Question

During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, the bags produce a maximum acceleration of 60g, but lasting for only 36 ms (or less).

Answer 1

solution:

maximum acceleration produce by bag =60 g

time to come to stop = 36 ms

applying equation

$V=u-at$

inserting values

$0=u-(60)(36*10^-^3)$

$u=(600*36)/1000=21.6 m/s$

now to find distance of penctration:

$S=ut-1/2at^2$

inserting values

$S= (21.6)(36*10^-^3)-1/2(600)(0.036)^2$

$S=0.38m$

hence distance traveled by person before coming to rest is 0.38 m