Braking and Stopping Distance
For initial speed u and constant acceleration a opposite the motion, setting v = 0 in v² = u² + 2as gives stopping displacement s_stop = u²/(2|a|). Reaction distance, if present, must be added separately.
Why this shows up in the exam
Vehicle braking · Penetration under constant resistance · Comparing stopping distances at different speeds
Learn the idea
With fixed braking deceleration, stopping distance grows as the square of the initial speed. Doubling speed carries four times the kinetic scale that uniform braking must remove, so the stopping distance becomes four times as large when the deceleration magnitude is unchanged.
🧠 Memory hook: Same brakes: stopping distance follows speed squared.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- s_stop = u²/(2|a|) — braking distance under constant deceleration
- s2/s1 = (u2/u1)² — stopping-distance scaling for unchanged brakes
How to approach it
- 1Set final velocity to zero
- 2Use the time-free equation
- 3Compare squared speeds before calculating full values
Common slip-ups that cost marks
- •Assuming stopping distance is proportional to speed
- •Using positive a without signed consistency
- •Including reaction time when none is stated
🌟 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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