Piecewise & Inverse Motion

A breakpoint is any instant at which the rule for the acceleration changes: a motor cuts out, a brake engages, a speed limiter starts holding the speed, a rope goes slack, a ball leaves a hand. Between breakpoints, use the constant-acceleration equations with a local clock τ = t − tᵢ and the state at the start of that interval. The classic slip is reusing global time. A car accelerates from rest at 2.5 m s⁻² for 4.0 s, reaching 10.0 m s⁻¹ at x = 20.0 m, then brakes at 1.5 m s⁻². At t = 6.0 s it has braked for τ = 2.0 s, so v = 10.0 − 1.5 × 2.0 = 7.0 m s⁻¹ and x = 20.0 + 10.0 × 2.0 − ½ × 1.5 × 2.0² = 37.0 m. Feed t = 6.0 s into the braking equation instead and you get v = 1.0 m s⁻¹ and x = 53.0 m: a car that has braked for six seconds when it has braked for two.

Integrate across a breakpoint to see what must match. The change in velocity over a short window of width ε around tᵢ is ∫a dt over that window. If the acceleration stays below some finite bound A, that change is at most Aε and vanishes as ε → 0, so v(tᵢ⁺) = v(tᵢ⁻). The same argument applied to ∫v dt makes position continuous whenever velocity is finite, which for a real object is always. Acceleration has no such protection: switch a force on and the a–t graph steps. So the graphs carry signatures. An a–t graph may have steps, a v–t graph may have corners but never gaps, and an x–t graph has neither. A corner in an x–t record therefore means a velocity jump, and a velocity jump means an impulse. A ball striking a wall at +6.0 m s⁻¹ and leaving at −4.0 m s⁻¹ after 5.0 ms of contact averages −10.0/0.0050 = −2.0 × 10³ m s⁻². Over a motion lasting seconds you may treat that as an instantaneous jump, provided you say so.

A chain of N constant-acceleration intervals involves 4N + 2 quantities: an acceleration and a duration for each interval, plus a position and a velocity at each of the N + 1 breakpoints. The update rules supply 2N equations, so the problem itself must hand you 2N + 2 independent facts. A station-to-station run has N = 3 and needs eight: it starts at x = 0 from rest, ends at rest at x = D, has a known speeding-up rate, a known braking rate, zero acceleration while cruising, and a timetabled total time. Delete the timetable and a whole family of trips fits, one for every cruise speed. Add a measured top speed as a ninth fact and the data are over-determined: they agree, or they expose a mistake. The count is necessary but not sufficient. The equations are nonlinear in the unknowns, so even a correctly counted system can have two solutions or none, and choosing between them is a separate step.

With the count done, eliminate the breakpoint times. In an accelerate–cruise–brake trip the two ramps last v_c/a₁ and v_c/a₂, and adding the three areas under v–t gives D = v_c T − (v_c²/2)(1/a₁ + 1/a₂), a quadratic in v_c. Take a lift that must rise 24.0 m in 12.0 s, speeding up and slowing down at 1.00 m s⁻². The quadratic is v_c² − 12.0v_c + 24.0 = 0, with roots 2.54 m s⁻¹ and 9.46 m s⁻¹. Both are real; only one can happen. The larger root spends 18.9 s on its ramps, more than the whole 12.0 s trip, leaving a cruise of −6.93 s. The smaller root leaves +6.93 s of cruising. The two cruise times are equal and opposite because each is ±√(T² − Tₘᵢₙ²), with Tₘᵢₙ = √(2 × 24.0 × 2.00) = 9.80 s. Squeeze the timetable below 9.80 s and the discriminant turns negative: no cruise speed exists, and the correct answer is that the lift cannot do it.

Sensors report derivatives and leave out the constants. An accelerometer trace fixes velocity only up to v₀ and position only up to x₀, so it needs two anchors; a velocity log needs one; a position log needs none but is noisy to differentiate. An anchor need not sit at t = 0, because an integral runs backwards as happily as forwards. A cart's velocity log reads a steady 0.50 m s⁻¹ from 0 to 2.0 s, then rises linearly to 1.10 m s⁻¹ at 5.0 s. Areas give 1.0 m for the first stretch and ½(0.50 + 1.10)(3.0) = 2.4 m for the second, but not where the cart is. A photogate sees it pass x = 2.0 m at t = 2.0 s. Backwards, x(0) = 2.0 − 1.0 = 1.0 m; forwards, x(5.0 s) = 2.0 + 2.4 = 4.4 m. A gap in a log is bridged the same way: whatever happened inside it must carry the state at its start to the state at its end, and a single constant acceleration in the gap is fixed by the velocities on either side.

A candidate solution is finished only when each piece passes its own checks. Every event time must fall inside the interval whose equation produced it; a catch-up time found from the accelerating-phase equation that lands after the acceleration phase ended belongs to a parabola the object never followed. Stops are breakpoints too: x = x₀ + v₀τ − ½aτ² keeps going past τ = v₀/a and describes a braked car reversing, which friction brakes never do, so the model must switch to v = 0 there. Limits give quick tests: let a₁ and a₂ grow without bound in the trip formula and it becomes D = v_c T, a constant-speed trip; set T = Tₘᵢₙ and the cruise time vanishes. Finally, name the idealization. Real controllers change acceleration over a finite jerk time, a fraction of a second in a lift. When the intervals themselves are that short, the corners in v–t round off, and the seam has to be modelled as a stretch of steadily changing acceleration.