Graphical Integration in Motion Analysis — Question 9

Question 9

A curved graph exposes the error in a straight-chord estimate

An exact velocity curve is $$v(t)=\left[\left(\frac{t}{\mathrm{s}}\right)^2-4\frac{t}{\mathrm{s}}+3\right]\mathrm{m/s}, \qquad 0\leq t\leq4\ \mathrm{s}.$$ The marker moves on a fixed straight axis and starts at $x(0)=0$. The plotted curve is the full record, not an interpolation of sparse observations.

1. Find both reversals and the exact displacement by integration. Then split at the reversals to find total distance.

2. Estimate displacement first with one endpoint trapezoid over 0–4 s, then with four equal 1 s trapezoids. Calculate the errors relative to the exact answer and explain their signs using graph curvature.

3. A student takes the absolute value of the exact signed integral as the distance. Identify the missing traveled length.

4. Can the curved graph be replaced by a constant average velocity while preserving both endpoint position and traveled distance? Explain which information the replacement preserves and loses.

Question 9 - Solution

A curved graph exposes the error in a straight-chord estimate

(a) Zeros delimit the signed regions.

With time numerically in seconds, $v=(t-1)(t-3)$, so $\boxed{t=1,3\ \mathrm{s}}$ are reversals. An antiderivative for displacement in metres is $$F(t)=\frac{t^3}{3}-2t^2+3t.$$ Its values are $F(0)=0$, $F(1)=4/3$, $F(3)=0$, and $F(4)=4/3$. Thus the three signed areas are $+4/3,-4/3,+4/3\ \mathrm{m}$: $$\boxed{\Delta x=4/3\ \mathrm{m}},\qquad\boxed{D=4\ \mathrm{m}}.$$

(b) Chords lie above this convex velocity curve.

For $N$ equal trapezoids, the quadratic has $v^{\prime\prime}=2$ in numerical SI units; integrating each chord error gives $(4/N)^3/6$ metres per panel. $$E_N=\frac{32}{3N^2}\ \mathrm{m};\qquad E_1=\frac{32}{3}\ \mathrm{m},\quad E_4=\frac23\ \mathrm{m}.$$

The single trapezoid uses $v(0)=v(4)=3\ \mathrm{m/s}$ and gives 12 m, an overestimate of $\boxed{32/3\ \mathrm{m}}$. At integer seconds the heights are $3,0,-1,0,3\ \mathrm{m/s}$. The four-panel estimate is $$\Delta x_4=\tfrac12(3+0)+\tfrac12(0-1)+\tfrac12(-1+0)+\tfrac12(0+3) =\boxed{2\ \mathrm{m}}.$$ Its error is $\boxed{+2/3\ \mathrm{m}}$. Since $v''>0$, every chord lies above the curve, even where both heights are negative. The trapezoidal signed area is therefore too large. Refinement reduces this error here; exact samples alone do not make straight connectors exact.

(c) Absolute value must be taken before accumulating distance.

The negative region contains $4/3\ \mathrm{m}$ of backward travel. Its cancellation with equal positive travel hides $2(4/3)=8/3\ \mathrm{m}$ in the net displacement. Taking $|\int v\,dt|=4/3\ \mathrm{m}$ misses that amount; $\int|v|\,dt=4\ \mathrm{m}$ counts it.

(d) An average preserves one signed area, not the path.

The constant average is $\bar v=(4/3)/4=1/3\ \mathrm{m/s}$. Over 4 s it gives the same final position, but only $4/3\ \mathrm{m}$ of distance and no reversal. It cannot preserve both quantities here. Endpoint equivalence is weaker than equivalence of the full motion.