For a particle moving along a straight line through a fixed point \(O\), the sign of each quantity tells you where the particle is and how it is moving. Positive means the positive direction of the \(s\)-axis.
Sign of the displacement \(s\)
| Value | What it tells you |
|---|---|
| \(s>0\) | The particle is on the positive side of \(O\). |
| \(s=0\) | The particle is at the fixed point \(O\). This is its starting position only if it started at \(O\). |
| \(s<0\) | The particle is on the negative side of \(O\). |
Sign of the velocity \(v\)
| Value | What it tells you |
|---|---|
| \(v>0\) | The particle moves in the positive direction. |
| \(v=0\) | The particle is instantaneously at rest. A change of direction requires a change in the sign of \(v\); zero velocity alone does not guarantee an extremum of displacement. |
| \(v<0\) | The particle moves in the negative direction. |
Sign of the acceleration \(a\)
| Value | What it tells you |
|---|---|
| \(a>0\) | The signed velocity is increasing. Speed increases when \(v>0\) and decreases when \(v<0\). |
| \(a=0\) | The velocity is stationary at that instant. This alone proves neither constant velocity on an interval nor a velocity extremum. |
| \(a<0\) | The signed velocity is decreasing. Speed decreases when \(v>0\) and increases when \(v<0\). |
Reading a displacement-time curve: \(s=t^3-4t^2\)
At the lowest point of the curve the gradient is zero: the particle is instantaneously at rest (\(v=0\)) and its displacement is a minimum — it changes direction there. Where the curve meets the \(t\)-axis again, \(s=0\): the particle is back at \(O\).
\(v=0\) does not mean \(a=0\). At a change of direction the particle is instantaneously at rest, yet its acceleration is usually non-zero.
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