s))\hat\jmath]$ m for times around zero. The position is continuous at zero.

1. Find velocity immediately before and after zero.

2. Does instantaneous velocity exist at zero? Does a common limiting speed settle that question?

3. Find the average velocity over $[-h,h]$ for any $h>0$ and explain why it is not a velocity at the corner.

Question 5 - Solution

A position can be continuous while velocity fails to exist

(a) Differentiate the two branches separately.

$$\vec v_- =(-1,1)\,\mathrm{m/s},\quad\vec v_+=(1,1)\,\mathrm{m/s}$$ For $t<0$, $\vec v=(-1,1)$ m/s; for $t>0$, $\vec v=(1,1)$ m/s. The horizontal component jumps, whereas the vertical component remains 1 m/s.

(b) Require agreement of vector limits.

$$\vec v_-\ne\vec v_+\ \Rightarrow\ \boxed{\vec v(0)\text{ does not exist}},\quad|\vec v_-|=|\vec v_+|=\sqrt2\,\mathrm{m/s}$$ The one-sided derivatives disagree, so $d\vec r/dt$ is undefined at zero. Both one-sided speeds equal $\sqrt2$ m/s, but equality of magnitudes cannot make two different velocity directions equal. The ideal sharp corner has no unique tangent velocity.

(c) A symmetric secant can conceal a corner.

$$\frac{\vec r(h)-\vec r(-h)}{2h}=\frac{(0,2h)\,\mathrm{m/s}}{2h}=(0,1)\,\mathrm{m/s}\ne\vec v_\pm$$ The endpoints have equal x coordinates and vertical separation $(2\ \mathrm{m/s})h$, giving average velocity $(0,1)$ m/s. This symmetric ratio exists for every h, yet neither one-sided instantaneous velocity equals it. A derivative requires the same limit from both sides, not just symmetric averaging.