Acceleration — Question 7

Question 7

Estimate acceleration from three positions with explicit error bounds

Near a selected time $t_0$, the exact position is known independently to have the local quadratic form $$x(t_0+\tau)=A+B\tau+\tfrac12 a_0\tau^2.$$ Here $A$ is in metres, $B$ in m/s, $a_0$ in $\mathrm{m/s^2}$, and $\tau$ in seconds. This model is supplied only for the local measurement interval. Its acceleration is $a_0$.

At $\tau=-h,0,+h$, with $h=0.20\ \mathrm{s}$, measured positions are $1.620,2.000,2.460\ \mathrm{m}$. Each true position is within $0.005\ \mathrm{m}$ of its reading. Every combination within these hard bounds is allowed; times are exact.

1. Eliminate $A$ and $B$ to derive an expression for $a_0$ using the three true positions. Evaluate the corresponding estimate from the readings.

2. Find the tight interval of allowed accelerations, accounting for the coefficient of the central position. Is the acceleration’s sign established under the given model and bounds?

3. If all three readings have the same additional constant position offset, does it alter this estimator? Explain algebraically.

4. With the same position-error bounds but $h=0.10\ \mathrm{s}$, what would be the error allowance for this estimator? Why does closer spacing not automatically improve a guaranteed acceleration estimate?

5. Without the supplied local quadratic model, would three position readings alone establish this exact instantaneous-acceleration interval? Distinguish a measured finite difference from a derivative.

Question 7 - Solution

Estimate acceleration from three positions with explicit error bounds

(a) A symmetric second difference cancels the constant and linear terms.

Adding the two outer positions and subtracting twice the central one gives $a_0h^2$. Therefore $$\boxed{a_0=\frac{x_- -2x_0+x_+}{h^2}},\qquad \widehat a=\frac{1.620-4.000+2.460}{0.20^2}=\boxed{2.00\ \mathrm{m/s^2}}.$$ The squared time denominator gives the correct acceleration unit.

(b) The central position contributes twice its error bound.

For each true-minus-measured error $e_i\in[-0.005,0.005]\ \mathrm{m}$, $$a_0-\widehat a=\frac{e_- -2e_0+e_+}{h^2}.$$ The numerator lies in $[-0.020,0.020]\ \mathrm{m}$, so $$\boxed{a_0\in[1.50,2.50]\ \mathrm{m/s^2}}.$$ The upper endpoint uses both outer errors $+0.005$ and the middle error $-0.005\ \mathrm{m}$; reversing these choices attains the lower endpoint. Each resulting triplet determines a quadratic, so the bounds are feasible and tight. Acceleration is certainly positive under these assumptions.

(c) A shared fixed offset cancels.

Replacing every reading by $x_i+c$ adds $c-2c+c=0$ to the numerator. Thus an extra shared constant offset does not affect the estimate. Independent errors need not cancel this way.

(d) Smaller spacing amplifies a fixed position-error bound.

Let $\varepsilon_x=0.005\ \mathrm{m}$ be each position-error bound. $$E_a(h)=\frac{4\varepsilon_x}{h^2},\qquad\frac{E_a(h/2)}{E_a(h)}=4.$$

At $h=0.10\ \mathrm{s}$, the error allowance would be $0.020/0.10^2=\boxed{2.0\ \mathrm{m/s^2}}$, four times the original allowance. The same bounded position error is divided by a smaller squared time interval. This comparison concerns the uncertainty allowance; new readings would be needed to obtain a new central estimate.

(e) Exact differentiation requires a model or limiting information.

Without the quadratic assumption, the finite difference need not equal the second derivative at $t_0$. Smooth functions can agree at three sampled times but have different second derivatives at the center. Thus the calculated bounds are conditional on the supplied local model, not a model-free certification of instantaneous acceleration.