Distinguish stopping from reversing

An instant at rest is not a promise to stay there

In the default investigation, the initial velocity is +2 m/s and constant acceleration is −1 m/s². At two seconds, velocity is zero. If the acceleration continues unchanged, the next velocity is negative. The object reverses direction instead of remaining stopped.

This distinction matters when interpreting a formula as a model of braking. A vehicle controller that slows a vehicle to rest and then holds it there does not maintain the same negative acceleration forever. The motion after stopping needs a changed rule. The current widget deliberately keeps acceleration constant so you can see why an unqualified extrapolation fails.

Solve for the zero-velocity time

For nonzero constant acceleration, set v₀ + at = 0 and solve for t₀ = −v₀/a. Here t₀ denotes a candidate time of zero velocity, v₀ is initial velocity in m/s, and a is acceleration in m/s². Their ratio has units of seconds. A negative candidate time refers to a time before the chosen start, not a future stop.

With v₀ = +2 m/s and a = −1 m/s², t₀ = −2/(−1) = 2 s. Position at that time is 2×2 + ½×(−1)×2² = +2 m. It is a maximum of position because velocity changes from positive to negative. At t = 4 s the object returns to zero, giving zero displacement but 4 m distance: two metres outward plus two metres back.

The standard constant-acceleration equations apply only while their assumptions hold. A zero-velocity result is one consequence of those equations; it does not automatically replace the acceleration with zero.

Distinguish four kinds of evidence

Your action: keep a short comparison record for these four settings. Before each change, predict the displayed classification.

• Initial +2 m/s, acceleration −1 m/s²: zero velocity occurs at 2 s, strictly inside the four-second interval. There is a direction reversal.
• Initial +4 m/s, acceleration −1 m/s²: zero velocity occurs exactly at 4 s. The object travels 8 m in the displayed interval and does not reverse strictly inside it. If the model continued unchanged, reversal would follow.
• Initial +4 m/s, acceleration −0.5 m/s²: the candidate stop is at 8 s, beyond the displayed interval. At 4 s velocity is still +2 m/s. “No zero in 0–4 s” does not mean it can never stop.
• Initial 0 m/s, acceleration 0 m/s²: the object remains at rest for the entire interval. There is no unique stopping instant. Dividing by zero in −v₀/a would be invalid, so this case is handled separately.

Another boundary case is an object beginning at rest with nonzero acceleration. Zero velocity occurs at time zero, followed by motion. It has not reversed direction within the displayed interval because no earlier direction is part of that interval. Do not invent a prehistory from an endpoint value.

Split travel at the turning point

For a one-dimensional path that reverses once, total distance is the magnitude of the displacement before the turn plus the magnitude after it. If x₀, x_turn and x_f denote initial, turning and final positions, the distance is |x_turn − x₀| + |x_f − x_turn|. Vertical bars mean magnitude, so each segment contributes a nonnegative length. Net displacement remains x_f − x₀.

A turning point may lie between the integer-second table readings. With initial +1 m/s and acceleration −1.5 m/s², it occurs at 2/3 s, not at the nearest display sample. The maximum position is 1/3 m. Final position at 4 s is −8 m. Total distance is 1/3 + 8⅓ = 8⅔ m, while displacement is −8 m. The model splits at the analytic turning time and only rounds the displayed answer; it does not miss the short outward journey by looking solely at integer-second positions.

Use stopping distance within its valid interval

For an initially positive velocity and a constant negative acceleration, the distance to the first stop is v₀²/(2|a|). This follows by substituting t₀ = −v₀/a into the position equation. The denominator uses acceleration magnitude so the reported distance is nonnegative. If initial speed doubles while the acceleration magnitude stays fixed, stopping distance increases by a factor of four; stopping time doubles.

This result excludes reaction delay, changes in friction, slopes, obstacles, tire behavior and a controller’s response. It is a model relationship, not a real-world safe stopping-distance recommendation. The widget also has a four-second observation window; it may not contain the stop for a particular setting.

Transfer: A teaching model starts at +3 m/s with constant acceleration −0.75 m/s². Find the first zero-velocity time and distance traveled before it. Then state what rule a model would need after that time to represent an object staying at rest rather than reversing.

Check the stopping transfer

The first zero is at 4 s. Position there is 3×4+½×(−0.75)×16 = 6 m, so travel to the stop is 6 m. To remain at rest in this one-dimensional kinematic model, velocity must stay zero and acceleration must become zero afterward. Continuing −0.75 m/s² would instead produce negative velocity. A physical force/controller model must justify that rule change.