Force on Driver in Example Car Crash
For the car crash scenario
where a car stops in 1 foot
from a speed of 30 mi/hr,
what is the force on the
driver? Assume a 160 lb
(mass = 5 slugs) driver.
If firmly held in nonstretching seatbelt harness: Stopping distance 1 ft.
If not wearing seatbelt, stopping distance determined by nature of collision with windshield, steering column, etc. : stopping distance 0.2 ft.
 Deceleration = 4836 ft/s^{2} = 1474 m/s^{2} = 150 g's
 Force = 24068 lb = 107059 N = 12 tons!!

No seatbelt! 
If seat belt harness stretches, increasing stopping distance by 50%: 1.5 ft.
 Deceleration = 645ft/s^{2} = 197 m/s^{2} = 20 g's
 Force = 3209 lb = 14274 N = 1.6 tons

Stretching seatbelt 
These calculated numbers assume constant deceleration, and are therefore an estimate of the average force of impact.

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