Turning Points and Direction Reversal
At an interior extremum of differentiable x(t), v(t*) = 0 is necessary. A reversal requires a sign change of v across t*; acceleration need not be zero there.
Why this shows up in the exam
Polynomial position laws · Finding maximum or minimum position · Separating distance from displacement
Learn the idea
A one-dimensional turning point occurs when velocity passes through zero and changes sign. A particle may pause without turning, so v = 0 alone is not enough. The positions just before and after the instant must be traversed in opposite directions.
🧠 Memory hook: Zero velocity is a checkpoint; a sign change proves the turn.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- v(t*) = 0 — candidate condition for a smooth turning point
- v(t*-)v(t*+) < 0 — velocity changes sign across a genuine reversal
How to approach it
- 1Solve v(t)=0
- 2Test signs on both sides
- 3Substitute valid turning times into x(t)
Common slip-ups that cost marks
- •Calling every zero of velocity a turn
- •Setting acceleration to zero instead
- •Ignoring additional roots or allowed time domains
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A particle has initial speed 2 m/s and constant acceleration 3 m/s^2 for 4 s. What distance does it cover?
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