Graphical Integration in Motion Analysis — Question 7

Question 7

An acceleration envelope yields bounds on two accumulated quantities

A marker begins with $x(0)=0$ and $v(0)=2\ \mathrm{m/s}$. Throughout 0–4 s its acceleration is integrable and obeys the hard pointwise bounds $$\left(0.75-0.50\frac{t}{\mathrm{s}}\right)\mathrm{m/s^2} \leq a(t)\leq \left(1.25-0.50\frac{t}{\mathrm{s}}\right)\mathrm{m/s^2}.$$ Every acceleration history inside the envelope is allowed; no probability model is given. The centre line is $a_c=(1.00-0.50t/\mathrm{s})\mathrm{m/s^2}$.

1. Find tight bounds on $v(4)$. Explain why taking the acceleration to be its centre line gives a possible value rather than a guaranteed value.

2. Use integration of the velocity envelopes, or the weight $4\ \mathrm{s}-t$ on acceleration, to find tight bounds on $x(4)$.

3. Prove that no allowed history reverses during the run. What bounds then apply to total distance?

4. Show that the stated bounds are attained by allowed histories. Explain why treating the $0.25\ \mathrm{m/s^2}$ half-width as random noise that averages away is unjustified.

Question 7 - Solution

An acceleration envelope yields bounds on two accumulated quantities

(a) Integrate the extreme histories.

Write $t$ numerically in seconds in the following SI expressions. Integrating the centre gives $v_c=2+t-t^2/4$. The envelope half-width integrates to $t/4$, so $$2+\tfrac34t-\tfrac14t^2\leq v(t)\leq2+\tfrac54t-\tfrac14t^2.$$ At 4 s this gives $\boxed{v(4)\in[1,3]\ \mathrm{m/s}}$. The central prediction is 2 m/s, but a sustained error of either sign is allowed.

(b) Integrate again, keeping the positive remaining-time weight.

The centre displacement is $$x_c(4)=\left[2t+\tfrac12t^2-\tfrac1{12}t^3\right]_0^4 =\frac{32}{3}\ \mathrm{m}.$$ The displacement half-width is $\int_0^4(t/4)\,dt=2\ \mathrm{m}$. Equivalently it is $\int_0^4(4-t)(1/4)\,dt$ in SI numerical values. Thus $$\boxed{\frac{26}{3}\ \mathrm{m}\leq x(4)\leq\frac{38}{3}\ \mathrm{m}}.$$

(c) Check the whole lower velocity envelope.

$$v(t)\geq2+\frac34t-\frac14t^2\geq\min\{v_{\mathrm{low}}(0),v_{\mathrm{low}}(4)\}=1\ \mathrm{m/s}>0.$$

The lower quadratic $2+3t/4-t^2/4$ is concave down. Its minimum on the closed interval occurs at an endpoint: the endpoint values are 2 and 1 m/s. Therefore every allowed velocity is at least 1 m/s throughout. No stop or reversal is possible, and $D=x(4)-x(0)$ has the same bounds as above.

(d) Extremal histories establish tightness.

Taking $a(t)$ equal to the lower boundary throughout attains both lower final bounds; taking it equal to the upper boundary attains both upper bounds. Both histories are allowed and continuous. A hard bound supplies no independence, mean-zero assumption or likely cancellation. Treating it as random noise would exclude permitted sustained deviations without evidence.