Instantaneous Velocity and Speed — Question 6

Question 6

Identical samples can conceal opposite instantaneous velocities

A sensor records exact positions $x=0,1,2\ \mathrm{m}$ at $t=0,1,2\ \mathrm{s}$, respectively. It provides no other observations. Consider the family of smooth candidate histories $$x_a(t)=t+a\,t(t-1)(t-2),\qquad 0\leq t\leq2.$$ Here $t$ is numerical time in seconds, $x_a$ numerical position in metres, and $a$ a dimensionless numerical parameter under this convention. Use polynomial differentiation to obtain velocity in m/s. No bound on velocity or traveled distance is supplied.

1. Show that every member of this family matches all three samples. Find the average velocities over 0–1 s, 1–2 s and 0–2 s.

2. Expand the polynomial and derive $v_a(t)$. Evaluate $v_a(1\ \mathrm{s})$ as a function of $a$.

3. Give one member moving east at 1 s, one moving west, and one with zero velocity at that instant. Give the corresponding instantaneous speeds.

4. Can the three samples impose any finite upper bound on the magnitude of the instantaneous velocity at 1 s within this family? Justify algebraically.

5. A software display joins the samples by a straight line and labels its slope “measured instantaneous velocity.” What extra model assumption has the software introduced? What can the samples establish without it?

Question 6 - Solution

Identical samples can conceal opposite instantaneous velocities

(a) The added term vanishes at every observation time.

The product $t(t-1)(t-2)$ is zero at 0, 1 and 2. Thus all candidates match the three recorded positions. Each of the three requested endpoint ratios is $\boxed{+1\ \mathrm{m/s}}$, independent of $a$.

(b) Agreement of function values does not force agreement of derivatives.

Expanding gives $x_a=t+a(t^3-3t^2+2t)$, so $$\boxed{v_a(t)=1+a(3t^2-6t+2)\ \mathrm{m/s}},\qquad \boxed{v_a(1)=(1-a)\ \mathrm{m/s}}.$$

(c) Choose parameters to exhibit three different local behaviors.

$$\begin{aligned}a=0 &: (v_a(1),s_a(1))=(+1,1)\ \mathrm{m/s}\quad\text{eastward},\\a=2 &: (v_a(1),s_a(1))=(-1,1)\ \mathrm{m/s}\quad\text{westward},\\a=1 &: (v_a(1),s_a(1))=(0,0)\ \mathrm{m/s}\quad\text{stationary instant}.\end{aligned}$$ All are smooth polynomials and match every measured sample. A stationary instant does not by itself imply rest over a nonzero interval.

(d) The permitted derivative is unbounded by these samples.

$$a=1-V\quad\Rightarrow\quad v_a(1)=V\ \mathrm{m/s},\qquad\sup_{a\in\mathbb R}|v_a(1)|=\infty.$$

For any desired numerical velocity $V$, choose $a=1-V$. Then $v_a(1)=V\ \mathrm{m/s}$ while all three samples remain unchanged. Thus no finite upper bound on $|v_a(1)|$ follows within the supplied model class. This is a statement about the information in the data and the absence of additional constraints, not a claim about unlimited speeds in a physical apparatus.

(e) Interpolation is an assumption about unobserved times.

A single straight connector selects the particular history $x=t$ and its constant derivative. The samples alone establish positions at the recorded instants and the three interval-average velocities. They do not prove that the position function is linear between observations, nor determine its instantaneous derivative at 1 s.