Graphical Integration in Motion Analysis — Question 10

Question 10

A closed trip must satisfy two different area constraints

A marker begins at $x(0)=0$ with $v(0)=-2\ \mathrm{m/s}$ on an east-positive line. A controller applies constant acceleration $+A$ during 0–2 s, zero during 2–4 s, and $-B$ during 4–6 s, where $A,B>0$ are in $\mathrm{m/s^2}$. Velocity is continuous; acceleration jumps are idealized.

The required final state is $x(6)=0$ and $v(6)=-2\ \mathrm{m/s}$. A report claims this motion travels exactly 4 m and remains within $-1\leq x\leq1\ \mathrm{m}$ throughout.

1. Use the final-velocity condition to relate $A$ and $B$. Why does that condition alone not ensure return to the initial position?

2. Accumulate velocity areas to solve for $A$ and $B$ using the final-position condition. Are the required accelerations feasible if both magnitudes are limited to $2\ \mathrm{m/s^2}$?

3. Find both reversal times and the positions there. Verify the two final-state requirements directly.

4. Calculate the traveled distance and assess both claims in the report. Could a different positive pair $A,B$ under this same three-stage schedule satisfy both final-state requirements and repair the report?

Question 10 - Solution

A closed trip must satisfy two different area constraints

(a) Zero net acceleration area restores velocity.

The total area is $2A-2B$, so $v(6)=v(0)$ requires $\boxed{A=B}$. This only restores velocity. Return to position additionally requires zero signed area under the resulting velocity graph.

(b) Impose zero velocity area over the entire run.

Use numerical SI values. With $A=B$, the stage-boundary velocities are $-2,-2+2A,-2+2A,-2$. The two outer trapezoids each have area $-4+2A$; the middle rectangle has area $-4+4A$. Thus $$\Delta x=-12+8A=0,\qquad\boxed{A=B=1.5\ \mathrm{m/s^2}}.$$ Both satisfy the 2 m/s$^2$ limit. The boundary velocities are now $-2,+1,+1,-2\ \mathrm{m/s}$.

(c) Check reversals and accumulated positions.

The first zero occurs at $-2+1.5t=0$, giving $\boxed{t_1=4/3\ \mathrm{s}}$. The initial negative triangle gives $\boxed{x(t_1)=-4/3\ \mathrm{m}}$. The first full trapezoid gives $x(2)=-1\ \mathrm{m}$; the middle rectangle gives $x(4)=+1\ \mathrm{m}$. On the final stage $v=1-1.5(t-4)$, so $$\boxed{t_2=14/3\ \mathrm{s}},\qquad \boxed{x(t_2)=1+\tfrac12(2/3)(1)=4/3\ \mathrm{m}}.$$ The final-stage signed area is $2(1-2)/2=-1\ \mathrm{m}$, restoring $x(6)=0$. Its acceleration area is $-3\ \mathrm{m/s}$, restoring $v(6)=-2\ \mathrm{m/s}$.

(d) Audit the whole path, not only its endpoints.

$$x_{\min}=-\frac43<-1,\qquad x_{\max}=\frac43>1,\qquad D=\frac43+\frac83+\frac43=\frac{16}{3}\ \mathrm{m}.$$

The successive monotone lengths are $4/3$, $8/3$ and $4/3\ \mathrm{m}$, so $$\boxed{D=16/3\ \mathrm{m}}.$$ The motion travels more than 4 m and exceeds each corridor boundary by $1/3\ \mathrm{m}$. Both claims fail despite correct final position and velocity. The final-velocity relation fixes $B=A$, and the final-position equation then fixes a unique $A=1.5$. No different positive pair within this specified schedule can repair the report while retaining both final-state requirements.