Graphical Integration in Motion Analysis — Question 6

Question 6

Samples give an estimate only after an interpolation choice

A sensor samples eastward velocity at four instants: The sampled $(t,v)$ pairs are $(0\ \mathrm{s},1\ \mathrm{m/s})$, $(1\ \mathrm{s},2\ \mathrm{m/s})$, $(3\ \mathrm{s},4\ \mathrm{m/s})$ and $(6\ \mathrm{s},4\ \mathrm{m/s})$. Velocity is known independently to be continuous and nondecreasing throughout the run. No acceleration bound is supplied. Straight connectors between the sample dots are an optional model, not observed data.

1. Find the trapezoidal estimate of displacement using the actual unequal time intervals. Under what extra interpolation assumption is it exact?

2. Use monotonicity to obtain lower and upper rectangle bounds on displacement. Are the endpoint bounds attained by a continuous compatible velocity, or only approached? Explain.

3. On each of the first two intervals, consider $$v(t)=v_i+(v_{i+1}-v_i)\left(\frac{t-t_i}{t_{i+1}-t_i}\right)^2,$$ with $v=4\ \mathrm{m/s}$ on 3–6 s. Verify compatibility with the data and assumptions, then calculate its displacement. What does this example establish about the estimate?

4. What can be said about distance versus displacement here? Would endpoint samples alone, without monotonicity, give these rectangle bounds?

Question 6 - Solution

Samples give an estimate only after an interpolation choice

(a) Weight every endpoint pair by its actual duration.

The interval lengths are 1, 2 and 3 s. The estimate is $$\boxed{\Delta x_{\mathrm{trap}}=1(1+2)/2+2(2+4)/2+3(4+4)/2=19.5\ \mathrm{m}}.$$ It is exact for piecewise-linear velocity between the samples. Continuity and monotonicity alone do not establish that interpolation.

(b) Enclose each velocity segment between its endpoint heights.

Lower rectangles sum to $1(1)+2(2)+3(4)=17\ \mathrm{m}$; upper rectangles sum to $1(2)+2(4)+3(4)=22\ \mathrm{m}$. Continuity with different endpoint heights on the first two intervals prevents equality at either extreme. Hence $$\boxed{17\ \mathrm{m}<\Delta x<22\ \mathrm{m}}.$$ These are sharp limiting bounds: a continuous increase can be confined to an arbitrarily short portion near either endpoint when no acceleration bound is imposed. The final interval is exactly constant because its endpoints agree and velocity is nondecreasing.

(c) Exhibit another permitted history.

Use $z=(t-t_i)/\Delta t_i$, with $\Delta t_i=t_{i+1}-t_i$ and $0\leq z\leq1$: $$\int_{t_i}^{t_{i+1}}v(t)\,dt=\Delta t_i\int_0^1[v_i+(v_{i+1}-v_i)z^2]\,dz=\Delta t_i\left[v_i+\frac{v_{i+1}-v_i}{3}\right].$$

The quadratic interpolation matches both endpoints and has nonnegative slope on each interval. Joined to the constant final piece, it is continuous and nondecreasing. Its mean height on a segment is $v_i+(v_{i+1}-v_i)/3$, so $$\Delta x=1(1+1/3)+2(2+2/3)+12 =\boxed{56/3\ \mathrm{m}}.$$ This differs from 19.5 m by $5/6\ \mathrm{m}$. The sample data therefore do not determine the exact trapezoidal value.

(d) Separate sign information from interpolation information.

Here $v\geq1\ \mathrm{m/s}$, so $D=\Delta x$. Without monotonicity, continuous curves can make arbitrarily large excursions between fixed samples; the rectangle bounds need not hold. Continuity alone places no finite bound on those excursions.