A particle is at instantaneous rest when its velocity is zero. It reverses direction at that moment only if the velocity changes sign, which you test by checking v on either side of the rest time.
This lesson is part of kinematics of linear motion in SPM Additional Mathematics. It pairs with total distance versus displacement, which uses these turning points.
Worked example: two rest points
A particle moves on a line with velocity v = t² − 6t + 5 m/s, for t ≥ 0. Describe its motion.
Find the rest points. v = (t − 1)(t − 5) = 0 at t = 1 and t = 5.
Test the sign of v.
| Time range | Test value | v | Motion |
|---|---|---|---|
| 0 ≤ t < 1 | t = 0 | 5 | positive direction |
| 1 < t < 5 | t = 3 | −4 | negative direction |
| t > 5 | t = 6 | 5 | positive direction |
The sign changes at both times, so the particle reverses at t = 1 and at t = 5.
Check with acceleration. a = dv/dt = 2t − 6. At t = 1, a = −4, so the particle next moves in the negative direction.
At t = 5, a = 4, so it next moves in the positive direction. Both match the table.
Rest without a reversal
Not every rest point is a turning point. Take v = (t − 2)², which is zero at t = 2.
Test values: at t = 1, v = 1, and at t = 3, v = 1. The sign is positive on both sides, so the particle pauses at t = 2 and then carries on in the same direction.
The acceleration is a = 2(t − 2), which is zero at t = 2. A rest with a = 0 and no sign change is a pause, not a reversal.
The mistake that costs marks
The slip is to say “v = 0, so the direction changes” without testing either side. It works for a parabola crossing the axis and fails for one that only touches it.
| Claim | Wrong | Right |
|---|---|---|
| v = 0 at t = 2 | Direction changes | Only if v changes sign |
| a = 0 | The particle is at rest | v is not changing at that instant |
| v negative | The particle is slowing | It moves in the negative direction |
A related slip is finding the times when a = 0 and calling them rest points. In the first example a = 0 at t = 3, where v = −4, so the particle is moving steadily at that instant and not at rest.
A method you can reuse
- Solve v = 0 and keep roots within the allowed time range.
- Test v on each side of every root.
- State a reversal only where the sign changes.
- Use the sign of a at the root as a second check on the next direction.
- Write the answer in words, with the time and the direction.
Check yourself
A particle has velocity v = t² − 4t + 3 m/s, for t ≥ 0. Find when it is at rest, say whether it reverses each time, and find its acceleration at t = 2.
Answer
v = (t − 1)(t − 3) = 0 at t = 1 and t = 3.
Signs: at t = 0, v = 3 (positive). At t = 2, v = −1 (negative). At t = 4, v = 3 (positive). The sign changes at both times, so the particle reverses at t = 1 and t = 3.
a = 2t − 4, so at t = 2, a = 0. At that moment v = −1, so the particle is not at rest.
What to study next
Now put the chapter together. Try the kinematics practice set, which mixes distance, displacement and rest points.
If the link between s, v and a is not yet secure, return to relating displacement, velocity and acceleration. For a teacher to test rest points with you, see online one-to-one Additional Mathematics tuition.