Constant Acceleration - A Special Case — Question 10

Question 10

Plan a rest-to-rest transfer within a speed cap

A point carriage must move 60 m east from rest and finish at rest. It follows this prescribed family of idealized protocols: accelerate at $+2\ \mathrm{m/s^2}$ to a chosen peak speed $V>0$, cruise at $V$ for a nonnegative duration, then brake at $-3\ \mathrm{m/s^2}$ to rest. Velocity remains continuous at the switches; the acceleration switches take negligible time. The speed cap is $V\leq9\ \mathrm{m/s}$.

1. Derive the acceleration and braking times and distances as functions of $V$. Find the peak speed for a no-cruise 60 m transfer, and test it against the cap.

2. For $V=9\ \mathrm{m/s}$, find each stage’s time and distance and the total transfer time. Verify both endpoint conditions and the speed cap.

3. Derive total time $T(V)$ for any allowed peak speed in this family. Prove that the largest allowed peak speed gives the shortest time within the stated family, using algebra or differentiation.

4. Explain why the full-trip endpoint-velocity mean gives an incorrect distance, and why one constant acceleration cannot by itself start at rest, finish at rest later, and produce the required nonzero displacement.

Question 10 - Solution

Plan a rest-to-rest transfer within a speed cap

(a) Use the appropriate constant acceleration on each moving stage.

With $V$ numerical in m/s, the times are $T_1=V/2$ and $T_3=V/3$ seconds. From $v_f^2-v_i^2=2a\Delta x$, the distances are $D_1=V^2/4$ and $D_3=V^2/6$ metres. With no cruise, $$60=\frac{5V^2}{12},\qquad\boxed{V=12\ \mathrm{m/s}}.$$ This exceeds the cap. A permitted protocol must instead include cruising.

(b) Evaluate all three stages at the cap.

At $V=9\ \mathrm{m/s}$, $$\boxed{D_1=20.25\ \mathrm{m},\quad D_3=13.50\ \mathrm{m},\quad D_2=26.25\ \mathrm{m}},$$ $$\boxed{T_1=4.50\ \mathrm{s},\quad T_3=3.00\ \mathrm{s},\quad T_2=35/12\ \mathrm{s}}.$$ The total time is $\boxed{T=125/12\ \mathrm{s}\approx10.4167\ \mathrm{s}}$. Distances sum to 60 m; velocity rises from 0 to 9, stays at 9, then falls to 0 m/s. It never reverses or exceeds the cap.

(c) Compare all allowed members of this specific family.

$$\frac{dT}{dV}=-\frac{60}{V^2}+\frac5{12}\leq-\frac{60}{81}+\frac5{12}=-\frac{35}{108}<0\quad(0<V\leq9).$$

For $0<V\leq9$, remaining cruise distance $60-5V^2/12$ is positive. Total numerical time in seconds is $$T(V)=\frac V2+\frac{60-5V^2/12}{V}+\frac V3 =\boxed{\frac{60}{V}+\frac{5V}{12}}.$$ For $0<V<9$, $$T(V)-T(9)=(9-V)\left(\frac{60}{9V}-\frac5{12}\right)>0,$$ since $60/(9V)>60/81>5/12$. Thus $V=9\ \mathrm{m/s}$ is the unique fastest choice within the specified protocol family. This proof does not assume arbitrary unlisted control strategies.

(d) A trip with acceleration switches is not one constant-acceleration interval.

Both endpoint velocities are zero, so their arithmetic mean would predict zero displacement if misapplied to the whole trip. Acceleration is not constant across the three stages. If one constant $a$ acted for $T>0$ with both endpoint velocities zero, $0=0+aT$ would require $a=0$, and displacement would also be zero. The nonzero rest-to-rest transfer requires changes in acceleration.