Position s is signed displacement from a fixed origin. Velocity is v=ds/dt; acceleration is a=dv/dt. For metres and seconds, their units are m, m,s-1 and m,s-2. Why: a positive displacement gradient means position increases; a negative gradient means it decreases. Differentiating again describes how that gradient changes. Method: differentiate; solve v=0 within the time domain; check the sign of v on both sides. Solve s=0 separately to locate visits to the origin. For s=t2-4t, v=2t-4, a=2. At t=2, velocity changes from negative to positive, so the particle reverses at position -4 m. It is at the origin at t=0,4, when it is moving. Check: v=(t-2)2 is zero at 2 but positive on both sides, so a rest time need not be a reversal. Never infer direction from the sign of position.

Position \(s\) is signed displacement from a fixed origin. Velocity is \(v=\dfrac{\mathrm{d}s}{\mathrm{d}t}\); acceleration is \(a=\dfrac{\mathrm{d}v}{\mathrm{d}t}\). For metres and seconds, their units are \(\mathrm{m}\), \(\mathrm{m\,s^{-1}}\) and \(\mathrm{m\,s^{-2}}\).

Why: a positive displacement gradient means position increases; a negative gradient means it decreases. Differentiating again describes how that gradient changes.

Method: differentiate; solve \(v=0\) within the time domain; check the sign of \(v\) on both sides. Solve \(s=0\) separately to locate visits to the origin.

For \(s=t^2-4t\), \(v=2t-4\), \(a=2\). At \(t=2\), velocity changes from negative to positive, so the particle reverses at position \(-4\) m. It is at the origin at \(t=0,4\), when it is moving.

Check: \(v=(t-2)^2\) is zero at \(2\) but positive on both sides, so a rest time need not be a reversal. Never infer direction from the sign of position.

Position, velocity and reversal

Position, velocity and reversal