SOLVETUTORMATH SOLVER

Instrument MI-03-078 · Physics

Car Crash Calculator

A wall removes every joule a moving car carries. Divide that energy by how far the front end folds, and you have the force everyone inside feels.

Instrument MI-03-078
Sheet 1 OF 1
Rev A
Verified
Type 03 — Mechanics SER. 2026-03078

Average impact force

337,500.0000 N

F = m·v² ⁄ 2d (work-energy)

22.9436 Deceleration (g)
The working Every figure verified twice
  1. F = 1500·15^2 ⁄ (2·0.5) = 337,500.0000
  2. gForce = 15^2 ⁄ (2·0.5·9.80665) = 22.9436
Worksheet log
  1. No entries yet — change an input to log a scenario.

How this instrument works

Every crash is an energy problem before it is a force problem. Any moving car carries ½mv² joules of kinetic energy, and rigid barriers take all of it away. Divide that energy by the distance over which the front structure folds, and you have the average force pressing back through the chassis. Gaspard-Gustave de Coriolis set this bookkeeping down in 1829, halving Leibniz's vis viva to define what we now call kinetic energy and pairing it with travail — work, force times distance. Two centuries later, that same product is what every barrier test measures.

One term in the denominator is settled by design, and it is the only one anybody controls after the fact. Mass is set by the market, speed by whoever is driving, but crumple distance belongs to whoever drew the car. Béla Barényi patented that insight for Mercedes-Benz in 1952: a rigid passenger cell flanked by deliberately weak front and rear sections engineered to collapse in controlled sequence. Stretching that collapse from 30 cm to 60 cm halves what reaches the cabin. Belts and airbags stretch it further, because a restrained body rides down over a longer path than any bumper does.

F is mean force, not peak force. Genuine crash pulses ramp as rails and engine mounts engage, spike, then fall away; instrumented barrier tests routinely log instantaneous decelerations two to three times higher than this average. The arithmetic also assumes an immovable obstacle absorbing everything, with no rebound and no rotation about either corner. For car-into-car impacts, settle momentum first to find how much speed each vehicle actually loses, then bring that number back here.

Fd=12mv2F d = \tfrac{1}{2} m v^{2}F=mv22dF = \frac{m v^{2}}{2d}ag=v22dg\frac{a}{g} = \frac{v^{2}}{2 d g}
F — average impact force, newtons (N) · m — vehicle mass, kilograms (kg) · v — impact speed, metres per second (m/s) · d — crumple distance, metres (m) · a — mean deceleration (m/s²) · g — 9.80665 m/s², standard gravity. Notice that mass cancels out of the g figure entirely.
  • Enter Vehicle mass — kerb weight plus occupants and luggage is the honest figure. Tonnes and pounds are available on the unit menu.
  • Set Impact speed to whatever the car was doing at the moment of contact, not the posted limit. Accepts m/s, km/h or mph.
  • Give Crumple distance: how far the structure collapses. Roughly 0.5 m suits most modern saloons meeting rigid walls; steel bollards give far less.
  • Read Average impact force in newtons or kilonewtons, and weigh Deceleration (g) against injury thresholds rather than the force alone.

Worked example — 1500 kg at 54 km/h into a wall

A 1500 kg hatchback meets a rigid barrier at 15 m/s — 54 km/h, an ordinary urban speed — and its front structure folds through 0.5 m. Kinetic energy first: ½ × 1500 × 15² = 168,750 J. Spread across half a metre, F = 168,750 ⁄ 0.5 = 337,500 N.

That is 337.5 kN, about what 34 tonnes would press down with if parked on the bumper, and Deceleration (g) reads 22.94 — all of that speed shed inside roughly 67 milliseconds. Rerun the same crash with only 0.25 m of crumple and both figures double, to 675 kN and 45.9 g. Half a metre of folded sheet metal is the entire margin between a bad day and a fatal one.

Questions

Is this a peak force or an average?

An average, taken across the whole crumple distance. Work-energy divides total kinetic energy by stopping distance and returns whatever constant force would do an identical job. Real crash pulses are jagged, and peaks of two to three times this mean are normal once longitudinal rails and powertrain mounts load up. Treat what you read here as a floor for what a structure sees, not a ceiling.

Why does doubling my speed quadruple the force?

Because energy scales with v², not v. At 30 m/s a 1500 kg car carries 675 kJ instead of 168.75 kJ, so an unchanged 0.5 m of crumple must dispose of four times as much, and force rises fourfold. Urban speed limits move injury statistics sharply for exactly this reason: cutting speed by a fifth removes 36% of what has to be absorbed.

Does deceleration in g depend on vehicle mass?

No. Deceleration (g) works out to v² ⁄ (2dg), and mass cancels out completely — a loaded van and an empty city car stopping over identical distances from identical speeds pull identical g. Mass only scales how much force a structure must carry. That is precisely why occupant injury criteria are written in g while body-in-white specifications are written in kilonewtons.

How many g can a person actually survive?

Rather more than most people expect, provided a load is brief and spread by a belt across bone rather than soft tissue. Colonel John Stapp rode a rocket sled to 46.2 g in 1954 and climbed off with burst capillaries in his eyes. Tolerance drops steeply as duration grows. The 22.94 g of our worked example is survivable with modern restraints and frequently lethal without them.

What crumple distance should I put in?

For a passenger car meeting a rigid full-width barrier, 0.4 to 0.8 m of usable front-end collapse covers most of a modern fleet: small city cars near a lower bound, long-nosed saloons near an upper one. Against something that deforms too — another car's nose, a guardrail, a hedge — add whatever that object gives, since both stopping distances add together.

Can I apply this to a two-car collision?

Not directly. Solve momentum first, find each vehicle's own change in velocity, then enter that delta-v as Impact speed alongside that same car's crumple distance. Typing in raw closing speed badly overstates what either driver experiences, because neither vehicle stops dead in a real exchange — they push each other, and lighter cars lose far more speed than heavy ones.

References