A Car Brakes Three Times Harder Than It Accelerates
Published 10/13/2026 · 4 min read · Everyday calculators
Lena Hoffmann — Science & education writer at OneKitly
Mathematics · Physics
Checked against 2 sources
Acceleration is the change in velocity divided by the time it took: (27 − 0) ÷ 10 = 2.7 metres per second squared. Twenty-seven metres per second is 97 km/h, so this is a car reaching motorway speed from rest in ten seconds — a brisk but ordinary figure, and about 0.28 of gravity. Now stop the same car from that speed in three seconds: (0 − 27) ÷ 3 = −9 m/s², which is 0.92 of gravity and more than three times the magnitude. Every road vehicle is like this. Four brakes acting on four contact patches can shed speed far faster than an engine can add it, which is why a stopping distance is measured in tens of metres while the distance needed to overtake is measured in hundreds.
Reaching 27 m/s from rest in ten seconds is 2.7 m/s². Stopping from the same speed in three seconds is −9 m/s² — a third of gravity against nine tenths of it.
Constant acceleration is the assumption nothing meets
The figure this calculation returns is an average over the interval, not a value the vehicle ever holds. A car pulls hardest in first gear, drops at every change, and fades further as air resistance grows with the square of the speed; the last few kilometres per hour take longer than the first twenty. Braking is closer to constant, because friction at the contact patch changes little with speed, which is another reason the two halves of this comparison are not symmetric. Where the distance covered matters rather than the time, the average is not enough — the same average acceleration with a different profile puts the car in a different place.
The sign is information, not decoration
The −9 is negative because the velocity fell, and the minus carries a direction rather than a judgement. A car reversing and gathering speed has a negative velocity and a negative acceleration and is speeding up; a car rolling forwards with the brakes on has a positive velocity and a negative acceleration and is slowing down. What tells the two apart is whether the signs of the velocity and the acceleration match, not whether the acceleration is negative. Confusing the two is the reason a question about a ball thrown upwards feels harder than it is: on the way up the velocity is positive and the acceleration negative, at the top the velocity is zero and the acceleration is still exactly the same as it was.
| Manoeuvre | Time | Acceleration | In g |
|---|---|---|---|
| 0 to 27 m/s | 10 s | +2.7 m/s² | 0.28 |
| 27 m/s to 0 | 3 s | −9.0 m/s² | 0.92 |
Worked with our own calculator
Acceleration calculator
Given
- Initial velocity (m/s)
- 1
- Final velocity (m/s)
- 14
- Time (s)
- 5
Result
- Acceleration (m/s²)
- 2.6
These figures are produced by the calculator below, not typed in by hand — they are recomputed whenever the tool changes.
Run it on your own figures →Frequently asked questions
- How do I turn this into a distance?
- For constant acceleration, the distance is the average velocity times the time, and the average of a straight ramp is simply the midpoint of the two speeds. Going from 0 to 27 m/s in ten seconds averages 13.5 m/s and covers 135 metres; stopping from 27 to 0 in three seconds averages the same 13.5 and covers 40.5. Those two numbers are the reason a braking distance feels short and an overtake feels long — and both are before any reaction time, which adds whatever the driver's delay multiplied by the full speed happens to be.
- Is one g a lot?
- It is the acceleration you already feel standing still, so it is a useful unit precisely because everyone has a reference for it. A braking figure near 0.9 g is close to what a road tyre on dry tarmac can deliver, which is why the number appears again and again in test data and why it stops improving once the limit is grip rather than brakes. Sideways it is the same limit — a car cannot corner at more than roughly the friction it can brake at, which is why hard braking and hard cornering at once is where a vehicle runs out of tyre.
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