Constant Acceleration - A Special Case — Question 4

Question 4

Equal-time position increments test a constant-acceleration model

Exact position samples on a straight line are The sampled $(t,x)$ pairs are $(0\ \mathrm{s},2\ \mathrm{m})$, $(2\ \mathrm{s},14\ \mathrm{m})$, $(4\ \mathrm{s},18\ \mathrm{m})$ and $(6\ \mathrm{s},14\ \mathrm{m})$, plotted below. Test the hypothesis $x(t)=x_0+v_0t+\tfrac12at^2$ with a single constant $a$. No statement about motion between samples is otherwise supplied.

1. Show that for equal spacing $h$, successive displacement increments differ by $ah^2$. Apply this test to all four samples and recover $x_0,v_0,a$.

2. Under the hypothesis, predict position and velocity at 8 s. Explain how zero net displacement over 0–8 s can coexist with motion.

3. Do the four exact samples prove that acceleration was constant between them? In numerical seconds/metres, consider adding $\lambda t(t-2)(t-4)(t-6)$ to the fitted position polynomial, where $\lambda\ne0$ is a numerical coefficient. Check the samples and identify why acceleration is no longer constant.

4. State what an additional sample inconsistent with the fitted polynomial would establish, and what an additional consistent sample would not establish by itself.

Question 4 - Solution

Equal-time position increments test a constant-acceleration model

(a) Compare consecutive equal-duration displacements.

Expanding the quadratic gives $$x(t+2h)-2x(t+h)+x(t)=\boxed{ah^2}.$$ The observed increments are $12,4,-4\ \mathrm{m}$, whose consecutive differences are both $-8\ \mathrm{m}$. With $h=2\ \mathrm{s}$, $\boxed{a=-2\ \mathrm{m/s^2}}$. The first sample gives $x_0=2\ \mathrm{m}$, and $12=2v_0+\tfrac12(-2)(2)^2$ gives $\boxed{v_0=8\ \mathrm{m/s}}$.

(b) Make the prediction conditional on the fitted model.

The model is $x=2+8t-t^2$ in metres with $t$ numerical seconds, and $v=8-2t\ \mathrm{m/s}$. It predicts $$\boxed{x(8)=2\ \mathrm{m},\qquad v(8)=-8\ \mathrm{m/s}}.$$ The point moves east, turns at 4 s, and returns to its initial coordinate at 8 s. The net displacement is zero despite nonzero traveled distance.

(c) Finite samples constrain values, not a whole continuous history.

$$x_\lambda(t)=2+8t-t^2+\lambda t(t-2)(t-4)(t-6),\quad a_\lambda(t)=-2+\lambda(12t^2-72t+88).$$

The added product vanishes at each sampled time, so every sampled value is preserved. Its expansion is $\lambda(t^4-12t^3+44t^2-48t)$. Its second derivative is $\lambda(12t^2-72t+88)$, which varies with time for $\lambda\ne0$. Thus smooth variable-acceleration histories also fit all four samples.

The constant-acceleration model is compatible with the data; the data alone do not prove it throughout the intervals or justify extrapolation to 8 s without that hypothesis.

(d) Distinguish rejecting a model from uniquely verifying it.

An exact inconsistent sample would reject this single fitted quadratic as a complete description of all the observations. Another consistent sample would add support, but a finite set can still be matched by other smooth functions. Stronger continuous-motion assumptions or independent evidence are needed for a unique model claim.