Car Crash Impact Force Calculator
Car Crash Impact Force: the Stopping Distance Decides Everything
A crash has no force of its own. It has whatever force is needed to take all the kinetic energy out over the distance you are given to stop in, and that distance is what a seat belt and a crumple zone exist to lengthen. Spread the same 50 km/h stop over 0.70 m instead of 0.07 m and the average force falls to a tenth. The page does both sums side by side, because the ratio is the lesson.
Speed and stopping distance to average force
A 70 kg passenger is in a car travelling at 50 km/h when it strikes a solid obstacle. Belted, the chest travels about 0.70 m while the belt stretches, the seat gives and the front of the car folds. Find the average force on the passenger, and compare it with the same crash stopped in 0.07 m.
One equation, two derivations, and the quantity that does all the work
- the work–energy route
- the car arrives with kinetic energy ½mv² and leaves with none, so something did that much negative work. Work is force times distance, so F d = ½mv² and F = mv²/2d. This is the route to use when the question gives you a distance
- the impulse route
- momentum mv has to go to zero, and impulse is force times time, so F t = mv. With a constant deceleration the average speed is v/2, so t = 2d/v, and substituting gives the same mv²/2d. Two different conservation statements, one answer — see impulse and change of momentum for that half worked properly
- d
- the distance over which the thing you named comes to rest, in metres. The whole page turns on this one number, and it is the one most often taken from the wrong place. It is not the braking distance before the impact and it is not the length of the car
- v
- the speed at the moment of impact, in m/s, which is km/h divided by 3.6. It appears squared, so it dominates everything: at the same crush distance, twice the speed is four times the force
- what cancels, and what does not
- the mass cancels out of the deceleration — a = v²/2d has no m in it — but it does not cancel out of the force on you, which is your own mass times that deceleration. A heavier car does not give its occupants a gentler stop unless it also crushes over a longer distance
- what this is not
- not a peak force, not a car-to-car collision, and not an injury prediction. For two vehicles sharing momentum, go to collisions and restitution; for the braking that happens before any of this, stopping distance
Worked example
A 70 kg passenger is in a car travelling at 50 km/h when it strikes a solid obstacle. Belted, the chest travels about 0.70 m while the belt stretches, the seat gives and the front of the car folds. Find the average force on the passenger, and compare it with the same crash stopped in 0.07 m.
Convert the speed first, before anything else. v = 50 / 3.6 = 13.89 m/s. Write it down. Every error on this page that is not a sign error is a km/h left in a formula that wanted m/s.
Find the energy. ½mv² = 0.5 × 70 × 13.89² = 0.5 × 70 × 192.9 = 6752 J. That is the whole problem in one number: 6752 joules have to be taken out of the passenger, and the only question left is over what distance.
Divide by the distance. F = 6752 / 0.70 = 9646 N, so about 9.6 kN. Sanity check: the passenger weighs 70 × 9.81 = 687 N, so this force is 14 times their own weight. Equivalently the deceleration is v²/2d = 192.9/1.4 = 138 m/s², which is 14 g. The two ways of saying it are the same arithmetic.
Now do it again with 0.07 m. The energy has not changed — same mass, same speed — so F = 6752 / 0.07 = 96 460 N, about 96 kN, or 140 g. One tenth of the distance, ten times the force, exactly. Nothing about the collision changed except where the passenger was allowed to stop.
Check it by impulse. At a constant deceleration the average speed is v/2 = 6.94 m/s, so stopping in 0.70 m takes t = 0.70/6.94 = 0.101 s, and F = mv/t = 70 × 13.89 / 0.101 = 9626 N. The same 9.6 kN to within rounding, from momentum rather than energy. Two independent routes agreeing is the cheapest check in mechanics.
Read the comparison, not the absolute number. The 9.6 kN depends on a crush distance chosen to make the sum clean. The ratio does not: it is 0.70/0.07 = 10 whatever the real distances turn out to be, and it would still be 10 at 30 km/h or in a lorry. That is why the useful sentence is “the force goes as one over the stopping distance” and not any particular number of kilonewtons.
The same crash at 50 km/h, stopped over different distances
| Stopping distance | Average force on a 70 kg person | Deceleration | How long the stop lasts |
|---|---|---|---|
| 0.02 m | 338 kN | 492 g | 2.9 ms |
| 0.07 m | 96 kN | 141 g | 10 ms |
| 0.25 m | 27 kN | 39 g | 36 ms |
| 0.70 m | 9.6 kN | 14 g | 101 ms |
| 2.00 m | 3.4 kN | 4.9 g | 288 ms |
Four questions that sound the same and are not
| The question | What it needs | Which page |
|---|---|---|
| How far to stop before I hit it? | reaction time and tyre grip, before any impact | stopping distance |
| What force while I am being stopped? | ½mv²/d — the crush distance | this page |
| What force if I know the contact time instead? | mv/t, the impulse form | impulse |
| What happens when two vehicles hit each other? | momentum sharing and how much energy survives | collisions |
Why the crumple zone is the whole answer, and what the number above is not
There is no such thing as the force of a crash. That sentence sounds like a quibble and it is the entire topic. A collision is the business of getting rid of kinetic energy, and force only enters because work is force times distance. Fix the energy and fix the distance and the average force is decided; change the distance and it changes in exact inverse proportion. So the question “how big is the force in a 50 km/h crash” has no answer until someone says over what distance.
Which is why the distance is the thing engineers buy. A crumple zone does not absorb the energy in any sense that a belt does not; both of them simply let you travel further before you are stationary. The front of the car folds, the seat belt stretches, the seat deforms, and together they turn a few centimetres into a few tens of centimetres. From 0.07 m to 0.70 m is a factor of ten in distance and therefore a factor of ten off the force, with the same mass and the same speed. Nothing else on this page is worth as much as that one ratio.
The speed is squared and that is not a detail. Kinetic energy goes as v², so at the same crush distance 60 km/h is not twenty per cent worse than 50 — it is forty-four per cent worse, because 60²/50² = 1.44. Doubling the speed quadruples the energy and quadruples the average force. It also halves nothing useful: the stopping time t = 2d/v gets shorter as the speed rises, so the same distance has to do its work faster.
What cancels and what does not. The deceleration a = v²/2d contains no mass at all, so in a crash of a given speed and a given crush distance a lorry driver and a child decelerate identically. The force is mass times that, so it is proportional to the mass of whatever you asked about — which is why the page asks you to say whether you mean yourself or the whole car, and why the two answers differ by the ratio of the masses and nothing else. In a collision between two vehicles the masses stop cancelling for a different reason: momentum is shared, and the lighter vehicle changes velocity more. That is a separate calculation.
An average is not a peak, and a standard knows it. Real impact pulses rise, peak and fall over something like a tenth of a second, so the peak force is higher than a single division can tell you. Vehicle regulation does not set a force limit at all. UN Regulation 94 asks for a head performance criterion below 1,000, a resultant head acceleration that does not exceed 80 g for more than 3 ms, a thorax compression under 50 mm and a viscous criterion under 1.0 m/s. Notice what those are: accelerations with durations attached, and compressions. A brief 80 g is treated differently from a sustained one, and no formula on this page can make that distinction.
Where this goes wrong in exams. Four slips, in order of frequency. One: km/h left in the formula, which multiplies the energy by about thirteen. Two: the braking distance used as d — that is the distance before the impact, and putting 25 m into this page gives a few hundred newtons and a confident wrong answer. Three: mass in grams. Four: answering with the force on the car when the question asked for the force on the passenger, or the reverse. None of the four is a physics error and all four cost the same mark. And one honest point that is not an exam slip at all: the reason to wear the belt is in the fourth column of the table above, not in any sentence on this page.
Frequently asked questions
Why does the force depend on the stopping distance at all? The crash is the crash.
Because force is what appears when you ask how the energy left, and energy leaves through force times distance. The 6752 joules of a 70 kg passenger at 50 km/h are fixed by the mass and the speed. If they are removed over 0.70 m the average force is 9.6 kN; over 0.07 m it is 96 kN. The collision did not change. The distance over which the passenger was allowed to come to rest did, and that is the only free variable there is.
Is a bigger, heavier car safer then?
Not for the reason people usually mean. The deceleration v²/2d has no mass in it, so extra mass on its own buys nothing in a crash against a fixed obstacle — what helps is a longer crush distance and a structure that keeps the cabin intact, and those are design, not weight. Mass does matter when two vehicles collide, because momentum is shared and the lighter one changes velocity more. Those are two different questions and this page only answers the first.
The page gives 9.6 kN. Is that what a crash test would measure?
No, and it is worth being clear about why. It is the average force implied by one speed and one distance, with the deceleration assumed constant. A real test measures a pulse that varies throughout the impact, from instrumented dummies, and reports head and thorax criteria rather than a force. The value here is the right size and it scales correctly; it is not a measurement, and the comparison between two distances is the part of it you can rely on completely.
Why do airbags and belts not just absorb the energy?
They do absorb energy — that is what a stretching belt and a venting airbag do — but the energy is not the problem the body has. The problem is the rate, and both devices work by extending the distance and the time over which the energy goes. An airbag adds a few more centimetres in front of your head and face and spreads the load over a wide area, which reduces both the force and the pressure. Same mechanism as the crumple zone, applied to the last few centimetres.
Can I use mv/t instead? My question gives a time, not a distance.
Yes, and it is the same physics. Impulse says F t = mv, so F = mv/t. With a constant deceleration the two are linked by t = 2d/v, and substituting one into the other returns mv²/2d. Use whichever the question gives you, and if it gives you both, compute both and check they agree — if they do not, the deceleration was not constant, which is itself a fair thing to say in an answer.
What is the “height you would fall from” line for?
It converts the speed into something you can picture. Arriving at 50 km/h is arriving at the speed you would reach by falling 9.8 m, because v²/2g is the same algebra as this whole page with g in place of the deceleration. It is not a claim about what a fall does to anybody. It is the one comparison that needs no equipment and no data: the same speed arrives the same way, whichever direction it came from.
Why does the page refuse to answer when I enter a stopping distance of zero?
Because the answer is infinite and printing a number there would be a lie. F = ½mv²/d has a division by d in it, and nothing stops in zero distance — not a car against a bridge pier, not a hammer on an anvil. Something always deforms by some amount. Put the real distance in, even if it is five millimetres, and the page will tell you how large the force gets.
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References
- United Nations Regulation No. 94, Uniform provisions concerning the approval of vehicles with regard to the protection of the occupants in the event of a frontal collision. Read 12 November 2026 from the official mirror at
mlit.go.jp/jidosha/un/UN_R094.pdf. Quoted verbatim: the front face of the test barrier “consists of a deformable structure”, “the vehicle shall overlap the barrier face by 40 per cent +/− 20 mm”, and “the mass of the vehicle submitted shall be the unladen kerb mass”. The pass conditions in paragraph 5.2 include “The head performance criterion (HPC) shall not exceed 1,000” with the resultant head acceleration not to exceed 80 g for more than 3 ms, rebound excluded; “The thorax compression criterion (ThCC) shall not exceed 50 mm”; “The viscous criterion (V*C) for the thorax shall not exceed 1,0 m/s”; a residual steering wheel displacement of no more than 80 mm upward and 100 mm rearward; and that no door may open during the test. Worth reading for one reason above all: a real standard does not set a limit on force. It sets limits on acceleration with a time attached, on compression and on compression rate — which is why the figures below are given in multiples of g as well as in kilonewtons. The paragraph giving the impact speed was not inside the portion of the document that could be retrieved, so no test speed is quoted here. - PHYSICS (042) Curriculum, Class XI–XII, Central Board of Secondary Education, session 2025–26. Unit III, Laws of Motion, is where this material sits; the line this project holds verbatim from that unit reads “Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicle on a level circular road, vehicle on a banked road)”. Quoted to establish what is in scope at this level, never to claim any endorsement by the board.
- NIST, Fundamental Physical Constants — Complete Listing, 2022 CODATA adjustment, Standard Reference Database 121. The line both of these pages lean on is “standard acceleration of gravity 9.806 65 (exact) m s−2”. The word exact is the point: gn is a conventional value fixed by the 3rd CGPM in 1901, so it carries no uncertainty and it is not the g where you are standing.
physics.nist.gov/cuu/Constants/ - BIPM, The International System of Units (SI), 9th edition (2019), Table 4: the newton is kg m s−2 and the joule is kg m² s−2. That is the whole reason a force times a distance is an energy, and it is the dimensional check that catches a kilojoule written where a kilonewton was meant.