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A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 g’s.
A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 g’s. (a) Calculate the force on a 65-kg person accelerating at this rate.
(b) What distance is traveled if brought to rest at this rate from 95 km/h?
Thedistancethepersonmoves beforecoming to restdepends on thedeceleration.(a) Theforceon the person is19,129.5 N(b) Thedistance traveledbefore coming to rest is approximately1.183 mReasons:(a) Theforceon the person = Mass × AccelerationThedecelerationof the person = 30 g'sThe person'smass,m= 65 - kgTherefore;The force on the person = 65 kg × 30 × 9.81 m/s² =19,129.5 N(b)Initial speed,u= 95 km/h = 26.38889 m/sDeceleration,a= 30 g'sWe get;v² = u² - 2·a·sWhen v = 0, we get;0² = 26.38889² - 2 × 30 × 9.81 × s2 × 30 × 9.81 × s = 26.38889²Thedistance traveledbefore coming to rest, s ≈1.183 mLearn more here:brainly.com/question/16675698...
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