0.2 Proper Pizza Delivery

In 0.1 a constant force got the driver to the door — and often through it — with leftover speed. That fails a quiet requirement of real delivery:

Arrive at the customer, stop, and stop pushing. The pizza (and the doorframe) appreciate all three.

Why constant force is not enough

With $a = F/m$ held positive the whole trip, speed only climbs:

At the door you still have $v > 0$. You cannot satisfy $v(T) = 0$ without changing the force — usually by braking, $F \to -F$ (same magnitude, opposite direction, in this ideal model).

Once you are at rest at $x = D$, set $F = 0$. Then $a = F/m = 0$ as well. That is the “proper” end state.

A simple open-loop plan: accel, then brake

Keep the vehicle from 0.1: no drag, $|F|$ fixed when thrusting or braking. One classic plan:

1. Apply $+F$ from the shop until a switch mark $x_s$.
2. Apply $-F$ (brake) until speed hits zero.
3. Hold $F = 0$ at rest.

If $|a|$ is the same on both phases, the symmetric choice $x_s = D/2$ stops exactly at the door. Switch earlier and you stop short; switch later and you overshoot or still have speed at $D$.

That plan is still open-loop — the switch is chosen ahead of time, not from a sensor reading the door. Feedback comes later in Section 1.1. Here the goal is only to feel the boundary conditions $v = 0$, $a = 0$.

Try it

Use Ideal halfway switch for a clean stop, then try braking early or late. Data shows $a$ going negative during the brake phase, then flat at zero.

Discussion

Fix the halfway switch $x_s = D/2$ and turn up $|F|$. Acceleration magnitude $|a| = |F|/m$ rises with it. For the same shop-to-door distance $D$, the trip finishes sooner: you climb to a higher peak speed in the middle, then scrub that speed over a shorter brake phase. The end conditions do not change. If the switch is right, you still arrive with

at $x = D$. Larger force compresses the schedule; it does not rewrite the boundary conditions.

Why push as hard as the throttle (and brakes) allow? In this idealized vehicle — a double integrator, $m\ddot{x} = F$ with $|F| \le F_{\max}$ — the fastest stop-at-the-door plans use the extremes: full accel, then full brake. That pattern is often called bang-bang. Intuition beats formal optimal control here. Every moment spent below $|F_{\max}|$ is a moment you could have been building speed faster or dumping it faster. Coasting at constant $v$, or easing the pedal, leaves distance on the table while the clock runs. For minimum time with a hard force limit, you want to get moving ASAP and stop moving ASAP. (That is the classic time-optimal idea for a double integrator; Pontryagin’s principle says the same thing more carefully later, if you go there.)

Time is not the only score. Larger $|F|$ means harsher accel and brake for the stack of boxes — and for whoever is riding. Soft pizza and sore wrists care about peak $|a|$ as much as the stopwatch cares about $T$. Control design is usually a compromise among speed, comfort, wear, and safety, not a single objective. 0.3 keeps the same stop and scores gas instead.

Use the widget to feel both sides. Run Ideal halfway switch with a small $F$, then again with a large one; open Data and compare how tall the $v$ peak gets and how quickly $a$ flips from $+F/m$ to $-F/m$. Then deliberately switch early or late: a bigger force does not rescue a bad plan. Magnitude sets how hard you push; the switch mark decides whether you stop short, cleanly at the door, or still roll through it.