Graphical Integration in Motion Analysis — Question 3

Question 3

Equal acceleration areas can produce different displacements

Two independent runs on an east-positive track both start at $x=0$ with $v=1\ \mathrm{m/s}$ and last 6 s. In run E, acceleration is $+2\ \mathrm{m/s^2}$ for the first 2 s and zero afterward. In run L, it is zero for the first 4 s and $+2\ \mathrm{m/s^2}$ for the final 2 s. The shown profiles are exact, with continuous velocity at switching times.

1. Use acceleration areas to find both final velocities. Reconstruct both velocity graphs and find both displacements.

2. Explain the displacement difference using the area between the velocity graphs. Why does equal total acceleration area not imply equal displacement?

3. A useful identity is $$x(T)-x(0)=v(0)T+\int_0^T (T-t)a(t)\,dt.$$ Explain the factor $T-t$ physically and check the units. Use the centroids of the two rectangular acceleration pulses to recover the displacements.

4. If a fixed nonnegative acceleration pulse of area $J$ is shifted later by $\delta$ without leaving the time window, how much does displacement change? State what must be held fixed for this comparison.

Question 3 - Solution

Equal acceleration areas can produce different displacements

(a) Integrate twice, retaining the time ordering.

Each pulse has area $2(2)=4\ \mathrm{m/s}$, so both final velocities are $\boxed{5\ \mathrm{m/s}}$. E’s velocity rises from 1 to 5 during 0–2 s, then stays at 5. L stays at 1 during 0–4 s, then rises to 5. Their velocity areas give $$\boxed{\Delta x_E=2(1+5)/2+4(5)=26\ \mathrm{m}}, \qquad\boxed{\Delta x_L=4(1)+2(1+5)/2=10\ \mathrm{m}}.$$ Both velocities stay positive, so these are also the traveled distances.

(b) Compare how long the velocity gain is available.

E’s velocity exceeds L’s by a triangle of area 4 m during 0–2 s, a rectangle of area 8 m during 2–4 s, and a triangle of area 4 m during 4–6 s. The total is $\boxed{16\ \mathrm{m}}$. Equal acceleration areas fix equal velocity changes, while the earlier gain contributes to position for longer.

(c) Weight each velocity increment by its remaining time.

An acceleration contribution $a(t)\,dt$ adds a small velocity increment that persists for $T-t$. Its position contribution is $(T-t)a(t)\,dt$, with units $\mathrm{s}(\mathrm{m/s^2})\mathrm{s}=\mathrm{m}$. Here the initial-velocity contribution is 6 m. The pulse area is 4 m/s and its time centroid is 1 s for E or 5 s for L: $$6+4(6-1)=26\ \mathrm{m},\qquad 6+4(6-5)=10\ \mathrm{m}.$$

(d) A time shift changes the weight of every pulse element.

Integrate over the unshifted pulse; both the original and delayed pulse remain inside the observation interval: $$\int_0^T (T-s-\delta)a(s)\,ds-\int_0^T(T-s)a(s)\,ds=-\delta\int_0^T a(s)\,ds=-J\delta.$$

Delaying the pulse lowers every remaining-time weight by $\delta$. Thus $\boxed{\Delta x_{\mathrm{late}}-\Delta x_{\mathrm{early}}=-J\delta}$. The pulse shape and area, initial velocity, final observation time and all other acceleration contributions must be held fixed. The shift must keep the entire pulse inside the observation window.