The deceleration sustained by the safety belt is 300m/s²
110 km/h = 30.55 m/s
The distance travelled (through the soft barrier) while stopping the vehicle at a deceleration of 30.6 g is
(v⊃-u²)/2a
=(0-30.55²)/(2*300)
=1.56 m
highway saftey engineers build soft barriers so that cars hittingthem will slow down at a safe rate. a person wearing a saftey belt can withstand an acceleration of 300 m/s how thick should barriers be to safely stop a car that hits a barrier at 110 km/h
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