Stopping Distance Calculator: Reaction plus Braking
Stopping Distance Calculator: Reaction Distance plus Braking Distance
Total stopping distance is two different functions of speed added together: a reaction distance that grows in proportion to speed and a braking distance that grows as the square of it. Double your speed and the first doubles while the second quadruples — which is the whole reason the total is not proportional to anything.
Stopping distance: thinking plus braking
A car travelling at 60 km/h brakes to a stop on dry asphalt. Take the driver’s reaction time as 1.5 s, the coefficient of friction as 0.70 and g as 9.81 m/s².
Two terms, two different powers of speed
- v
- speed at the moment the hazard appears, in m/s. Divide a km/h figure by 3.6 before anything else.
- tr
- reaction time, in s. The car travels at full speed for the whole of it, so this term is linear in v.
- μ
- coefficient of friction between tyre and road, dimensionless. Take it from the question; it is a property of the pair of surfaces, not a constant of nature.
- μg
- the greatest deceleration the tyres can deliver, in m/s². The maximum friction force is μmg and the mass cancels against ma, which is why the vehicle’s mass does not appear.
- v²/(2μg)
- braking distance, in m, straight out of v² = u² + 2as with the final velocity zero. Quadratic in v, which is the point of the page.
Worked example
A car travelling at 60 km/h brakes to a stop on dry asphalt. Take the driver’s reaction time as 1.5 s, the coefficient of friction as 0.70 and g as 9.81 m/s².
Convert the speed first. 60 km/h ÷ 3.6 = 16.667 m/s. Everything else is in metres and seconds, so this cannot be left until later.
Reaction distance. No braking has started yet, so the car covers v tr = 16.667 × 1.5 = 25.0 m at undiminished speed. This term is linear in speed.
Deceleration from the friction. The largest friction force the tyres can give is μmg, so the largest deceleration is μg = 0.70 × 9.81 = 6.867 m/s². The mass has cancelled.
Braking distance. From v² = u² + 2as with v = 0: s = u²/(2 × 6.867) = 277.78/13.734 = 20.2 m. This term is quadratic in speed.
Add them. 25.0 + 20.2 = 45.2 m, taking 1.5 + 2.43 = 3.93 s in total. Note that at this modest speed the thinking distance is the larger of the two; by 100 km/h the braking distance has overtaken it.
Check the scaling. At 120 km/h the thinking distance doubles to 50.0 m while the braking distance quadruples to 80.9 m, for a total of 130.9 m — not twice 45.2 m but nearly three times it. That non-proportionality is the single most useful thing on this page.
The same driver and the same road at six speeds, with tr = 1.5 s, μ = 0.70, g = 9.81
| Speed (km/h) | Speed (m/s) | Thinking (m) | Braking (m) | Total (m) | Total ÷ total at 60 |
|---|---|---|---|---|---|
| 20 | 5.56 | 8.3 | 2.2 | 10.6 | 0.23 |
| 40 | 11.11 | 16.7 | 9.0 | 25.7 | 0.57 |
| 60 | 16.67 | 25.0 | 20.2 | 45.2 | 1.00 |
| 80 | 22.22 | 33.3 | 36.0 | 69.3 | 1.53 |
| 100 | 27.78 | 41.7 | 56.2 | 97.8 | 2.16 |
| 120 | 33.33 | 50.0 | 80.9 | 130.9 | 2.90 |
| 140 | 38.89 | 58.3 | 110.1 | 168.5 | 3.72 |
| 160 | 44.44 | 66.7 | 143.8 | 210.5 | 4.65 |
What Rule 126 of the Highway Code actually says, and what it does not
| The rule’s own wording | What it means for this calculation |
|---|---|
| Never get closer to the vehicle in front than the overall stopping distance. | The overall stopping distance is the figure this page computes — thinking plus braking. |
| Allow at least a two-second gap on high-speed roads and in tunnels where visibility is reduced. | A two-second gap is 2v, linear in speed. The stopping distance grows faster than that, so the ratio of the two rises with speed — see the figures under the answer. |
| The gap should be at least doubled on wet roads and up to ten times greater on icy roads. | Consistent with the braking term being inversely proportional to μ: a lower μ multiplies the braking distance and leaves the thinking distance alone. |
| Allow a four-second gap for a large vehicle in a tunnel, and at least a five-metre gap when stopped in one. | Fixed figures with no speed in them; nothing to compute. |
| Typical stopping distances for 20 to 70 mph. | Published in the official text only as a diagram, so this page does not reproduce those figures. The model above is the physics, not the Highway Code table. |
Why the total is not proportional to speed, and why the mass drops out
Stopping distance is two completely different pieces of physics bolted together, and almost every mistake in the topic comes from treating it as one. During the reaction time, nothing has happened yet: the brakes are not on, the car has not slowed, and it covers speed × time at full speed. Then the brakes bite, and from that moment it is a constant-acceleration problem with a = −μg and a final velocity of zero.
The consequence is that the two terms scale differently. The thinking distance is proportional to v. The braking distance is proportional to v². So doubling the speed doubles one and quadruples the other, and the total grows by some factor between two and four depending on which term dominates. For a 1.5 s reaction on a dry road, going from 60 to 120 km/h takes the total from 45 m to 131 m — a factor of 2.9, which is neither 2 nor 4 and cannot be guessed from either. There is no single multiplier, and that is the useful insight rather than a complication.
Which term dominates shifts with speed. At 20 km/h the thinking distance is nearly four times the braking distance; by 100 km/h the braking distance is the larger. That crossover is why urban road safety campaigns talk about distraction and motorway campaigns talk about tailgating: at low speed the driver’s attention is most of the distance, and at high speed the tyres are.
The mass cancels, and this surprises people every time. The maximum friction force a tyre can exert is μN, and on level ground N = mg, so the force is μmg. Newton’s second law then gives a = F/m = μg, with the m gone. A fully laden lorry and an empty hatchback on identical tyres and identical tarmac have the same maximum deceleration and the same braking distance. Real lorries do stop in longer distances, but for reasons this model does not contain: brake system response time, load transfer, tyre compound, and the simple fact that a long vehicle’s driver is further from the front of it.
Where a number had to be left out. This page needs a coefficient of friction, and no authoritative published table of tyre-on-road values exists in the way that NIST publishes constants: μ depends on the tyre compound, the tread depth, the surface dressing, the temperature, the water depth and the slip ratio, and quoted figures for “dry asphalt” range widely. So the field takes the value from you, which is also what an exam question does. The same applies to reaction time, which varies between drivers and within one driver by more than any single figure can represent.
Three things this model leaves out, and all three make real stopping distances longer. There is no brake system delay: a real car takes a few tenths of a second between the pedal moving and full braking force, and that interval is covered at close to full speed, so it behaves like extra reaction time rather than extra braking distance. There is no gradient: going downhill adds a component of weight along the road, which is equivalent to reducing the deceleration by g sin of the slope angle, so a 10% downgrade cuts the deceleration by about 1 m/s². And there is no allowance for the brakes being anything less than able to reach the friction limit on every wheel at once. A question that mentions any of those three is asking you to extend the model, not to use this one.
The arithmetic generalises past cars. Any object sliding to a halt under friction stops in v²/(2μg), and the reaction term is whatever delay precedes the braking. A cyclist, a sledge, a curling stone and a box sliding across a factory floor all obey it, with μ the only thing that changes. It is also the reason a speed limit reduction has a disproportionate effect: cutting a limit from 40 to 30 km/h removes a quarter of the reaction distance but nearly half the braking distance, so the total falls by more than the speed does. That non-proportionality, not the arithmetic, is the thing worth taking away from the page.
The one official document this page does lean on is Rule 126 of the Highway Code, and it leans on the rule’s wording rather than its famous table of distances. That table is published in the official text only as a diagram, so its figures could not be read from the source and are not reproduced here. What the wording does give is the two-second gap, doubled on wet roads and up to ten times greater on ice — and those multipliers are exactly consistent with a braking distance inversely proportional to μ, which is a satisfying check that the model and the guidance are describing the same world. Note too that a two-second gap is not a stopping distance: it is a gap to a vehicle that is itself moving and will itself take time to stop.
Why the braking term is quadratic, in one sentence, because the answer is energy. The kinetic energy to be destroyed is ½mv², and the friction force μmg destroys it at a fixed rate per metre, so the distance needed is ½mv² ÷ μmg = v²/(2μg) — the same expression the equations of motion give, with the mass cancelling for the same reason. That view makes the scaling obvious: energy goes as the square of speed, so the distance over which it has to be dissipated does too. It also explains why brakes get hot rather than doing anything else: the kinetic energy has to go somewhere, and at 60 km/h a 1,200 kg car is carrying about 167 kJ of it.
Frequently asked questions
Why is the total stopping distance not proportional to speed?
Because it is the sum of two terms with different powers of speed. The thinking distance is v × tr, proportional to v. The braking distance is v²/(2μg), proportional to v². Doubling the speed doubles the first and quadruples the second, so the total grows by a factor between 2 and 4 that depends on which term is bigger — 2.9 for a 1.5 s reaction on a dry road going from 60 to 120 km/h.
Does a heavier vehicle take longer to stop?
Not in this model, and the reason is worth understanding. The maximum friction force is μmg and the deceleration is that force divided by m, so a = μg with the mass cancelled out. In practice heavy vehicles do stop in longer distances, but because of brake response times, load transfer, tyre and brake heating and vehicle length — none of which is in v²/(2μg). A school question asking whether the mass matters wants the cancellation.
What coefficient of friction should I use?
The one your question gives you. There is no single authoritative published value: μ for tyre on road depends on compound, tread, surface dressing, temperature, water depth and how close the tyre is to locking, and quoted figures vary widely even for the same nominal surface. That is why this page asks rather than assuming. What is reliable is the relationship: braking distance is inversely proportional to μ, so halving the grip doubles the braking distance exactly.
Is a two-second gap the same as a stopping distance?
No, and conflating them is a common confusion. A two-second gap is 2v metres, proportional to speed. A stopping distance has a quadratic term and grows faster, so at motorway speeds it is considerably longer than a two-second gap. The gap works because the vehicle in front is also moving and will also need time and distance to stop; it is not designed to protect you from a stationary obstruction.
Does the reaction distance depend on the road surface?
Not at all, and that is the cleanest way to see that the two terms are separate physics. During the reaction time the brakes are not applied, so the grip is irrelevant and the car simply travels at full speed. Ice does not lengthen the thinking distance by a single metre; it lengthens the braking distance enormously. A question that applies a wet-road factor to the whole total has made an error.
How do I do this in mph and feet?
Convert into metres and seconds, compute, and convert back. One mph is 1.609344 km/h, so multiply mph by 1.609344 to get km/h, or by 0.44704 to get m/s directly. One metre is 3.28084 feet. The figures under the answer give your speed in both km/h and mph so you can check you converted the right way; dividing when you should have multiplied gives a number about 2.6 times out, which is large enough to notice.
What does a hill do to the stopping distance?
It changes the deceleration, not the reaction distance. On a slope the component of weight along the road adds to or subtracts from the braking force, so the deceleration becomes μg cos α ± g sin α for a slope angle α. A 10% downgrade is about 5.7°, which costs roughly 1 m/s² of deceleration — around 14% of the 6.87 m/s² in the worked example, so the braking distance grows by about a sixth. This page assumes level ground and does not model it.
Related calculators
References
- Department for Transport, The Highway Code, Rule 126 (“Stopping Distances”). The wording gives the two-second gap, at least doubled on wet roads and up to ten times greater on ice, a four-second gap for large vehicles in tunnels and a five-metre gap when stopped in one. Its numerical table of typical stopping distances is published only as a diagram, so this page does not reproduce those figures.
- NIST, CODATA Internationally Recommended Values of the Fundamental Physical Constants — complete listing. The standard acceleration of gravity is tabulated as 9.806 65 m s−2, marked exact rather than carrying an uncertainty.
physics.nist.gov/cuu/Constants/ - OCR, Data, Formulae and Relationships Booklet for AS and A Level GCE Physics A (H156 and H556), 8 pages. Lists the acceleration of free fall as g = 9.81 m s−2 alongside the thirteen other physical constants the papers assume.
- BIPM, The International System of Units (SI), 9th edition (2019). Definitions of the metre and the second, and the convention that the radian is the coherent SI unit of plane angle with dimension one — which is why every trigonometric formula in mechanics is written in radians.